Sachin

Finding factors for an algebraic expression is simple when it consists of only a few terms. But when it comes to an expression that has more than two or three terms, students feel difficult to solve those problems. Students need a better process to solve algebraic expression factorization. Every student who is looking for the best method to solve algebraic expression factorization can follow the grouping method.

Factoring Terms by Grouping is the easy and best method to solve different expressions easily. Also, the process of Factoring by Grouping The Terms is very simple compared to other methods.

Procedure for Factoring Algebraic Expressions by Grouping

Follow the below steps to find the factorization of a given expression using the below steps.

(i) Take out a factor from each group from the groups of the given expression.
(ii) Factorize each group
(iii) Lastly, take out the common factor.

Factoring Terms by Grouping Examples

1. Factoring of algebraic expression

(i) 2ma + mb + 2na + nb

Solution:
Given expression is 2ma + mb + 2na + nb.
Group the first two terms and last two terms.
The first two terms are 2ma + mb and the second two terms are 2na + nb.
Take m common from the first two terms.
m (2a + b)
Take n common from the second two terms.
n (2a + b)
m (2a + b) + n (2a + b)
Then, take (2a + b) common from the above expression.
(2a + b) (m + n)

The final answer is (2a + b) (m + n).

(ii) 3xm – ym – 3xn + yn

Solution:
Given expression is 3xm – ym – 3xn + yn.
Group the first two terms and last two terms.
The first two terms are 3xm – ym and the second two terms are – 3xn + yn.
Take m common from the first two terms.
m (3x – y)
Take -n common from the second two terms.
-n (3x – y)
m (3x – y) – n (3x – y)
Then, take (3x – y) common from the above expression.
(3x – y) (m – n)

The final answer is (3x – y) (m – n).

(iii) 12a² + 6ab – 4ma – 2mb

Solution:
Given expression is 12a² + 6ab – 4ma – 2mb.
Group the first two terms and last two terms.
The first two terms are 12a² + 6ab and the second two terms are – 4ma – 2mb.
Take 6a common from the first two terms.
6a (2a + b)
Take -2m common from the second two terms.
-2m (2a + b)
6a (2a + b) – 2m (2a + b)
Then, take (2a + b) common from the above expression.
(2a + b) (6a – 2m)

The final answer is (2a + b) (6a – 2m).

(iv) am² – bm² + an² – bn² + ar² – br²

Solution:
Given expression is am² – bm² + an² – bn² + ar² – br².
Group the first two terms, middle two terms, and last two terms.
The first two terms are am² – bm², the middle two terms are + an² – bn², and the last two terms are + ar² – br².
Take m² common from the first two terms.
m² (a – b)
Take n² common from the middle two terms.
n² (a – b)
Take r² common from the middle two terms.
r² (a – b)
m² (a – b) + n² (a – b) + r² (a – b)
Then, take (a – b) common from the above expression.
(a – b) (m² + n² + r²)

The final answer is (a – b) (m² + n² + r²).

(v) ax – ay + bx – by

Solution:
Given expression is ax – ay + bx – by.
Group the first two terms and last two terms.
The first two terms are ax – ay and the second two terms are + bx – by.
Take a common from the first two terms.
a (x – y)
Take b common from the second two terms.
b (x – y)
a (x – y) + b (x – y)
Then, take (x – y) common from the above expression.
(x – y) (a + b)

The final answer is (x – y) (a + b).

2. Factoring the following algebraic expression

(i) 4a + 2ab + b + 2

Solution:
Given expression is 4a + 2ab + b + 2.
Group the first two terms and last two terms.
The first two terms are 4a + 2ab and the second two terms are b + 2.
Take 2a common from the first two terms.
2a (2 + b)
Take 1 common from the second two terms.
1 (b + 2)
2a (2 + b) + 1 (2 + b)
Then, take (2 + b) common from the above expression.
(2 + b) (2a + 1)

The final answer is (2 + b) (2a + 1).

(ii) 3m³ + 5m² + 3m + 5

Solution:
Given expression is 3m³ + 5m² + 3m + 5.
Group the first two terms and last two terms.
The first two terms are 3m³ + 5m² and the second two terms are 3m + 5.
Take m² common from the first two terms.
m² (3m + 5)
Take 1 common from the second two terms.
1 (3m + 5)
m² (3m + 5) + 1 (3m + 5)
Then, take (3m + 5) common from the above expression.
(3m + 5) (m² + 1)

The final answer is (3m + 5) (m² + 1).

(iii) b³ + 3b² + b + 3

Solution:
Given expression is b³ + 3b² + b + 3.
Group the first two terms and last two terms.
The first two terms are b³ + 3b² and the second two terms are b + 3.
Take b² common from the first two terms.
b² (b + 3)
Take 1 common from the second two terms.
1 (b + 3)
b² (b + 3) + 1 (b + 3)
Then, take (b + 3) common from the above expression.
(b + 3) (b² + 1)

The final answer is (b + 3) (b² + 1).

(iv) 1 + s + s²t + s³t

Solution:
Given expression is 1 + s + s²t + s³t.
Group the first two terms and last two terms.
The first two terms are 1 + s and the second two terms are s²t + s³t.
Take 1 common from the first two terms.
1 (1 + s)
Take s²t common from the second two terms.
s²t (1 + s)
1 (1 + s) + s²t (1 + s)
Then, take (1 + s) common from the above expression.
(1 + s) (1 + s²t)

The final answer is (1 + s) (1 + s²t).

(v) m – 1 – (m – 1)² + bm – b

Solution:
Given expression is m – 1 – (m – 1)² + bm – b.
Group the first two terms, middle and last two terms.
The first two terms are m – 1, the middle term is – (m – 1)², and the last two terms is bm – b.
Take 1 common from the first two terms.
1 (m – 1)
Take 1 common from the middle term.
– (m – 1)²
Take b common from the last two terms.
b (m – 1)
1 (m – 1) – (m – 1)²+ b (m – 1)
Then, take (m – 1) common from the above expression.
(m – 1) (1 – m + 1 + b)
(m – 1) (2 + b – m)

The final answer is (m – 1) (2 + b – m).

Factorize by grouping the terms consists of two or more product terms as resultant value. To divide the expression into product terms, we need to identify the greatest common factor which divides all the remaining terms in the expression. Step by step process on how to do Factorization by Grouping the Terms are clearly given in this article.

Procedure to Factorize by Grouping the Terms

Refer to the below-mentioned step by step process and learn the Factoring by Grouping Terms. They are along the lines

1. Note down the given expression.
2. Group the first two terms and last two terms.
3. Factor out the greatest common factor from each group.
4. Finally, you will get two or more product terms as result.

Solved Examples on Factorization by Grouping Terms

1. Factorize grouping the following expressions?

(i) 18x³y³ – 27x²y³ + 36x³y²

Solution:
The given expression is 18x³y³ – 27x²y³ + 36x³y²
Factor out the greatest common factor from the above expression.
That is, 9x²y²(2xy – 3y + 4x).

Therefore, the final solution for the expression 18x³y³ – 27x²y³ + 36x³y² is 9x²y²(2xy – 3y + 4x).

(ii) 12a²b³ – 21a³b²

Solution:
The given expression is 12a²b³ – 21a³b²
Factor out the greatest common factor from the above expression.
3a²b²(4b – 7a).

Finally, the solution for the expression 12a²b³ – 21a³b² is 3a²b²(4b – 7a).

(iii) a³ – a² + a – 1.

Solution:
The given expression is a³ – a² + a – 1
Group the first two terms and last two terms.
Here, first two terms are a³ – a² and the last two terms are a – 1.
So, (a³ – a²) + (a – 1).
Now, factor out the greatest common factor from each group.
That is, a²(a – 1) + (a – 1).
(a – 1) (a² + 1).

Therefore, solution for the expression a³ – a² + a – 1 is (a – 1) (a² + 1).

(iv) prs + qurs – pt – qut

Solution:
The given expression is prs + qurs – pt – qut.
Group the first two terms and last two terms.
Here, the first two terms are prs + qurs and last two terms are – pt – qut.
Then, (prs + qurs) – (pt + qut).
Now, factor out the greatest common factor from the above two groups.
That is, rs(p + qu) – t(p + qu).
(p + qu) (rs – t).

Therefore, the solution for the expression prs + qurs – pt – qut is (p + qu) (rs – t).

(v) a² – 3a – ab + 3b
Solution:

The given expression is
a² – 3a – ab + 3b.
Group the first two terms and last two terms.
Here, the first two terms are a²-3a and the last two terms are ab + 3b.
Then, (a² – 3a) – (ab – 3b).
Now, factor out the greatest common factor from the above two groups.
That is a(a – 3) – b(a – 3).
(a – 3) (a – b).

Therefore, solution for the expression is a² – 3a – ab + 3b is (a – 3) (a – b).

2. How to factorize by grouping the following expressions?
(i) 2q^4 – q³ + 4q – 2
Solution:
The given expression is
2q^4 – q³ + 4q – 2.
Group the first two terms and last two terms.
Here, the first two terms are 2q^4 – q³ and the last two terms are 4q – 2.
Then, (2q^4 – q^3) + (4q-2).
Now, factor out the greatest common factor from the above two groups.
That is, q^3(2q – 1) +2(2q – 1).
(2q – 1) (q^3 + 2).

Therefore, solution for the expression is 2q4 – q3 + 4q – 2 is (2q-1) (q^3+2).

(ii) ac + bc – ad – bd.

Solution:
The given expression is ac + bc – ad – bd.
Group the first two terms and last two terms.
Here, the first two terms are ac+bc and the last two terms are – ad-bd.
Then, (ac + bc) – (ad + bd).
Now, factor out the greatest common factor from the above two groups.
That is, c(a+b) -d(a+b).
(a+b) (c-d).

Therefore, solution for the expression is ac + bc – ad – bd is (a+b) (c-d).

(iii) pa – pb – qa –qb.

Solution:
The given expression is pa – pb – qa–qb.
Group the first two terms and last two terms.
Here, the first two terms are pa – pb, and the last two terms are qa – qb.
Then, (pa – pb) – (qa -qb).
Now, factor out the greatest common factor from the above two groups.
That is, p(a – b) – q(a – b).
(p – q) (a – b).

Therefore, the solution for the expression is pa – pb – qa – qb is (p – q) (a – b).

3. How to factorize by grouping the algebraic expressions?
(i)x^2z^2 + xzw + xyz + yw

Solution:
The given expression is
x^2z^2 + xzw + xyz + yw.
Group the first two terms and last two terms.
Here, the first two terms are x^2z^2 + xzw and the last two terms are xyz + yw.
Then, (x^2z^2 + xzw) + (xyz + yw).
Now, factor out the greatest common factor from the above two groups.
That is, xz(xz + w) + y(xz + w).
(xz + y) (xz + w).

Therefore, solution for the expression is x^2z^2 + xzw + xyz + yw is (xz + y) (xz + w).

(ii) 5x + xy + 5y + y^2

Solution:
The given expression is 5x + xy + 5y + y^2
Group the first two terms and last two terms.
Here, the first two terms are 5x + xy, and the last two terms are 5y + y^2.
Then, (5x + xy) + (5y + y^2).
Now, factor out the greatest common factor from the above two groups.
That is, x(5 + y) + y(5 + y).
(5 + y) (x + y).

Therefore, solution for the expression is 5x + xy + 5y + y^2 is (5 + y) (x + y).

(iii) xy – yz – xz + z^2

Solution:
The given expression is xy – yz – xz + z^2
Group the first two terms and last two terms.
Here, the first two terms are xy – yz and the last two terms are – xz + z^2.
Then, (xy – yz) – (xz – z^2).
Now, factor out the greatest common factor from the above two groups.
That is, y(x – z) – z(x – z).
(x – z) (y – z).

Therefore, solution for the expression xy – yz – xz + z^2 is (x – z) (y – z).

4. Factorize the expressions
(i) a^4 + a^3 + 2a + 2

Solution:
The given expression is a^4 + a^3 + 2a + 2.
Group the first two terms and last two terms.
Here, the first two terms are a^4 + a^3 and the last two terms are 2a + 2.
Then, (a^4 + a^3) + (2a + 2).
Now, factor out the greatest common factor from the above two groups.
That is, a^3(a + 1) + 2(a + 1).
(a + 1) (a^3 + 2).

Therefore, solution for the expression a^4 + a^3 + 2a + 2 is (a + 1) (a^3 + 2).

(ii) a²b² + d²b² – cd² – ca²

Solution:
The given expression is a²b² + d²b² – cd² – ca²
Group the first two terms and last two terms.
Here, the first two terms are a²b² + d²b² and the last two terms are – cd² – ca².
Then, (a^2b^2 + d^2b^2) – (cd^2 + ca^2).
Now, factor out the greatest common factor from the above two groups.
That is, b^2(a^2 + d^2) – c(d^2 + a^2).
(a^2 + d^2) (b^2 – c).

Therefore, solution for the expression a2b2 + d2b2 – cd2 – ca2 is(a^2 + d^2) (b^2 – c).

5. Factorize by grouping the terms
(x^2 + 3x)^2 – 2(x^2 + 3x) – y(x^2 + 3x) + 2y

Solution:
Thegiven expression is (x^2 + 3x)^2 – 2(x^2 + 3x) – y(x^2 + 3x) + 2y.
Group the first two terms and last two terms.
here, first two terms are (x^2 + 3x)^2 – 2(x^2 + 3x) and the last two terms are –y(x^2 + 3x) + 2y.
Then, [(x^2 + 3x)^2 – 2(x^2 + 3x)] –[y(x^2 + 3x) – 2y].
That is, (x^2 + 3x)(x^2 + 3x – 2) – y(x^2 + 3x – 2).
(x^2 + 3x – 2)(x^2 + 3x – y).

Therefore, solution for the expression (x2 + 3x)2 – 2(x2 + 3x) – y(x2 + 3x) + 2y is (x^2 + 3x – 2)(x^2 + 3x – y).

Are you looking for the different problems on Factorization by Grouping? Then, you are in the right place. We have given all types of factorization problems on our website. Students can learn to Factorize by Grouping the Terms in these articles. While solving Factorization by grouping problems, students need to group the terms with common factors before factoring.

How to Factor by Grouping?

Have a look at the Factoring by Grouping Steps and learn how to solve related problems easily. Follow the guidelines provided and perform factorization by grouping method. They are as under

• Note down the given expression, group the first two terms and last two terms.
• Find out the greatest common factor(GCF) from the first term and second term.
• Now, find the common factor from the above two groups.
• Finally factor out the terms in terms of product.

Factorization by Grouping Examples

1. Factor grouping the expressions?
1 + x + xy + x²y.

Solution: Given Expression is 1 + x + xy + x²y.
Group the first two terms and last two terms.
First two terms are 1 + x and the last two terms are xy + x²y.
(1+ x) + (xy + x²y).
Find out the common factor from the above two groups.
(1+x) + xy(1+x).
Factor out the terms in terms of product.
(1+x) (1+xy).

By factor grouping the expression 1 + x + xy + x²y., we will get (1+x) (1+xy).

2. How to factor by grouping the following algebraic expressions?

(i) x² – xy + xz – zy.

Solution:
Given Expression is x² – xy + xz – zy
Group the first two terms and last two terms.
First two terms are x² – xy and the last two terms are xz – zy.
(x² – xy) + (xz – zy)
Find out the common factor from the above two groups.
x(x – y) + z(x – y)
Factor out the terms in terms of product.
(x + z) (x – y)

Factor by grouping the expression x² – xy + xz – zy, we will get the result as (x+z) (x-y).

(ii) x² + 3x + xy + 3y.

Solution:
Given Expression is x² + 3x + xy + 3y.
Group the first two terms and last two terms.
First two terms are x² + 3x and the last two terms are xy + 3y.
(x² + 3x) + (xy + 3y)
Find out the common factor from the above two groups.
x(x + 3) + y(x + 3)
Factor out the terms in terms of product.
(x+y) (x+3).

Factor by grouping the expression x² + 3x + xy + 3y, we will get solution as (x+y) (x+3).

3. Factorize the algebraic expressions.

(i) 2a + ba + 2b + b²

Solution:
Given Expression is 2a + ba + 2b + b²
Group the first two terms and last two terms.
First two terms are 2a + ba and the last two terms are 2b + b².
(2a + ba)+ (2b + b²).
Find out the common factor from the above two groups.
a(2 + b) + b(2 + b).
Factor out the terms in terms of product.
(a + b) (2 + b).

By factorizing the expression 2a + ba + 2b + b², we will get (a + b) (2 + b).

(ii) b² – yb + 5b– 5y.

Solution:
Given Expression is b² – yb + 5b– 5y.
Group the first two terms and last two terms.
First, two terms are b² – yb and the last two terms are 5b – 5y.
(b² – yb)+ (5b – 5y).
Find out the common factor from the above two groups.
b(b – y) + 5(b – y).
Factor out the terms in terms of product.
(5 + b) (b – y).

By factorizing the expression b² – yb + 5b– 5y, we will get (5 + b) (b – y).

(iii) pq – rq – ps + rs.

Solution:
Given Expression is pq – rq – ps + rs.
Group the first two terms and last two terms.
First two terms are pq – rq and the last two terms are – ps + rs.
(pq – rq) – (ps – rs).
Find out the common factor from the above two groups.
q(p – r) -s(p – r).
Factor out the terms in terms of product.
(p – r) (q – s).

By factorizing the expression pq – rq – ps + rs, we will get (p – r) (q – s).

(iv) ab – 2ac – db + 2dc.

Solution:
The given expression is ab – 2ac – db + 2dc.
Group the first two terms and last two terms.
Fist two terms are ab – 2ac and the last two terms are – db + 2dc.
(ab – 2ac) – (db – 2dc).
Find out the common factor from the above two groups.
a(b – 2c) -d(b – 2c).
Factor out the terms in terms of product.
(b – 2c) (a – d).

By factorizing the expression ab – 2ac – db + 2dc., we will get (b – 2c) (a – d).

(v) pq² – 3rqs – pqs + 3rs²

Solution:
The given expression is pq² – 3rqs – pqs + 3rs²
Group the first two terms and last two terms.
First two terms are pq^2 – 3rqs and the last two terms are –pqs + 3rs².
(pq² – 3rqs) – (pqs – 3rs²).
Find out the common factor from the above two groups.
q(pq – 3rs) -s(pq – 3rs).
Factor out the terms in terms of product.
(pq – 3rs) (q – s).

By factorizing the expression pq² – 3rqs – pqs + 3rs², we will get (pq – 3rs) (q – s).

4. Factor each of the following expressions by grouping

(i) a² – 3a – ab + 3b.

Solution:
Given expression is a² – 3a – ab + 3b
Group the first two terms and last two terms.
First two terms are a² – 3a and the last two terms are – ab + 3b.
(a² – 3a) – (ab – 3b).
Find out the common factor from the above two groups.
a(a – 3) – b(a – 3).
Factor out the terms in terms of product.
(a – 3) (a – b).

By factorizing the expression a² – 3a – ab + 3b, we will get (a – 3) (a – b).

(ii) pq² + rq² + 2p + 2r.

Solution:
Given expression is pq² + rq² + 2p + 2r
Group the first two terms and last two terms.
First two terms are pq² + rq² and the last two terms are 2p + 2r.
(pq² + rq²) + (2p + 2r).
Find out the common factor from the above two groups.
q²(p +r) + 2(p+r).
Factor out the terms in terms of product.
(q²+ 2) (p + r).

By factorizing the expression pq² + rq² + 2p + 2r, we will get (q²+ 2) (p + r).

(iii) 2pq² + 3pqr – 2sqr – 3sr²

Solution:
Given expression is 2pq² + 3pqr – 2sqr – 3sr²
Group the first two terms and last two terms.
First two terms are 2pq² + 3pqr and the last two terms are – 2sqr – 3sr².
(2pq² + 3pqr) – ( 2sqr + 3sr²).
Find out the common factor from the above two groups.
pq(2q + 3r) – sr(2q + 3r).
Factor out the terms in terms of product.
(2q + 3r) (pq – sr).

By factorizing the expression 2pq² + 3pqr – 2sqr – 3sr², we will get (2q + 3r) (pq – sr).

(iv) par² + qars – pnrs – qns²

Solution:
The given expression is par² + qars – pnrs – qns²
Group the first two terms and last two terms.
First two terms are par² + qars and the last two terms are – pnrs – qns².
(par² + qars) – (pnrs – qns²).
Find out the common factor from the above two groups.
ar(pr + qs) – ns(pr + qs).
Factor out the terms in terms of product.
(pr + qs) (ar – ns).

By factorizing the expression par² + qars – pnrs – qns², we will get (pr + qs) (ar – ns).

5. Factorize
(i) (a + b) (2a + 5) – (a + b) (a + 3).

Solution:
The given expression is (a + b) (2a + 5) – (a + b) (a + 3).
Find out the common factor from the above expression.
(a + b) [(2a + 5) – (a + 3)].
Expand the terms.
(a + b)(2a + 5 – a – 3).
Simplify the second term.
(a + b) (a + 2).

By factorizing the expression (a + b) (2a + 5) – (a + b) (a + 3), we will get (a + b)(a + 2).

(ii) 6xy – y² + 12xz – 2yz.

Solution:
The given expression is 6xy – y² + 12xz – 2yz.
Group the first two terms and last two terms.
First two terms are 6xy – y² and the last two terms are 12xz – 2yz.
(6xy – y²) + (12xz – 2yz).
Find out the common factor from the above two groups.
y(6x – y)+ 2z(6x – y).
Factor out the terms in terms of product.
(6x – y) (y + 2z).

By factorizing the expression 6xy – y² + 12xz – 2yz, we will get (6x – y) (y + 2z).

Students can find the factors for an algebraic expression when one of its factors is binomial. Find given algebraic expression Factorization when Binomial is Common Factor. All related problems are included in this article along with a clear explanation. Therefore, students can practice every problem available in this article and improve their knowledge. The process of solving the factorization problem is very simple if you follow the procedure we explained below. Go through the complete article and learn different methods to solve factorization problems.

Factorization of Algebraic Expressions when Binomial is Common

Follow the simple and easy guidelines on Factorization of Algebraic Expressions When Binomial is Common. They are as follows

Step 1: In the first step, find the common binomial factor.
Step 2: Note down the given expression as the product of this binomial and the quotient obtained on dividing the given expression by this binomial.

Solved Examples on Factorization When Binomial is a Common

1. Factorize the algebraic expressions.

(i) 7b(3x – 4y) + 3a(3x – 4y)

Solution:
Given expression is 7b(3x – 4y) + 3a(3x – 4y)
In the given expression, the binomial factor is (3x – 4y) as it is common in both terms.
Take the (3x – 4y) common and multiply it with the remained terms.
(3x – 4y) (7b + 3a)

The final answer is (3x – 4y) (7b + 3a)

(ii) 12(9a + 6b)² – 4(9a + 6b)

Solution:
Given expression is 12(9a + 6b)² – 4(9a + 6b)
12 (9a + 6b) (9a + 6b) -4(9a + 6b)
In the given expression, the binomial factor is (9a + 6b) as it is common in both terms.
Take the (9a + 6b) common and multiply it with the remained terms.
(9a + 6b)(12(9a + 6b) – 4)
(9a + 6b)(108a + 72b – 48)
The final answer is (9a + 6b)(108a + 72b – 48)

2. Factorize the expression 10r(m – 2n) – 8m + 16n

Solution:
Given expression is 10r(m – 2n) – 8m + 16n
Lets take -8m  + 16n from the above equation.
Take -8 common from -8m + 16n
-8(m – 2n)
Place -8(m – 2n) in 10r(2m – 4n) – 8m + 16n equation.
10r(m – 2n) -8(m – 2n)
In the above expression, the binomial factor is (m – 2n) as it is common in both terms.
Take the (m – 2n) common and multiply it with the remained terms.
(m – 2n)(10r – 8)

The final answer is (m – 2n)(10r – 8)

3. Factorize (a – 4b)^2 – 7a + 28b

Solution:
(a – 4b)(a – 4b) – 7a + 28b
Given expression is (a – 4b)(a – 4b) – 7a + 28b
Lets take – 7a + 28b from the above equation.
Take -7 common from – 7a + 28b
-7(a – 4b)
Place -7(a – 4b) in (a – 4b)(a – 4b) – 7a + 28b equation.
(a – 4b)(a – 4b) – 7(a – 4b)
In the above expression, the binomial factor is (a – 4b) as it is common in both terms.
Take the (a – 4b) common and multiply it with the remained terms.
(a – 4b)(a – 4b – 7)

The final answer is (a – 4b)(a – 4b – 7)

All solved examples of Factorization problems when Binomial is a common factor are included in this article. Check how to find the binomial factor and practice all problems given here. Compare the terms in the given expression and factor out the greatest common factor term from the expression. Based on the greatest common factor, we will get the solution for the expression that is in terms of the product of two or more terms.

How to Factorize taking out the Binomial Common Factor?

Go through the below step by step procedure and do Factorization when a Binomial is a Common Factor. They are along the lines

1. Note down the given expression
2. Note down the first term and second term from the expression
3. Compare the two terms and observe the greatest common factor
4. Now, factor out the greatest common factor from the expression
5. Finally, we will get the result in the form of the product of two or more terms.

Solved Binomial Factor Examples

1. Factorize the expression (6x + 1)² – 5(6x + 1)?

Solution:
The given expression is (6x + 1)² – 5(6x + 1)
Here, the first term is (6x + 1)² and second term is – 5(6x + 1)
By comparing the above two terms, we can observe the greatest common factor and that is (6x + 1)
Now, factor out the greatest common factor from the expression
That is, (6x + 1) [(6x + 1) – 5]
(6x + 1)(6x – 4)

Therefore, the resultant value for the expression (6x + 1)² – 5(6x + 1) is (6x + 1)(6x – 4)

2. Factorize the algebraic expression 4x(y – z) + 3(y – z)?

Solution:
The given expression is 4x(y – z) + 3(y – z)
Here, the first term is 4x(y – z) and the second term is 3(y – z)
By comparing the above two terms, we can observe the greatest common factor and that is (y – z)
Now, factor out the greatest common factor from the expression
That is, (y – z)(4x + 3)

Therefore, the resultant value for the expression 4x(y – z) + 3(y – z) is (y – z)(4x + 3)

3. Factorize the expression (3x – 4y) (p – q) + (3x – 2y) (p – q)?

Solution:
The given expression is (3x – 4y) (p – q) + (3x – 2y) (p – q)
Here, the first term is (3x – 4y) (p – q) and the second term is (3x – 2y) (p – q)
By comparing the above two terms, we can observe the greatest common factor and that is (p – q)
Now, factor out the greatest common factor from the expression
That is, (p – q)[(3x – 4y) + (3x – 2y)]
= (p – q)(6x – 6y)

Therefore, the resultant value for the expression (3x – 4y) (p – q) + (3x – 2y) (p – q) is (p – q)(6x – 6y)

4. Factorize the expression 6(2x + y)² +8(2x + y)?

Solution:
The given expression is 6(2x + y)² +8(2x + y)
Here, the first term is 6(2x + y)² and the second term is 8(2x + y)
By comparing the above two terms, we can observe the greatest common factor and that is (2x + y)
Now, factor out the greatest common factor from the expression
That is, (2x + y)[6(2x + y) + 8]
(2x + y)(12x + 6y + 8)

Therefore, the resultant value for the expression 6(2x + y)² +8(2x + y) is (2x + y)(12x + 6y + 8)

While finding the factorization when monomial is a common factor, an algebraic expression can be expressed as the sum or the difference of monomials. Look at the different problems that have an algebraic expression with Monomial as a common factor. Also, find all Factorization concepts on our website with a detailed explanation.

How to Find Factorization of an Expression?

Below is the step by step guidelines on how to factorize an expression. They are along the lines

Step 1: Note Down the algebraic expression.
Step 2: Find the different terms of given algebraic expressions.
Step 3: Find the HCF of all the separate terms of the given algebraic expression.
Step 4: Express each term of the algebraic expression as the product of H.C.F and divide each term of the given expression by the HCF.
Step 5: Now use the distributive property of multiplication over addition or subtraction to express the algebraic expression as the product of H.C.F and the quotient of the expression divided by the H.C.F.
Step 6: In the final step, keep the H.C.F. outside the bracket and the quotients obtained within the bracket.

Solved Examples of Factorization when Monomial is Common

1. Factorize each of the following

(i) 4b + 20

Solution:
The given expression is 4b + 20
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 4 and 20 is 4.
HCF of literal coefficients:
The lowest power of b is 0.
Therefore, the HCF of literal coefficients is nothing.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4b + 20 is 4.
Multiply and divide each term of the given expression 4b + 20 with 4.
4(4b/4 + 20/4) = 4(b + 5)

The final answer is 4(b + 5)

(ii) 3m² + 2m

Solution:
The given expression is 3m² + 2m
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3 and 2 is nothing.
HCF of literal coefficients:
The lowest power of m is 1.
Therefore, the HCF of literal coefficients is m.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3m² + 2m is m.
Multiply and divide each term of the given expression 3m² + 2m with m.
m(3m²/m + 2m/m) = m(3m + 2)

The final answer is m(3m + 2)

(iii) 6a²b – 3ab²

Solution:
The given expression is 6a²b – 3ab²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 6 and 3 is 3.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
Therefore, the HCF of literal coefficients is ab.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 6a²b – 3ab² is 3ab.
Multiply and divide each term of the given expression 3m² + 2m with m.
3ab(6a²b/3ab – 3ab²/3ab) = 3ab(2a – b)

The final answer is 3ab(2a – b)

(iv) 9cd – 6bc

Solution:
The given expression is 9cd – 6bc
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 9 and 6 is 3.
HCF of literal coefficients:
The lowest power of b is 0.
The lowest power of c is 1.
The lowest power of d is 0.
Therefore, the HCF of literal coefficients is c.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 9cd – 6bc is 3c.
Multiply and divide each term of the given expression 9cd – 6bc with 3c.
3c(9cd/3c – 6bc/3c) = 3c(3d – 2b)

The final answer is 3c(3d – 2b)

2. Factorize 18x²y²z + 12xyz.

Solution:
The given expression is 18x²y²z + 12xyz
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 18 and 12 is 3.
HCF of literal coefficients:
The lowest power of x is 1.
The lowest power of y is 1.
The lowest power of z is 1.
Therefore, the HCF of literal coefficients is xyz.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 18x²y²z + 12xyz is 3xyz.
Multiply and divide each term of the given expression 18x²y²z + 12xyz with 3xyz.
3xyz(18x²y²z/3xyz + 12xyz/3xyz) = 3xyz(6xy + 4)

The final answer is 3xyz(6xy + 4)

3. Factorize the expression: 8m³ – 32m²n

Solution:
The given expression is 8m³ – 32m²n
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 8 and 32 is 8.
HCF of literal coefficients:
The lowest power of m is 2.
The lowest power of n is 0.
Therefore, the HCF of literal coefficients is m².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 8m³ – 32m²n is 8m².
Multiply and divide each term of the given expression 8m³ – 32m²n with 8m².
8m²(8m³/8m² – 32m²n/8m²) = 8m²(m – 4n)

The final answer is 8m²(m – 4n)

Every algebraic expression has factors as 1 and itself. If the Monomial is a Common Factor, then the product of two or more numbers or variables, then it will have factors of 1 and itself. Find different problems on the Factorization of algebraic expressions and know the process to solve each problem in an easy way. We have included tips and tricks to solve Factorization Problems that help students to learn the Factorization concept easily.

How to Find Factorization of an Expression when a Monomial is a Common Factor?

Follow the below process to find the Factorization of given problems.

(i) Write an algebraic expression.
(ii) Find the H.C.F. of all the individual terms of the expression.
(iii) Divide each individual term of the expression by the H.C.F.
(iv) Keep the H.C.F. outside the bracket and the quotients obtained within the bracket.

The greatest common factor (GCF) of two or more monomials is the product of the greatest common factor of the coefficients and the greatest common factors of the variables

Solved Examples on Monomial as a Common Factor

1. Factorize each of the following algebraic expressions.

(i) 15a + 10

Solution:
The given expression is 15a + 10
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 15 and 10 is 5.
HCF of literal coefficients:
The lowest power of a is 0.
Therefore, the HCF of literal coefficients is nothing.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 15a + 10 is 5.
Multiply and divide each term of the given expression 15a + 10 with 5a
5((15a/5) + (10/5)) = 5(3a + 2)

The final answer is 5(3a + 2)

(ii) 18mn² + 9m²n – 12mn

Solution:
The given expression is 18mn² + 9m²n – 12mn
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 18, 9, and 12 is 3.
HCF of literal coefficients:
The lowest power of m is 1.
The lowest power of n is 1.
Therefore, the HCF of literal coefficients is mn.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 18mn² + 9m²n – 12mn is 3mn.
Multiply and divide each term of the given expression 15a + 10 with 5a
3mn((18mn²/3mn) + (9m²n/3mn) – (12mn/3mn)) = 3mn (6n + 3m – 4)

The final answer is 3mn (6n + 3m – 4).

(iii) 20x²y – 8xy² + 32xy

Solution:
The given expression is 20x²y – 8xy² + 32xy
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 20, 8, and 32 is 4.
HCF of literal coefficients:
The lowest power of x is 1.
The lowest power of y is 1.
Therefore, the HCF of literal coefficients is xy.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 20x²y – 8xy² + 32xy is 4xy.
Multiply and divide each term of the given expression 20x²y – 8xy² + 32xy with 4xy
4xy(20x²y/4xy – 8xy²/4xy+ 32xy/4xy) = 4xy (5x – 2y + 8)

The final answer is 4xy (5x – 2y + 8).

(iv) 13a³ + 39a²b

The given expression is 13a³ + 39a²b
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 13 and 39 is 13.
HCF of literal coefficients:
The lowest power of a is 2.
The lowest power of b is 0.
Therefore, the HCF of literal coefficients is a².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 13a³ + 39a²b is 13a².
Multiply and divide each term of the given expression 13a³ + 39a²b with 13a².
13a²(13a³/13a² + 39a²b/13a²) = 13a² (a + 3b)

The final answer is 13a² (a + 3b).

Factors of Algebraic Expressions is the product of numbers, algebraic expressions, algebraic variables, etc. Here the numbers, algebraic expressions, algebraic variables are the factors of the algebraic expressions. Generally, factors of a number can be calculated with the product of its multiples. You can learn more about Factorization and related concepts by always seeking help from us. For example, by multiplying 2, 3, 7, we get 42. Therefore, 2, 3, 7 are the factors of 42. Factors of a given number are the product of two or more numbers.

Basics of Algebra Expressions

Expressions: Expressions are formed with constants and variables. 2x – 5 is an expression that formed with the variable x and numbers 2 and 5.

Terms, factors, and Coefficients: Algebraic expression is the combination of terms, coefficients, and factors. For example, in 5x + 7, 5x and 7 are terms, x and 7 are factors, and 5 and 7 are numeric coefficients.

Monomials, binomials, and Polynomials: An expression having only one term is known as a monomial, an expression having two terms are called binomial. Also, an expression with various terms and with a non-zero coefficient is called as Polynomials.

Factors of the Monomials

Product of variable and numbers can form a monomial that consists of different factors.

1. Write all the possible factors of 5ab²?

Solution:
The possible factors of 5 are 1 and 5.
The possible factors of ab² are a, b, b², ab, ab²
All possible factors of 5ab² are a, b, b², ab, ab², 1, 5, 5a, 5b, 5b², 5ab, and 5ab².

2. Write down all the factors of 7m²n?

Solution:
The possible factors of 7 are 1 and 7.
The possible factors of m²n are m, n, mn, m², m²n.
All possible factors of 7m²n are m, n, mn, m², m²n, 1, 7, 7m, 7n, 7mn, 7m², 7m²n.

3. Write all the factors of 3x²y²?

Solution:
The possible factors of 3 are 1 and 3.
The possible factors of x²y² are x, y, xy, x², y², x²y, xy², x²y²
All possible factors of 3x²y² are x, y, xy, x², y², x²y, xy², x²y², 1, 3, 3x, 3y, 3xy, 3x², 3y², 3x²y, 3xy², 3x²y²

4. Write down all the factors of 3xyz?

Solution:
The possible factors of 3 are 1 and 3.
The possible factors of xyz are x, y, z, xy, xz, yz, xyz
All possible factors of 3xyz are x, y, z, xy, xz, yz, xyz, 1, 3, 3x, 3y, 3z, 3xy, 3xz, 3yz, 3xyz

Highest Common Factor (HCF) of Monomials

The H.C.F. of two or more monomials is the product of the H.C.F. of the numerical coefficients and the common variables with the least powers.

1. Find the H.C.F. of 12a³b², 14a²b³, 6ab⁴.

Solution:
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 12, 14, and 6 is 2.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 2.
Therefore, the HCF of literal coefficients is ab².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 12a³b², 14a²b³, 6ab⁴ is 2ab².

The final answer is 2ab².

2. Find the H.C.F. of 6xy, 18x²y?

Solution:
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 6 and 18 is 6.
HCF of literal coefficients:
The lowest power of x is 1.
The lowest power of y is 1.
Therefore, the HCF of literal coefficients is xy.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 6xy, 18x²y is 6xy.

The final answer is 6xy.

3. Find the H.C.F. of 9abc² and 36ac?

Solution:
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 9 and 36 is 9.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 0.
The lowest power of c is 1.
Therefore, the HCF of literal coefficients is ac.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 9abc² and 36ac is 9ac.

The final answer is 9ac.

Students can solve the system of linear equations by using the comparison method. Here, you need to compare a pair of equations and find the value of the variables. You can find a detailed explanation of solving the simultaneous linear equations in the following sections. The comparison method is one of the easiest ways to solve a pair of equations. So, follow the below guidelines and solve the questions.

How to Solve Simultaneous Equations Using Comparison Method?

Simultaneous Equations are nothing but two or more equations having two or more quantities. There are many ways to solve the system of linear equations, but we are using the comparison method to get the answer easily. Check the steps and get the solution.

• Let us take (i) and (ii) as the equations numbers.
• Express both equations x in terms of y.
• And consider obtained equations as (iii), (iv).
• Equate the obtained equations i.e the values of x in the equation forming the equation in y.
• Solve the linear equation in y and simplify it.
• Substitute the obtained y value in any one equation and solve it.
• Thus, you will get the solution using the comparison method.

Solved Examples on System of Linear Equations

Example 1.

Solve the pair of simultaneous equations 5x + 3y = 9, 2x – 3y = 12 using the comparison method with detailed explanation.

Solution:

Given the system of linear equations are

5x + 3y = 9 —— (i)

2x – 3y = 12 ——- (ii)

Express of x in terms of y.

From Equation (i) 5x + 3y = 9, we get

Subtract 3y from both sides

5x + 3y – 3y = 9 – 3y

5x = 9 – 3y

Divide both sides of the equation by 5.

5x/5 = (9 – 3y)/5

x = (9 – 3y)/5

Therefore, x = (9 – 3y)/5 —— (iii)

From equation (ii) 2x – 3y = 12, we get

Add 3y to both sides of the equation.

2x – 3y + 3y = 12 + 3y

2x = 12 + 3y

Divide both sides by 2.

2x/2 = (12 + 3y)/2

x = (12 + 3y)/2

Therefore, x = (12 + 3y)/2 —– (iv)

Equate the values of x in equation (iii) and equation (iv) forming the equation in y

From equations (iii) and (iv), we get

(9 – 3y)/5 = (12 + 3y)/2

Solve the above linear equation

Cross multiply the fractions.

2(9 – 3y) = 5(12 + 3y)

18 – 6y = 60 + 15y

18 – 60 = 15y + 6y

-42 = 21y

y = -42/21

y = -2

Substituting the value of y in equation (iii) or equation (iv), find the value of x

Put y = -2 in equation (iii), we get

x = (9 – 3(-2))/5

x = (9 + 6)/5

= 15/5

x = 3

Required solution of the two equations

Therefore, x = 3, y = -2.

Therefore, we have compared the values of x obtained from equation (i) and (ii) and formed an equation in y, so this method of solving simultaneous equations is known as the comparison method. Similarly, comparing the two values of y, we can form an equation in x.

Example 2.

Solve the system of linear equations 5x + y = 10, 7x – 3y + 8 = 0 using comparison method.

Solution:

Given pair of system of linear equations are

5x + y = 10 —- (i)

7x – 3y + 8 = 0 —– (ii)

Express x in terms of y

From equation (i) 5x + y = 10, we get

5x = 10 – y

x = (10 – y)/5

From equation (ii) 7x – 3y + 8 = 0, we get

7x – 3y = -8

7x = 3y – 8

x = (3y – 8)/7

Equate the values of x obtained from both equations.

(10 – y)/5 = (3y – 8)/7

Solve the equation in y.

7(10 – y) = 5(3y – 8)

70 – 7y = 15y – 40

70 + 40 = 15y + 7y

110 = 22y

y = 110/22

y = 5

Putting the value of y in equation (i)

Substitute y = 5 in equation (i), we get

5x + 5 = 10

5x = 10 – 5

5x = 5

x = 5/5

x = 1

Required solution of the two equations

Therefore, x = 1, y = 5.

Example 3.

Solve the linear equations 2x + y = 10 and x + y = 4 by comparison method.

Solution:

Given simultaneous linear equations are

2x + y = 10 —- (i)

x + y = 4 —- (ii)

express y in terms of x.

From equation (i), we can write

y = 10 – 2x

From equation (ii), we can write

y = 4 – x

Equate both the values of y obtained from equations (i) and (ii)

10 – 2x = 4 – x

10 – 4 = 2x – x

x = 6

Substitute x = 6 in the equation (ii)

6 + y = 4

y = 4 – 6

y = -2

Required solution of the two equations

Therefore, x = 6, y = -2.

Example 4.

Find the solution to the following simultaneous equations.

4x + 3y = 14 and 5x + 7y = 11.

Solution:

Given linear equations are

4x + 3y = 14 —– (i)

5x + 7y = 11 —— (ii)

Express x in terms of y for both equations.

From equation (i) 4x + 3y = 14, we can write

4x = 14 – 3y

x = (14 – 3y)/4

From equation (ii) 5x + 7y = 11, we can write

5x = 11 – 7y

x = (11 – 7y)/5

Equate the values of x obtained from equations (i), (ii)

(14 – 3y)/4 = (11 – 7y)/5

5(14 – 3y) = 4(11 – 7y)

70 – 15y = 44 – 28y

28y – 15y = 44 – 70

13y = -26

y = -26/13

y = -2

Substitute y = -2 in the equation (i) and simplify it to get the x value.

Put y = -2 in 4x + 3y = 14

4x + 3(-2) = 14

4x – 6 = 14

4x = 14 + 6

4x = 20

x = 20/4

x = 5

Therefore the required solutions are x = 5, y = -2.

The equation is a mathematical statement where two things are equal to one another. Every equation contains two expressions, one at the right-hand side of the equality called R.H.S and the second at the left-hand side of the equality called the L.H.S. Among the different types of equations we have are linear equations, quadratic equations, and polynomial equations.

Simultaneous Linear Equations Definition

The Simultaneous Linear Equations are the equations that have two or more quantities related using two or equations. These equations are also called the system of equations, which has a fine set of equations for which the common solution is sought. The general form of the simultaneous equations are as follows:

ax + by + c = 0

dx + ey + f = 0.

Here each linear equation represents a straight line in the graph. Both x, y are variables, a, b, c, d are real numbers either a, b or c, d anyone is non zero. Every linear equation in two variables has an infinite number of solutions. Here, we will learn about two linear equations in 2 variables.

Examples

• 3x + 2y = 5
• x – y + 1 = 0
• y = 2
• x/2 + y = 6
• y – 8 = x

Methods to Solve Simultaneous Linear Equations

We have four important methods to solve simultaneous equations. They are comparison, substitution, elimination, and cross-multiplication. Out of all these methods, the simplest methods will effectively solve the equations and get the answers.

1. Comparison Method:

Express one variable like x in terms of another variable called y for both equations. Equate the values of x in the equation, solve the equation, and find the value of y. Substitute the y value in any of the given equation to get the x value. Thus, you can obtain the solutions for the linear equation by using the comparison method.

2. Substitution Method:

Get the value of one variable in terms of the other from any one of the equations. Substitute this value in another equation. Solve the equations to find variable values. Substitute the variable value to find another variable value.

3. Elimination Method:

First of all, observe the given linear equations and multiply the one equation by a suitable constant to make the variable coefficients equal and to be eliminated. Add or subtract the obtained new equations and cancel the terms having the same coefficient. Solve the equation, and substitute the obtained value to get the set of values of two variables.

4. Cross Multiplication Method:

Use the cross multiplication method to find the solution set for the given equations. You can refer to the Cross Multiplication page to know more details about this method.

Simultaneous Linear Equations Examples

Example 1.

Solve the system of equations 2x – 3y = 1, x + y + 1 = 0 by using the substitution method?

Solution:

Given linear equations are,

2x – 3y = 1 ——- (i)

x + y + 1 = 0 ——– (ii)

From (i) Express x in terms of y,

2x = 3y + 1

x = (3y + 1) / 2

Substitute x = (3y + 1) / 2 in (ii) equation

(3y + 1)/2 + y + 1 = 0

(3y + 1) + 2y + 2 = 0

5y + 3 = 0

5y = -3

y = -3/5

Put y = -3/5 in (i) equation

2x – 3(-3/5) = 1

2x + 9/5 = 1

(10x + 9)/5 = 1

10x + 9 = 5

10x = 5 – 9

10x = -4

x = -4/10

x = -2/5

Therefore, the required solution set for the given simultaneous equations is x = -2/5, y = -3/5.

Example 2.

Solve the equations 2x – y – 3 = 0, 3x + 2y – 8 = 0 by using the comparison method?

Solution:

Given simultaneous equations are,

2x – y – 3 = 0 —– (i)

3x + 2y – 8 = 0 —— (ii)

Express x in terms of y for both equations.

From equation (i) 2x – y – 3 = 0, we get

2x – y = 3

2x = 3 + y

x = (y + 3) / 2 —— (iii)

From equation (ii) 3x + 2y – 8 = 0, we get

3x + 2y = 8

3x = 8 – 2y

x = (8 – 2y) / 3 —– (iv)

Equate the values of x in equation (iii) and equation (iv) forming the equation in y

From equation (iii) and (iv), we get,

(y + 3) / 2 = (8 – 2y) / 3

Cross multiply the fractions.

3(y + 3) = 2(8 – 2y)

3y + 9 = 16 – 4y

3y + 4y = 16 – 9

7y = 7

y = 7/7

y = 1

Putting the value of y in equation (iii) or equation (iv), find the value of x

substitute y = 1 in equation (iii)

x = (1 + 3) / 2

x = 4/2

x = 2

The required solution for the equations is x = 2, y = 1.

Therefore, we have compared the values of x obtained from equation (i) and (ii) and formed an equation in y, so this method of solving simultaneous equations is known as the comparison method. Similarly, by comparing the two values of y, we can form an equation in x.

Example 3.

Solve the simultaneous equations 3x + 7y = 27, 5x + 2y = 16 by using the elimination method?

Solution:

Given equations are

3x + 7y = 27 —— (i)

5x + 2y = 16 ——- (ii)

Multiply the equation (i) by 5, we get

5(3x + 7y = 27)

5(3x + 7y) = 27 x 5

15x + 35y = 135 —– (iii)

Multiply the equation (ii) by 3, we get

3(5x + 2y = 16)

3(5x + 2y) = 16 x 3

15x + 6y = 48 —– (iv)

Subtracting (iii) from (iv), we get

15x + 35y = 135

15x + 6y = 48
–      –        –
_________________

0 + 29y = 87

y = 87/29

y = 3

Substitute y = 3 in equation (i)

3x + 7(3) = 27

3x + 21 = 27

3x = 27 – 21

3x = 6

x = 6/3

x = 2

Therefire, required solution set is x = 2, y = 3.

Example 4.

Solve the equations 7x + 3y + 15 = 0, 12y – 5x = 39 by using the cross multiplication method?

Solution:

Given two equations are

7x + 3y + 15 = 0

12y – 5x = 39

On transposition, we get

7x + 3y + 15 = 0

-5x + 12y – 39 = 0

Writing the co-efficient in the following way, we get:

3        15        7          3
12      -39       -5         12

By cross-multiplication method:

x / [(3(-39) – (15)(12)] = y / [(15)(-5) – (-39)(7)] = 1/[(7)(12) – (3)(-5)]

x/[-117 – 180] = y/[-75 + 273] = 1/[84 + 15]

x/(-297) = y/(198) = 1/99

-x/297 = 1/99

-x = 297/99

x = -3

y/198 = 1/99

y = 198/99

y = 2

Therefore, required solution is x = -3, y = 2.

This page contains various linear equations word problems. Students have to practice all these questions to score better marks in the exam. To solve these questions on linear equations, you need to understand the given details and make an equation using one variable. And finally, solve that equation. For better understanding, you can check our Linear Equations Practice Problems with Solutions.

Example 1.

Solve the following linear equations:

(a) x/4 – 1/2 = x/5 + 2/3

(b) x – (x + 1)/3 = 1 – (x – 1)/4

Solution:

(a) x/4 – 1/2 = x/5 + 2/3

(x – 2) / 4 = (3x + 10) / 15

Cross multiply the fractions.

15(x – 2) = 4(3x + 10)

15x – 30 = 12x + 40

Transferring 12x from R.H.S to L.H.S becomes -12x, -30 from L.H.S to R.H.S becomes + 30.

15x – 12x = 40 + 30

3x = 70

Divide both sides of the equation by 3.

3x/3 = 70/3

x = 70/3

Therefore, the required solution set is x = 70/3.

(b) x – (x + 1)/3 = 1 – (x – 1)/4

[3x – (x + 1)] / 3 = [4 – (x – 1)] / 4

Cross multiply the fractions.

4(3x – x – 1) = 3(4 – x + 1)

4(2x – 1) = 3(5 – x)

8x – 4 = 15 – 3x

Transferring -3x from R.H.S to L.H.S becomes 3x, -4 from L.H.S to R.H.S becomes +4.

8x + 3x = 15 + 4

11x = 19

x = 19/11

Therefore, the required solution set is x = 19/11.

Example 2.

Solve the following equations and verify them.

(a) 4(3y + 2) – 5(6y – 1) = 2(y – 8) – 6(7y – 4) + 4y

(b) 2/(3x + 1) + 4/(3x + 1) = 5/3x

Solution:

(a) 4(3y + 2) – 5(6y – 1) = 2(y – 8) – 6(7y – 4) + 4y

12y + 8 – 30y + 5 = 2y – 16 – 42y + 24 + 4y

13 – 18y = 8 – 36y

Transferring -36y from R.H.S to L.H.S becomes 36y, 13 from L.H.S to R.H.s becomes -13.

-18y + 36y = 8 – 13

18y = -5

Divide both sides of the equation by 18.

18y/18 = -5/18

y = -5/18

Verification:

L.H.S = 4(3y + 2) – 5(6y – 1)

Substitute y = -5/18

L.H.S = 4(3(-5/18) + 2) – 5(6(-5/18) – 1)

= -60/18 + 8 + 150/18 + 5

= 90/18 + 13

= 5 + 13

= 18

R.H.S = 2(y – 8) – 6(7y – 4) + 4y

Substitute y = -5/18

R.H.S = 2(-5/18 – 8) – 6(7(-5/18) – 4) + 4(-5/18)

= -10/18 – 16 + 210/18 + 24 -20/18

= 180/18 + 8

= 10 + 18

= 18

L.H.S = R.H.S

Hence, proved.

(b) 2/(3x + 1) + 4/(3x + 1) = 5/3x

6/(3x + 1) = 5/3x

Cross multiply the fractions.

6 x 3x = 5 x (3x + 1)

18x = 15x + 5

Transferring 15x from R.H.S to L.H.S becomes -15x.

18x – 15x = 5

3x = 5

Divide both sides of the equation by 3

3x/3 = 5/3

x = 5/3

Verification,

L.H.S = 2/(3x + 1) + 4/(3x + 1)

Put x = 5/3

L.H.S = 2/(3(5/3) + 1)) + 4/(3(5/3) + 1))

= 6/(15/3 + 1)

= 6/((15 + 3)/3)

= 6/(18/3)

= 6/6

= 1

R.H.S = 5/3x

Put x = 5/3

R.H.S = 5/(3 x 5/3)

= 5/(15/3)

= 5/5

= 1

L.H.S = R.H.S

Hence proved.

Example 3.

The sum of the digits of a 2 digit number is 8. The number obtained by interchanging the digits exceeds the original number by 18.

Solution:

Let the 2 digit number at ones place is x, then

Tens place of the number = 8 – x

Original number = 10(8 – x) + x

= 80 – 10x + x

= 80 – 9x

By interchanging the digits

New number = 10x + 1(8 – x)

= 10x + 8 – x

= 9x + 8

According to question,

New number – Original number = 18

9x + 8 – (80 – 9x) = 18

9x + 8 – 80 + 9x = 18

18x – 72 = 18

18x = 18 + 72

18x = 90

x = 90/18

x = 5

Hence, the digits at ones place is 5.

The digits at tens place = (8-5) = 3.

So, the original number is 35 and the new number is 53.

Example 4.

The mother’s age is three times her son’s age. Four years ago, she was 4 times her son’s age. Find their present ages.

Solution:

Let the present age of son be x, then

Present mother’s age = 3x

Four years ago,

Son age = (x – 4)

Mothers age = (3x – 4)

According to the question,

Mothers age = 4 x son’s age

3x – 4 = 4(x – 4)

3x – 4 = 4x – 16

-4 + 16 = 4x – 3x

x = 12

Hence, the son’s present age is 12 years, mothers present age is 12 x 3 = 36 years.

Example 5.

If a rectangle has a width of 8 inches and has a perimeter of 36 inches, then what is the length?

Solution:

Given that,

Rectangle width w = 8 inches

Rectangle perimeter P = 36 inches

Perimeter of a rectangle = 2(length + width)

36 = 2(length + 8)

36 = 2length + 16

36 – 16 = 2length

20 = 2length

length = 20/2

length = 10

Hence, the rectangle length is 10 inches.

Example 6.

A bicycle and a bicycle helmet cost 240 dollars. How much did each cost, if the bicycle cost 5 times as much as the helmet?

Solution:

Let the cost of the bicycle helmet be x, then

Cost of the bicycle = 5x

According to the question,

x + 5x = 240

6x = 240

x = 240/6

x = 40

Hence, the cost of a bicycle helmet is 40 dollars, the cost of the bicycle is 40 x 5 = 200 dollars.

Example 7.

The total cost for tuition plus room and board at State University is 2,584 dollars. Tuition costs 704 dollars more than room and board. What is the tuition fee?

Solution:

Let the room and board cost be x, then

Tuition cost = 704 + x

According to the question,

704 + x + x = 2584

704 + 2x = 2584

2x = 2584 – 704

2x = 1880

x = 1880/2

x = 940

Hence, the tuition fee is 940 + 704 = 1664.

Example 8.

A sum of $2700 is to be given in the form of 63 prizes. If the prize is of either $100 or $25, find the number of prizes of each type.

Solution:

Let x be the number of $100 and y be the number of $25.

x + y = 63

100x + 25y = 2700

Multiply both sides of the equation x + y = 63 by 25.

25(x + y) = 63 x 25

25x + 25y = 1575

Subtract 100x + 25y = 2700 from 25x + 25y = 1575

100x + 25y – (25x + 25y) = 2700 – 1575

100x + 25y – 25x – 25y = 1125

75x = 1125

x = 1125/75

x = 15

Substitute x = 15 in x + y = 63

15 + y = 63

y = 63 – 15

y = 48.

Hence 15 number of $100, 43 number of $25 prizes.

Example 9.

Of 240 stamps that harry and his sister collected, Harry collected 3 times as many as his sisters. How many did each collect?

Solution:

Let x be the number of stamps collected by harry’s sister, then

Number of stamps collected by harry = 3x

According to the question,

x + 3x = 240

4x = 240

x = 240/4

x = 60

Hence, the number of stamps collected by harry is 3 x 60 = 180 and his sister is 60.

Example 10.

A lab technician cuts a 12 inch piece off tubing into two pieces in such a way that one piece is 2 times longer than the other.

Solution:

Let, the smaller piece length is, longer piece length is y.

y = 2x

According to the question,

x + y = 12

x + 2x = 12

3x = 12

x = 12/3

x = 4

y = 2 x 4 = 8

Hence, the length of shorter piece is 4 inches, longer piece is 8 inches.

Students who are looking for the practice material on linear equations can stay tuned to this page. Here, we are giving several problems on linear equations. Interested students can solve them by using addition, subtraction, multiplication, and division operation. You can also find the step by step solution guide for each and every question. While practicing this Linear Equations Word Problems just have a look at the important notes mentioned below.

How to Solve Linear Equations?

Go through the below steps on solving linear equations. Follow them and arrive at the solutions easily. They are listed as under

• A linear equation in one variable is one that contains only one variable and its highest power is 1.
• One can add or subtract the same number to both sides of the equation.
• One can divide or multiply by both sides of the equation by the same non-zero integer.
• The process in which any term in the equation can be moved to the other side of the equal symbol by changing its sign from (+ to -), (- to +), (x to ÷), and (÷ to x) is called the transposition.
• Cross multiplication means the process of multiplying the numerator of L.H.S with the denominator of the R.H.S and multiplying the denominator of L.H.S with the numerator of R.H.S.

Linear Equations Questions and Answers

Example 1.

Solve the following linear equations.

(a) (3 – 7x)/(15 + 2x) = 1

(b) 8x + 9 – 3x = 8 + 4x + 1

(c) 3x – 12 = 0

Solution:

(a) (3 – 7x)/(15 + 2x) = 1

Multiply both sides by (15 + 2x).

(3 – 7x) = 1(15 + 2x)

3 – 7x = 15 + 2x

Transfer – 7x to R.H.S becomes 7x

3 = 15 + 2x + 7x

3 = 15 + 9x

Transfer 15 from R.H.S to L.H.s becomes -15.

3 – 15 = 9x

-12 = 9x

Divide both sides by 9

9x / 9 = -12/9

x = -4/3

Therefore required solution is x = -4/3.

(b) 8x + 9 – 3x = 8 + 4x + 1

5x + 9 = 9 + 4x

Transfer 4x from R.H.S to L.H.S becomes -4x

5x – 4x + 9 = 0

x + 9 = 0

Subtract both sides from -9

x + 9 – 9 = 0 – 9

x = -9

Therefore, required solution set is x = -9.

(c) 3x – 12 = 0

3x = 12

Divide both sides of the equation by 3.

3x/3 = 12/3

x = 4

Therefore, required solution set is x = 4.

Example 2.

Solve the following equations and represent them on a graph.

(a) (3y – 2)/3 + (2y + 3)/3 = (y + 7)/6

(b) 5x – 11 = 3x + 9

(c) (0.5y – 9)/0.25 = 4y – 3

Solution:

(a) (3y – 2)/3 + (2y + 3)/3 = (y + 7)/6

[(3y – 2) + (2y + 3)] / 3 = (y + 7) / 6

(5y + 1) = (y + 7) / 2

Multiply both sides by 2.

2(5y + 1) = (y+7) / 2 x 2

10y + 2 = y + 7

Transferring y from R.H.S to L.H.S becomes -y, 2 from L.H.S to R.H.S becomes -2.

10y – y = 7 – 2

9y = 5

y = 5/9

Therefore, the required solution set is y = 5/9.

(b) 5x – 11 = 3x + 9

Transferring 3x from R.H.S to L.H.S becomes -3x, -11 from L.H.S to R.H.S becomes 11.

5x – 3x = 9 + 11

2x = 20

Divide both sides by 2.

2x/2 = 20/2

x = 10.

Therefore, the required solution set is x = 10

(c) (0.5y – 9)/0.25 = 4y – 3

Multiply both sides by 0.25.

(0.5y – 9)/0.25 x 0.25 = (4y – 3)0.25

0.5y – 9 = y – 0.75

Transferring 0.5y from L.H.S to R.H.S becomes -0.5y, -0.75 from R.h.s to L.H.S becomes 0.75.

-9 + 0.75 = y – 0.5y

-8.25 = 0.5y

Divide both sides by 0.5.

0.5y/0.5 = -8.25/0.5

y = -16.5

Therefore, the required solution set is y = -16.5

Example 3.

Solve the equations and verify them.

(a) (x – 3)/4 + (x – 1)/5 – (x – 2)/3 = 1

(b) y/2 – 1/2 = y/3 + 1/4

Solution:

(a) (x – 3)/4 + (x – 1)/5 – (x – 2)/3 = 1

L.C.M of 4, 5, 3 is 60.

[15(x – 3) + 12(x – 1) – 20(x – 2)] / 60 = 1

Multiply both sides of the equation by 60.

[15(x – 3) + 12(x – 1) – 20(x – 2)] / 60 x 60 = 1 x 60

[15(x – 3) + 12(x – 1) – 20(x – 2)] = 60

15x – 45 + 12x – 12 – 20x + 40 = 60

27x – 20x – 57 + 40 = 60

7x – 17 = 60

7x = 60 + 17

7x = 77

x = 77/7

x = 11

Therefore, the required solution set is x = 11.

Verification:

L.H.S = (x – 3)/4 + (x – 1)/5 – (x – 2)/3

Substitute x = 11

L.H.S = (11 – 3)/4 + (11 – 1)/5 – (11 – 2)/3

= 8/4 + 10/5 – 9/3

= 2 + 2 – 3

= 4 – 3 = 1

Hence, L.H.S = R.H.S

(b) y/2 – 1/2 = y/3 + 1/4

(y – 1)/2 = (4y + 3)/12

Cross multiply the fractions.

12(y – 1) = 2(4y + 3)

12y – 12 = 8y + 6

12y – 8y = 6 + 12

4y = 18

y = 18/4

y = 4.5

Verification:

L.H.S = y/2 – 1/2

Substitute y = 4.5

L.H.S = 4.5/2 – 1/2

= (4.5 – 1)/2

= 3.5/2 = 1.75

R.H.S = y/3 + 1/4

Substitute y = 4.5

R.H.S = 4.5/3 + 1/4

= (4.5 x 4 + 3) / 12

= (18 + 3) / 12

= 21/12 = 1.75

Hence, L.H.S = R.H.S

Example 4.

Solve the below-mentioned linear equations.

(a) 8a – (4a + 32) = 16

(b) 4(x + 5) = 3(x – 2) – 2(x + 2)

Solution:

(a) 8a – (4a + 32) = 16

8a – 4a – 32 = 16

4a – 32 = 16

Transferring -32 from L.H.S to R.H.S becomes +32.

4a = 16 + 32

4a = 48

Divide both sides of the equation by 4.

4a/4 = 48/4

a = 12

Therefore, the required solution set a = 12.

(b) 4(x + 5) = 3(x – 2) – 2(x + 2)

4x + 20 = 3x – 6 – 2x – 4

4x + 20 = x – 10

Transferring x from R.H.S to L.H.S becomes -x, 20 from L.H.S to R.H.S becomes -20.

4x – x = -10 – 20

3x = -30

Divide both sides by 3.

3x/3 = -30/3

x = -10

Therefore, the required solution set x = -10.

Example 5.

Solve the following equations and verify them.

(a) ⅓ (21 – 3x) = ½ (8 – 4x)

(b) (0.4y – 3)/(1.5y + 9) = -7/5

Solution:

(a) ⅓ (21 – 3x) = ½ (8 – 4x)

Cross-multiply the fractions.

2(21 – 3x) = 3(8 – 4x)

42 – 6x = 24 – 12x

Transferring -12x from R.H.S to L.H.S becomes 12x, 42 from L.H.S to R.H.S becomes -42.

-6x + 12x = 24 – 42

6x = -18

Divide both sides by 6.

6x/6 = -18/6

x = -3

Verification:

L.H.S = ⅓ (21 – 3x)

Put x = -3

L.H.S = ⅓ (21 – 3(-3))

= ⅓ (21 + 9)

= ⅓ (30)

= 10

R.H.S = ½ (8 – 4x)

Put x = -3

R.H.S = ½ (8 – 4(-3))

= ½ (8 + 12)

= ½ (20)

= 10

L.H.S = R.H.S

Hence proved.

(b) (0.4y – 3)/(1.5y + 9) = -7/5

Cross multiply the fractions.

5(0.4y – 3) = -7(1.5y + 9)

2y – 15 = -10.5y – 63

Transferring -10.5y from R.H.S to L.H.S becomes 10.5y, -15 from L.H.S to R.H.S becomes 15.

2y + 10.5y = 15 – 63

12.5y = -48

Divide both sides by 12.5

12.5y/12.5 = -48/12.5

y = -3.84

Verification:

L.H.S = (0.4y – 3)/(1.5y + 9)

Substitute y = -3.84

L.H.S = (0.4(-3.84) – 3) / (1.5(-3.84) + 9)

= (-1.536 – 3) / (-5.76 + 9)

= -4.536 / 3.24

= -1.4

R.H.S = -7/5

= -1.4

L.H.S = R.H.S.

Hence proved.

We have provided several problems that involve relations among known and unknown numbers and can be put in the form of linear equations. Those equations can be stated in words and it is the main reason we prefer these Word Problems on Linear Equations. You can practice as many types of questions as you want to get an expert in this concept. For better understanding, we even listed linear equations examples with solutions.

Steps to Solve Word Problems on Linear Equations

Below are the simple steps to solve the linear equations word problems. Follow these instructions and solve the questions carefully.

• Read the problem carefully and make a note of what is given in the question and what is required.
• Denote the unknown things as the variables like x, y, z, a, b, . . .
• Translate the given word problem into mathematical statements.
• Form the linear equations in one variable by using the conditions provided in the question.
• Solve the unknown parameters from the equation.
• Verify the condition with the obtained answer ad cross check whether it is correct or not.

Linear Equations Examples with Answers

Example 1.

A motorboat goes downstream in the river and covers a distance between two coastal towns in 5 hours. It covers this distance upstream in 6 hours. If the speed of the stream is 3 km/hr, find the speed of the boat in still water?

Solution:

Let the speed of the boat in still water = x km/hr

Speed of the boat down stream = (x + 3) km/hr

Time taken to cover the distance = 5 hrs

Therefore, distance covered in 5 hrs = (x + 3) x 5

Speed of the boat upstream = (x – 3) km/hr

Time taken to cover the distance = 6 hrs

Therefore, distance covered in 6 hrs – (x – 3) x 6

Therefore, the distance between the two coastal towns is fixed, i.e., the same.

As per the question

5(x + 3) = 6(x – 3)

5x + 15 = 6x – 18

15 + 18 = 6x – 5x

33 = x

x = 33

Required speed of the boat is 33 km/hr

Example 2.

The perimeter of a rectangular swimming pool is 144 m. Its length is 2 m more than twice its width. What are the length and width of the pool?

Solution:

Let l be the length of the swimming pool, w be the width of the swimming pool.

According to the question,

length l = 2w + 2

The perimeter of swimming pool = 144 m

2l + 2w = 144

Substitute l = 2w + 2

2(2w + 2) + 2w = 144

4w + 4 + 2w = 144

6w = 144 – 4

6w = 140

w = 140 / 6

w = 23.3

Then, the length is

l = 2(23.3) + 2

= 46.6 + 2

= 48.6

Hence, the length and width of the rectangular swimming pool is 48.6 m, 23.3 m.

Example 3.

The sum of three consecutive even numbers is 126. What are the numbers?

Solution:

Let the first even number be x, the second number be (x + 2), the third number be (x + 4).

According to the question, the sum of consecutive even numbers is 126.

First Number + Second Number + Third Number = 126

x + (x + 2) + (x + 4) = 126

3x + 6 = 126

Subtract 6 from both sides.

3x + 6 – 6 = 126 – 6

3x = 120

Divide both sides by 3.

3x / 3 = 120 / 3

x = 40

The first number is 40, the second number is (x + 2) = 40 + 2 = 42, third number is (x + 4) = 40 + 4 = 44.

Hence, the three consecutive even numbers are 40, 42, 44.

Example 4.

When five is added to three more than a certain number, the result is 19. What is the number?

Solution:

Let us take the number as x.

According to the question,

Add 5 to the three more than a certain number.

5 + x + 3 = 19

x + 8 = 19

Subtract 8 from both sides of the equation.

x + 8 – 8 = 19 – 8

x = 11.

So, the number is 11.

Example 5.

Eleven less than seven times a number is five more than six times the number. Find the number?

Solution:

Let the number be x.

According to the question,

11 less than the seven times a number is five more than six times the number.

7x – 11 = 6x + 5

7x – 6x = 5 + 11

x = 16

Hence the required number is 16.

Example 6.

Two angles of a triangle are the same size. The third angle is 12 degrees smaller than the first angle. Find the measure of the angles.

Solution:

Let the triangle be ∆ABC.

So, ∠A = ∠B and ∠C = ∠A – 12 degrees

The sum of three angles of a triangle = 180 degrees

∠A + ∠B + ∠C = 180

∠A + ∠A + ∠A – 12 = 180

3∠A – 12 = 180

Add 12 to both sides of the equation.

3∠A – 12 + 12 = 180 + 12

3∠A = 192

Divide both sides by 3.

3∠A / 3 = 192/3

∠A = 64

Hence, the first and second angles of the triangle are 64 degrees, 64 degrees and the third angle is 64 – 12 = 52 degrees.

Example 7.

The perimeter of a rectangle is 150 cm. The length is 15 cm greater than the width. Find the dimensions.

Solution:

Let the rectangle width is w.

Length of rectangle l = w + 15 cm

Given that, the perimeter of a rectangle is 150 cm

2l + 2w = 150

Substitute l = w + 15 cm in above equation.

2(w + 15) + 2w = 150

2w + 30 + 2w = 150

4w + 30 = 150

Subtract 30 from both sides.

4w + 30 – 30 = 150 – 30

4w = 120

Divide both sides by 4.

4w/4 = 120/4

w = 30

Hence, the rectangle width is 30 cm, the length is (30 + 15) = 45 cm.

Example 8.

If Mr. David and his son together had 220 dollars, and Mr. David had 10 times as much as his son, how much money had each?

Solution:

Let Mr. David’s son has x dollars.

The amount at Mr. David = 10x dollars

Given that, Mr. David and his son together had 220 dollars

x + 10x = 220

11x = 220

Divide both sides by 11.

x = 220 / 11

x = 20

Hence, Mr. David has 20 x 10 = 200 dollars and his son has 20 dollars.

Solved questions on linear equations in one variable are provided below with a detailed explanation. You have to eliminate brackets and expand the given linear equation. Find the variable and translate the problem to the mathematical statement. Use the conditions and solve the equation. After that verify whether the answer satisfies the condition or not.

How To Solve Linear Equation in One Variable?

Go through the step by step procedure listed for Solving Linear Equations in One Variable. They are mentioned below

• Read the given question twice or many times until you observe the parameters like what is given and what you need to find.
• To make a strategy, represent the unknown values as variables.
• Convert the word problem into mathematics.
• Portray the problem as a linear equation in one variable using the conditions provided.
• Solve the equation for unknown.
• Verify to be sure whether the answer satisfies the conditions of the problem.

Linear Equations in One Variable Word Problems with Solutions

Example 1.

The sum of two numbers is 44. If one exceeds the other by 6, find the numbers?

Solution:

Let x be one of the two numbers.

Then, another number is (x + 6).

The Sum of two numbers is 44.

x + x + 6 = 44

2x + 6 = 44

2x = 44 – 6

2x = 38

x = 38/2

x = 19.

The second number is x + 6 = 19 + 6 = 25

So, the two numbers are 19 and 25.

Example 2.

The sum of four consecutive multiples of 5 is 650. Find these multiples?

Solution:

If x is a multiple of 5, the next multiple is x + 5, third number is x + 10, and fourth multiple is x + 15.

The sum of four consecutive multiples of 5 is 650

x + x + 5 + x + 10 + x + 15 = 650

4x + 30 = 650

Subtract 30 from both sides

4x + 30 – 30 = 650 – 30

4x = 620

Divide both sides by 4.

4x/4 = 620/4

x = 155

The first multiple is 155, the second multiple is 155 + 5 = 160, the third multiple is 155 + 10 = 165, fourth multiple is 155 + 15 = 170.

So, the consecutive multiple of 5 is 155, 160, 165, and 170.

Example 3.

Two numbers are in the ratio 8 : 1. If they differ by 126, what are those numbers?

Solution:

Given ratio is 8:1

From the ratio two numbers can be assumed as 8x, x.

The difference between numbers is 126

8x – x = 126

7x = 126

Divide both sides by 7.

7x/7 = 126/7

x = 18

8x = 8 x 18 = 144.

Hence, two numbers are 18, 144.

Example 4.

The ratio of the three numbers is 5: 6: 7. If the sum of those three numbers is 54, find those numbers?

Solution:

The given ratio of three numbers are 5: 6: 7

From the ratio, the numbers can be assumed as 5x, 6x, and 7x.

The sum of three numbers is 54

5x + 6x + 7x = 54

18x = 54

Divide both sides by 18.

18x / 18 = 54 / 18

x = 3

So, the numbers be

5x = 5 x 3 = 15

6x = 6 x 3 = 18

7x = 7 x 3 = 21.

Hence, the three numbers are 15, 18, and 21.

Example 5.

If you subtract 2/3 from a number and multiply the result by 2/3, you will obtain 1/6. What is the number?

Solution:

Let us say x is the required number.

From the given information, we can write

(x – 2/3) * 2/3 = 1/6

2x/3 – 4/9 = 1/6

2x/3 = 1/6 + 4/9

2x/3 = (3 + 8)/18

2x/3 = 11/18

Cross multiply the fractions.

2x * 18 = 11 * 3

36x = 33

x = 33/36

So, the number is 33/36.

Example 6.

A total of $1500 is distributed among 100 persons as compensation for the work. The given compensation is either of $50 or $100. Find the number of compensations of each type?

Solution:

Total number of compensations = 100

Let the number of compensations of $50 is x

Then the number of compensations of $100 is ($100 – x)

Amount spend on x compensations of $50 = $50x

Amount spend on (150 – x) compensations of $100 = $100(100 – x)

The total amount spent for compensation = $1500

According to the question,

50x + 100(100 – x) = 1500

50x + 10000 – 100x = 1500

-50x + 10000 = 1500

-50x = 1500 – 10000

-50x = -8500

50x = 8500

x = 8500 / 50

x = 170

100 – x = 100 – 170 = 70

Therefore, compensation of $50 are 170, and compensations of $100 are 70.

Example 7.

Divide 28 into two parts in such a way that 6/5 of one part is equal to 2/3 of the other.

Solution:

Let one part be x.

Then other art = 28 – x

It is given 6/5 of one part = 2/3 of the other.

6x/5 = 2/3(28 – x)

6x/5 = (28 x 2)/3 – 2x/3

6x/5 + 2x/3 = 56/3

(18x + 10x) / 15 = 56/3

28x/15 = 56/3

Cross multiply the fractions.

28x * 3 = 56 * 15

84x = 840

x = 840 / 84

x = 10

Then the two parts are 10 and 28 – 10 = 18.

Example 8.

The numerator of a rational number is less than its denominator by 3. If the denominator is increased by 7 and the numerator is decreased by 1, the new number becomes 3/2. Find the original number?

Solution:

Let the denominator of a rational number = x

Then the numerator of the rational number = x – 3

When denominator is increased by 7, then new denominator = x + 7

When the numerator is decreased by 1, then new numerator = x – 3 – 1 = x – 4

The new number formed = 3/2

According to the question,

(x – 4) / (x + 7) = 3/2

Cross multiply.

2(x – 4) = 3(x + 7)

2x – 8 = 3x + 21

-8 – 21 = 3x – 2x

-29 = x

x = -29

The original number i.e., (x – 3) / x = (-29 – 3) / -29 = (-32)/-29 = 32/29.

Do you want to learn how to solve linear equations? Then stay on this page, here we are giving the detailed steps of solving the linear equations. Go through the simple steps and some worked example problems to get a clear idea on the topic. You can easily solve the linear equations by using addition, subtraction, multiplication, and division operations.

Steps to Solve a Linear Equation

Have a look at the detailed steps and solve the linear equations effortlessly. Follow the guidelines and arrive at the solutions easily. They are as follows

• Transfer variables to one side and constants to the other side of the equation.
• Use arithmetic operations to transfer those.
• Find the value of a variable.
• The variable value becomes the required solution.

Linear Equations Questions with Solutions

Example 1.

Solve the equation (x + 4) / 2 = 8 + (x – 6) / 5 and verify your answer?

Solution:

Given linear equation is (x + 4) / 2 = 8 + (x – 6) / 5

(x + 4) / 2 = [40 + (x – 6)] / 5

Cross multiply the fraction.

5(x + 4) = 2(40 + x – 6])

5x + 20 = 2(x + 34)

5x + 20 = 2x + 68

Transferring 2x to L.H.S. changes to negative 2x and 20 to R.H.S. changes to -20.

5x – 2x = 68 – 20

3x = 48

x = 48/3

x = 16

Therefere, x = 16 is the solution for the equation.

Verification:

L.H.S = (x + 4) / 2

Put x = 16

L.H.S = (16 + 4) / 2

= 20/2 = 20

R.H.S = 8 + (x – 6) / 5

Put x = 16

R.H.S = 8 + (16 – 6) / 5

= 8 + 10/5

= (40 + 10) / 5

= 50/5 = 10

Since, L.H.S = R.H.S, verified.

Example 2.

Solve the equation 12 – x/6 = 7?

Solution:

Given linear equation is 12 – x/6 = 7

(72 – x) / 6 = 7/1

Cross multiply.

72 – x = 7 x 6

72 – x = 42

72 – 42 = x

30 = x

x = 30

Therefore, x = 30 is the solution for the equation.

Example 3.

Solve the equation 3(x – 3) = 4(2x + 1) and represent the solution graphically?

Solution:

Given linear equation is 3(x – 3) = 4(2x + 1)

3x – 9 = 8x + 4

-9 – 4 = 8x – 3x

-15 = 5x

x = -15/5

x = -3

Therefore, x = -3 is the solution for the equation.

Example 4.

Solve (x + 5) / (x – 5) = 1/2?

Solution:

Given linear equation is (x + 5) / (x – 5) = 1/2

Cross multiply the fraction

2(x + 5) = x – 5

2x + 10 = x – 5

2x – x = -5 – 10

x = -15

Therefore, x = -15 is the solution for the equation.