Common Core Math

Factorization of Perfect Square is the process of finding factors for an equation which is in the form of a² + 2ab + b² or a² – 2ab + b². Get to know the step by step procedure involved for finding factors of a perfect square. Have a look at the different examples taken to illustrate the Factorization of Perfect Square Problems. By following this article, you will better understand the concept and solving process of perfect square factorization.

(i) a² + 2ab + b² = (a + b)² = (a + b) (a + b)
(ii) a² – 2ab + b² = (a – b)² = (a – b) (a – b)

Factorization of Perfect Square Solved Examples

1. Factorize the perfect square completely
(i) 16a² + 25b² + 40ab

Solution:
Given expression is 16a² + 25b² + 40ab
The given expression 16a² + 25b² + 40ab is in the form a² + 2ab + b².
So find the factors of given expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = 4a, b = 5b
Apply the formula and substitute the a and b values.
16a² + 25b² + 40ab
(4a)² + 2 (4a) (5b) + (5b)²
(4a + 5b)²
(4a + 5b) (4a + 5b)

Factors of the 16a² + 25b² + 40ab are (4a + 5b) (4a + 5b)

(ii) 9x² – 42xy + 49y²

Solution:
Given expression is 9x² – 42xy + 49y²
The given expression 9x² – 42xy + 49y² is in the form a² – 2ab + b².
So find the factors of given expression using a² – 2ab + b² = (a – b)² = (a – b) (a – b) where a = 3x, b = 7y
Apply the formula and substitute the a and b values.
9x² – 42xy +49y²
(9x)² – 2 (9x) (7y) + (7y)²
(9x – 7y)²
(9x – 7y) (9x – 7y)

Factors of the 9x² – 42xy + 49y² are (9x – 7y) (9x – 7y)

(iii) 25m² + 80m + 64

Solution:
Given expression is 25m² + 80m + 64
The given expression 25m² + 80m + 64 is in the form a² + 2ab + b².
So find the factors of given expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = 5m, b = 8
Apply the formula and substitute the a and b values.
25m² + 80m + 64
(5m)² + 2 (5m) (8) + (8)²
(5m + 8)²
(5m + 8) (5m + 8)

Factors of the 25m² + 80m + 64 are (5m + 8) (5m + 8)

(iv) a² + 6a + 8

Solution:
Given expression is a² + 6a + 8.
The Given expression is a² + 6a + 8 is not a perfect square.
Add and subtract 1 to make the given expression a² + 6a + 8 is not a perfect square.
a² + 6a + 8 + 1 – 1
a² + 6a + 9 – 1
The above expression a² + 6a + 9 is in the form a² + 2ab + b².
So find the factors of the expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = a, b = 3
Apply the formula and substitute the a and b values.
a² + 6a + 9
(a)² + 2 (a) (3) + (3)²
(a + 3)²
(a + 3)² – 1
(a + 3)² – (1)²
(a + 3 + 1) (a + 3 – 1)
(a + 4) (a + 2)

Factors of the a² + 6a + 8 are (a + 4) (a + 2)

2. Factor using the identity

(i) 4x⁴ + 1

Solution:
Given expression is 4x⁴ + 1.
The Given expression is 4x⁴ + 1 is not a perfect square.
Add and subtract 4x² to make the given expression 4x⁴ + 1 is not a perfect square.
4x⁴ + 1 + 4x² – 4x²
4x⁴ + 4x² + 1 – 4x²
The above expression 4x⁴ + 4x² + 1 is in the form a² + 2ab + b².
So find the factors of the expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = 2x², b = 1
Apply the formula and substitute the a and b values.
4x⁴ + 4x² + 1
(2x²)² + 2 (2x²) (1) + (1)²
(2x² + 1)²
(2x² + 1)² – 4x²
(2x² + 1)² – (2x)²
(2x² + 1 + 2x) (2x² + 1 – 2x)
(2x² + 2x + 1) (2x² – 2x + 1)

Factors of the 4×4 + 1 are (2x² + 2x + 1) (2x² – 2x + 1)

(ii) (a + 2b)² + 2(a + 2b) (3b – a) + (3b – a)²

Solution:
Given expression is (a + 2b)² + 2(a + 2b) (3b – a) + (3b – a)²
The given expression (a + 2b)² + 2(a + 2b) (3b – a) + (3b – a)² is in the form a² + 2ab + b².
So find the factors of given expression using a² + 2ab + b² = (a + b)² = (a + b) (a + b) where a = a + 2b, b = 3b – a
Apply the formula and substitute the a and b values.
(a + 2b)² + 2(a + 2b) (3b – a) + (3b – a)²
(a + 2b)² + 2 (a + 2b) (3b – a) + (3b – a)²
(a + 2b + 3b – a)²
(5b)²
25b²

Factors of the (a + 2b)² + 2(a + 2b) (3b – a) + (3b – a)² are 25b²

Looking for help on the concept Train Passes through a Bridge then you can find all of it here. Refer to Solved Examples on How to find the Speed Time and Distance when a Train Crosses a Bridge. We have provided solutions for all the Problems covered on the concept of Train Passing through a Bridge. or Tunnel or any Stationary Object.

In this case, the Length of the Train and Length of the Bridge are added to get the Distance. Practice the Train Passes through a Bridge Problems with Answers and get the elaborate explanation provided to understand the concept better.

How to Solve Train Passing through a Bridge or Tunnel Problems?

Let us consider the Length of the Train = x meters

Length of the Stationary Object = y meters

The Speed of the Train is z km/hr

Time Taken by the Train to Cross the Bridge = (Length of Train + Length of Bridge)/Speed of Train

= (x+y)m /z km/hr

To change between km/hr to m/sec multiply with 5/18

Solved Examples on Train Crossing a Stationary Object having Some Length

1. A train 250 m long crosses a bridge which is 100 m long in 50 seconds. What is the speed of the train?

Solution:

Length of the Train = 250 m

Length of the Bridge = 100 m

Speed of the Train = (Length of the Train+Length of Stationary Object)/Time

= (250 m+100 m)/50

= 350 m/50 sec

= 7 m/sec

2. A train 260 m long is running at a speed of 40 km/hr. What time will it take to cross an 80 m long tunnel?

Solution:

Length of the Train = 260 m

Speed of the Train = 40 km/hr

= 40 *5/18 m/sec

= 11.11 m/sec

Length of the Tunnel = 80 m

Time = Distance/Speed

= 260 m/11.11 m/sec

= 23.40 sec

Train takes 23.40 sec to travel the tunnel.

3. Find the time taken by a 180 m long train passes through a bridge which is 75 m long, running at a speed of 54 km/hr?

Solution:

Length of the Train = 180 m

Length of the Bridge = 75 m

Speed of the Train = 54 km/hr

Speed of train in m/sec = 54*5/18

= 15

Time = Distance/Speed

= 225 m/15 m/sec

= 9 sec

Therefore, train takes 9 sec to cross the bridge.

4. A 120 m long train is running at a speed of 45 km/hr. If it takes 20 seconds to cross a platform, find the length of the platform?

Solution:

Speed of the Train = 45 km/hr

Speed of Train in m/sec = 45*5/18

= 12.5 m/sec

Time taken to cross the platform = 20 sec

Length of the Train = 120 m

Consider the Length of Platform = x

Speed = (Length of Train +Length of Platform)/Time

12.5 m/sec = (120+x)/20

12.5*20 = 120+x

250 = 120+x

x = 250 – 120

= 130 m

Therefore, Length of Platform is 130 m.

Practice the Questions available below when Train Passes through a Pole and get full-fledged knowledge about the concept. Know How to Calculate Speed Time and Distance when a train crosses a stationary object in the coming modules. Train Passes a Pole, Man or Tree Problems with Solutions along with step by step explanation makes it easy for you to understand the concept. Refer to them and learn how to solve the Train Passing through a Pole Problems easily.

How to Solve Train Passes through a Pole Problems?

When a Train Passes through a Pole or Tree the formulas to find the Time Speed and Distance are given below

Suppose the Length of the Train = x mts

Speed of the Train = y km/hr

Time taken by the Train to Cross a Pole or Stationary Object = Length of the Train/Speed of the Train

= x meters/y km/hr

To Change km/hr to m/sec simply multiply with 5/18.

Solved Examples to Calculate When Train Passes through a Pole

1. A train 200 m long is running at a uniform speed of 75 km/hr. How much time will it take to cross a pole?

Solution:

Length of the Train = 200 m

Speed of the Train = 75 km/hr

To convert it to m/sec multiply with 5/18

= 75*5/18

= 20.83 m/sec

Time taken by the train to cross the pole = Length of the Train/Speed of the Train

= 200 m/20.83 m/sec

= 9.6 sec

Therefore, Train takes 9.6 sec to cross the pole.

2. Find the time taken by a train 350 m long, running at a speed of 60 km/hr in crossing the pole?

Solution:

Length of the Train = 350 m

Speed of the Train = 60 km/hr

Time taken by the Train to cross the pole = Length of the Train/Speed of the Train

= 350m/60 km/hr

To change from km/hr to m/sec multiply with 5/18 i.e. 60*5/18 = 16.66 m/sec

= 350 m/16.66m/sec

= 21.008 sec

Thus, the train takes 21.008 sec to cross the pole.

3. A train is running at a speed of 54 km/hr. It crosses a tower in 7 seconds. Find the Length of the train?

Solution:

Speed of the Train = 54 km/hr

Time taken by train to cross tower = 7 seconds

Length of the Train = Speed * Time

= 54*(5/18)*7

= 105 m

Therefore, the Length of the Train is 105 m

4. A train is running at a speed of 140 km/hr. if it crosses a pole in just 8 seconds, what is the length of the train?

Solution:

Speed of the Train = 140 km/hr

Time taken to cross the pole = 8 seconds

Length of the Train = Speed * Time

= 140 km/hr*8

= 140*(5/18)*8

= 311.11 m

Therefore, the length of the train is 311.11 m

Are you looking for help on the concept Train Passes a Moving Object in the Opposite Direction? Then you have reached the right place. Learn the Formulas for Speed Time and Distance in the case of a train crossing a moving body in the opposite direction. Get Solved Examples for finding the Train Crossing a Moving Object in Opposite Direction long with detailed solutions. Practice the Problems available and get a good hold of the concept.

How to calculate Time Speed and Distance for Train Crossing a Moving Object in Opposite Direction?

Let us assume the length of the train = l m

Speed of the train =  x km/hr

Speed of the object = y km/hr

Relative Speed = (x+y) km/hr

Time taken by train to cross the moving object = Distance/Speed

= l m/(x+y) km/hr

Simply rearrange the formula to obtain the other measures if few are known.

Solved Problems on Train Passes a Moving Object in the Opposite Direction

1. A train 200 m long is running at a speed of 50 km/hr. In what time will it pass a man who is running at the speed of 4 km/hr in the opposite direction in which the train is moving?

Solution:

Length of the Train = 200 m

Speed of the Train = 50 km/hr

Speed of Man = 4 km/hr

Relative Speed = (50+4) Km/hr

= 54 km/hr

Time taken by train to cross a man = Distance/Relative Speed

= 200 m/54 Km/hr

= 200 m/ (54*5/18) m/sec

= 200 m/15 m/sec

= 13.33 sec

Therefore, the train takes 13.33 sec to cross the man.

2. Two trains 125 meters and 170 meters long are running in the opposite direction with speeds of 60 km/hr and 45 km/hr. In how much time they will cross each other?

Solution:

Total Distance Covered = 125+170

= 295 m

Speed of first train = 60 kmph

Speed of second train = 45 kmph

Relative Speed = 60+45

= 105 kmph

Relative Speed in m/sec = 105*5/18

= 29.1 m/sec

Time taken by trains to cross each other = Distance/Speed

=295m/29.1 m/sec

= 10.1 sec

Therefore, it takes 10.1 sec for both the trains to cross each other.

3. Two trains running in opposite directions cross a man standing on the platform in 30 seconds and 21 seconds respectively and they cross each other in 25 seconds. The ratio of their speeds is?

Solution:

Let us consider the speed of two trains be x m/sec and y m/sec

Length of the first train = 30x m(since distance = Speed*Time)

Length of the second train = 21y m

From the given data

Time taken by both trains to cross each other = 25 sec

(30x+21y)/(x+y) = 25

30x+21y = 25x+25y

30x-25x = 25y-21y

5x=4y

x/y = 4/5

Therefore, the Ratio of Speeds is 4:5

Gain complete knowledge on the concept Train Passes a Moving Object in the Same Direction. Learn related formulas like Speed, Time and Distance when a train passes through a moving object in the Same Direction. Get the Step by Step Procedure along with a detailed explanation for the entire concept. Solve Problems on Train Passing a Moving Object in the Same Direction and understand the concept behind them easily.

How to find Time Speed and Distance when a Train Passes a Moving Object in the Same Direction?

Follow the guidelines for calculating the Time Speed and Distance when a Train Passes a Moving Body in the Same Direction. They are as such

Let us consider the length of the train as l mt and the speed of the train is x km/hr

Speed of the Object = y km/hr

Relative Speed = (x-y) km/hr

Time Taken by the train to cross a moving object in the same direction is = Distance/ Relative Speed

= l m/(x-y) km/hr

You can rearrange the equation and find whichever measure you need as a part of your work.

Solved Problems on Train Passes through a Moving Object in the Same Direction

1. A train 150 m long is running at a speed of 50 km/hr. At what time will it pass a man who is running at the speed of 5 km/hr in the same direction in which the train is moving?

Solution:

Length of the Train = 150 m

Speed of the Train = 50 Km/hr

Speed of the Man = 5 km/hr

Relative Speed = Speed of Train – Speed of Man

= 50 – 5

= 45 km/hr

= 45 *5/18

= 12.5 m/sec

Time Taken by Train to Cross the Man = Distance/Speed

= 150 m/12.5 m/sec

= 12 sec

Therefore, Train takes 12 sec to cross the man.

2. Two trains 110 meters and 140 meters long are running in the same direction with speeds of 70 km/hr and 55 km/hr. In how much time will the first train cross the second?

Solution:

Distance Covered = 110+140

= 250 meters

Speed of first train = 70 km/hr

Speed of second train = 55 km/hr

Relative Speed = (70 – 55)

= 15 km/hr

Relative Speed in m/sec = 15*5/18

= 4.1 m/sec

Time taken by first train to cross second = Distance/Speed

= 250 m/4.1 m/sec

= 60.9 sec

Therefore, the first train takes 60.9 sec to cross the second train.

3. A train running at 60 kmph takes 30 seconds to pass a platform. Next, it takes 10 seconds to pass a man walking at 5 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform?

Solution:

Let us consider the length of the train and length of the platform as x and y

Distance traveled by train while crossing the platform is x+y

Time taken to cross the platform is 30 sec

Speed of the train = 60 kmph

= 60 *5/18

= 16.66 m/sec

Time taken by train to cross the platform is

Time = Distance/Speed

30 =(x+y)/16.66  …….(1)

Time taken by train to cross the platform next = 10 sec

Speed of the man = 5 kmph

Relative Speed = 60 kmph – 5 kmph

= 55 kmph

= 55 *5/18

= 15.2 m/sec

Time taken by train to cross the man is

Time = Distance/Speed

10 sec = x/(15.2 m/sec)

x =  152.7 m

By applying the value of x in equation 1 we have

30 =(152.7+y)/16.66

499.8 = 152.7+y

y = 347.1 m

Hence the length of the train and platform are 152.7 m and 347.1 m respectively.

Get full-fledged knowledge when two objects move in opposite direction by going through the complete article. Learn Formula, Solved Examples on Two Bodies Moving in Opposite Direction towards each other. Get Step by Step Solutions for all the Problems on finding Relative Speed, Time and Distance. If two bodies move in opposite direction relative speed is obtained by adding the speeds of both the bodies.

How to Calculate Time Speed and Distance for Two Objects Moving in Opposite Direction?

Consider Two Objects Moving in Opposite Direction with different speeds

Speed of 1st Object = x km/hr

Speed of 2nd Object = y km/hr

Relative Speed = (x+y) km/hr

The speed of One Object with respect to another is called relative speed.

Distance between them = d km

Time after two objects meet = d km/(x+y) km/hr

Distance covered in t hours = relative speed * time

= (x+y) km/hr * t

Solved Examples on Two Object Moving in Opposite Direction

1. Two Athletes are running from the same place at the speed of 10 km/hr and 5 km/hr. find the distance between them after 20 minutes if they move in the opposite direction?

Solution:

Speed of 1st Athlet = 10 km/hr

Speed of 2nd Athlet = 5 km/hr

Relative Speed = (10+5) km/hr

= 15 km/hr

Time = 20 min = 20/60 = 1/3 hr

Distance = Relative Speed * Time

= 15 km/hr * 1/3 hr

= 5 km

Distance between Two Athlets is 5 Km.

2. Two cars travel from the same location at the speed of 10 km/hr and 5 km/hr respectively. Calculate the distance between the cars after 30 minutes given that both cars are traveling in the opposite direction?

Solution:

Speed of 1st Car = 10 km/hr

Speed of 2nd Car = 5 km/hr

Time = 30 minutes = 1/2 hr

Relative Speed of Cars = (10+5) km/hr

= 15 km/hr

Distance between Cars = Relative Speed * Time

= 15 km/hr * 1/2 hr

= 7.5 km

Both cars have a distance of 7.5 km

When Two Objects move in the same direction then their relative speed is calculated by the difference of their speeds. Get to know about the Time and Distance Formulas when two objects move in the same direction. Get to all about the concept objects that move in the same direction by going through the entire article. Learn, how to calculate the Speed when two objects travel in the same direction by checking the solved examples provided.

How to Calculate Speed Time and Distance when Two Objects Move in the Same Direction?

Let us assume two bodies or objects are moving in the same direction having different speeds.

Suppose the Speed of 1st Object is x km/hr

Speed of 2nd Object is y km/hr

Thus, the Relative Speed = (x – y) km/hr[if x>y]

Time taken by two objects to meet = Distance Traveled/Relative Speed

= d km/(x-y) km/hr

We know that Relative Speed is the Speed of the object with respect to one another

Consider the taken after both the bodies meet = t hrs

Distance covered in t hrs = Time * Relative Speed

= (x-y)km/hr *t hrs

For better understanding refer to the solved problems explaining how to calculate when two objects move in the same direction.

Solved Problems on Two Objects Moving in the Same Direction

1. Two athletes are running from the same place at the speed of 8 km/hr and 6 km/hr. find the distance between them after 20 minutes if they move in the same direction?

Solution:

Speed of Athlet A = 8 km/hr

Speed of Athlet B = 6 km/hr

Time = 20 mins = 1/3 hr

Relative Speed = (8-6) km/hr = 2 km/hr

Distance traveled by them = Relative Speed of Athlets/Time Taken

= 2 km/hr/(1/3) hr

= 6 km

Distance between after they travel for 20 mins is 6 Km.

2. Two vehicles are traveling from the same location at the speed of 10 km/hr and 7 km/hr respectively. Calculate the distance between the vehicles after 15 minutes given that both vehicles are traveling in the same direction?

Solution:

Speed of first Vehicle = 10 km/hr

Speed of second Vehicle = 7 km/hr

Relative Speed = (10-7) km/hr

= 3 km/hr

Time = 15 minutes = 15/60 = 1/4 hr

Distance between both the vehicles = Relative Speed * Time

= 3km/hr *1/4 hr

= 3/4 km

= 0.75 km

Therefore, the distance between both the vehicles is 0.75 km

Learn completely on how to convert from one unit of speed to others from here. Practice the Questions in Conversion of Units of Speed here and get a good hold of the concept. Convert from km/hr to m/sec and m/sec to km/hr easily by going through the further sections. To know more about Speed Time and Distance you can always look up to us. Check out Solved Examples on Speed Conversions so that you can solve related problems on your own. Step by Step solutions provided makes it easy for you to understand the Units of Speed Conversion in an effective manner.

How to Convert Km/hr to m/sec?

To convert Km/hr to m/sec follow the below listed guidelines. They are as such

We know 1 km = 1000m and 1 hour = 60 minutes and in turn 1 minute = 60 sec

1 hour = 60*60 = 3600 sec

Dividing km/hr we have = 1000m/3600sec

On simplifying we have km/hr = 5/18 m/sec

Therefore to convert from km/hr we simply multiply with 5/18.

Solved Problems on Converting Km/hr to m/sec

1. Convert 54 km/hr to m/sec?

Solution:

To convert, 54 km/hr to m/sec simply multiply with 5/18

= 54*5/18

= 15 m/sec

Therefore 54km/hr converted to m/sec is 15m/sec.

2. The speed of the bike is 108 km/hr, what is its speed in m/sec?

Solution:

Given Speed of Bike = 108 km/hr

To obtain the Speed in m/sec simply multiply with 5/18

= 108*5/18

= 30 m/sec

Therefore, the Speed of the Bike is 30 m/sec.

3. A car covers a distance of 120 km in the first three hours, 50 km in the next 1 hour, and 25 km in the next 1/2 hour. Convert the speed into m/sec?

Solution:

Total Distance traveled by the Car = (120+50+25) = 195 km

Total Time Taken = (3+1+1/2) = 4 1/2 hrs

Speed of the Car = 195km/4.5 hr = 43.3 km/hr

Speed of the Car in m/sec = 43.3*18/5

= 155.88 m/sec

How to Convert m/sec to km/hr?

Follow the below listed guidelines inorder to change between m/sec to km/hr

We know 1km = 1000m

thus, 1m = 1/1000 km

1 hour = 3600 sec

thus, 1 sec = 1/3600 hr

applying the same we have 1 m/1 sec = (1/1000km)/(1/3600 hr)

= 3600km/1000hr

= 18/5 km/hr

Therefore, to convert from m/sec to km/hr simply multiply with 18/5

Solved Problems on Converting m/sec to km/hr

1. Convert 60 m/sec to km/hr?

Solution:

Given 60 m/sec

to change to km/hr multiply with 18/5

= 60*18/5

= 216 km/hr

2. The speed of a cyclist is 15 m/sec. find the speed in km/hr?

Solution:

Speed of a Cyclist = 15 m/sec

To convert to km/hr multiply with 18/5

= 15*18/5

= 54 km/hr

Factorization is nothing but the breaking down an entity (a number, a matrix, or a polynomial) into the product of another entity or factors. When you multiply the product of the factors, you will get the original number or matrix. Difficult algebraic or quadratic equations can reduce into small forms using the Factorization method. The factors of equations can be a variable, an integer, or an algebraic expression itself.

In simple words, Factorization can be explained as the reverse process of multiplication.

Examples:
(i) Product: 4x (3x – 6y) = 12x² – 24xy; Factorization: 12x² – 24xy = 4x (3x – 6y)
(ii) Product: (x + 4)(x – 3) = x² + x – 12; Factorization: x² + x – 12 = (x + 4)(x – 3
(iii) Product: (4a + 6b)(4a – 6b) = 16a² – 36b²; Factorization: 16a² – 36b² = (4a + 6b)(4a – 6b)

Maths Factorization

Find out different Factorization concepts and concern links below. You can learn each individual concept with a clear explanation with the help of the below links. We have explained every topic details separately with the solved examples. Therefore, students can easily get a grip on Factorization concepts by referring to our detailed concepts.

• Factors of Algebraic Expressions
• Monomial is a Common Factor
• Factorization when Monomial is Common
• Binomial is a Common Factor
• Factorization when Binomial is Common
• Factorization by Grouping
• Factorize by Grouping The Terms
• Factoring Terms by Grouping
• Factorization by Regrouping
• Factorize by Regrouping The Terms
• Factoring Terms by Regrouping
• Factorization by Using Identities
• Factorization of Perfect Square
• Factorization of Perfect Square Trinomials
• Difference of Two Squares
• Factoring Differences of Squares
• Factorize the Difference of Two Squares
• Evaluate the Difference of Two Squares
• Factorize the Trinomial x^2 + px + q
• Factorize the Trinomial ax^2 + bx + c
• Factorization of Quadratic Trinomials

Simple Factorization

Simple Factorization can be easily understood by the below examples.

(i) HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
(ii) HCF of literal coefficients = product of each common literal raised to the lowest power.

Factor

The factor is something that is to be multiplied.
Sum = term + term
Product = factor × factor
For example, x = m(n + 1); m and n + 1 are the factors.

Examples:

1. Factorize 4x²y² – 2xy

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 4 and 2 is 2.
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 4x²y² – 2xy is 2xy.
Multiply and divide each term of the given expression 4x²y² – 2xy with 2xy
2xy((4x²y²/2xy) – (2xy/2xy)) = 2xy (2xy – 1)

The final answer is 2xy (2xy – 1)

2. Find the HCF of 24a³b²c³ and 27a⁴bc⁴.

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 24 and 27 is 3.
HCF of literal coefficients:
The lowest power of a is 3.
The lowest power of b is 1.
And, the lowest power of c is 3.
Therefore, the HCF of literal coefficients is a³bc³.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 24a³b²c³ and 27a⁴bc⁴ is 3a³bc³.

The final answer is 3a³bc³

3. Find the HCF of 2a²bc, 4a³b and 14bc.

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 2, 4, and 14 is 2.
HCF of literal coefficients:
The lowest power of a is 0.
The lowest power of b is 1.
And, the lowest power of c is 0.
Therefore, the HCF of literal coefficients is b.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 2a²bc, 4a³b and 14bc is 2b.

The final answer is 2b.

Factorization Solved Examples

1. Factorize 4y³ – 24y⁵

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 4 and 24 is 4.
HCF of literal coefficients:
The lowest power of y is 3
Therefore, the HCF of literal coefficients is y³.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4y³ – 24y⁵ is 4y³.
Multiply and divide each term of the given expression 4y³ – 24y⁵ with 4y³
4y³((4y³/4y³) – (24y⁵ /4y³) = 4y³(1 – 6y²)

The final answer is 4y³(1 – 6y²)

2. Factorize 21m²n⁵ – 7mn² + 28m⁵n

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 21, 7, and 28 is 7.
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 21m²n⁵ – 7mn² + 28m⁵n is 7mn.
Multiply and divide each term of the given expression 21m²n⁵ – 7mn² + 28m⁵n with 7mn
7mn((21m²n⁵ /7mn) – (7mn² /7mn) + (28m⁵n/7mn)) = 7mn(3mn⁴ – n + 4m⁴)

The final answer is 7mn(3mn⁴ – n + 4m⁴)

3. Factorize 3m(d + 5e) – 3n(d + 5e)

Solution:
Firstly, find the HCF of both given terms.
HCF of their numerical coefficients 3 and 3 is 3.
HCF of literal coefficients:
The lowest power of m is 0
The lowest power of n is 0
Therefore, the HCF of literal coefficients is d+ 5e.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3m(d + 5e) – 3n(d + 5e) is 3(d + 5e).
Multiply and divide each term of the given expression 3m(d + 5e) – 3n(d + 5e) with 3(d + 5e)
3(d + 5e)((3m(d + 5e)/3(d + 5e)) – (3n(d + 5e)/3(d + 5e))) = 3(d + 5e)(m – n)

The final answer is 3(d + 5e)(m – n)

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Quadratic Equation is a second-degree polynomial equation with one variable x. The graph of quadratic equations makes nice curves. For every equation, we have two values of the variables called the roots. Get the formula and simple step by step process to solve the roots of any quadratic equation in the following sections.

Quadratic Equation Definition

The quadratic equation is an equation where the highest exponent of the variable is square. The standard form of quadratic equations is ax² + bx + c = 0. Where x is the variable, a, b, c are the constants and a should not be equal to zero. The power of x should not be negative and must be 2.

Quadratic Equation Formula

The formulas to find the solution or roots of the quadratic equation are given below:

(α, β) = [-b ± √(b² – 4ac)] / 2a

The values of the variable satisfying the given quadratic equation are called its roots. The roots of the quadratic equation details are mentioned here.

• Both roots of the equation are zero when b and c are zero.
• One of the roots of the quadratic equation is zero and the other is -b/a if c = 0
• The two roots are reciprocal to each other when the coefficients a and c are equal.

Nature of Roots of Quadratic Equation

In the quadratic equation formula, the term (b² – 4ac) is called the discriminant of the equation and it gives information about the nature of the roots. Below listed are the conditions that define the nature of roots.

• If the discriminant value is zero, then the equation will have equal roots i.e α = β = -b/2a.
• If the discriminant value is greater than zero, then the equation will have real roots.
• If the discriminant value is less than zero, then the equation will have imaginary roots.
• If the discriminant is greater than zero and it is a perfect square, then the equation will have rational roots.
• If the discriminant is greater than zero and it is not a perfect square, then the equation will have irrational roots.
• If the discriminant is greater than zero, perfect square, and a = 1, b, c are integers, then the equation will have integral roots.

How to Solve Quadratic Equations?

By solving the quadratic equations, you will get two roots that satisfy the equation. The easy steps to solve the quadratic equation roots are given here. The two ways to find the quadratic equations roots are the algebraic method and the graphical method. Here, we will learn about those methods.

Algebraic Method:

• Convert the given equation in the form of ax² + bx + c = 0, get a, b, c values.
• Substitute the values in the quadratic formula.
• Perform all the required calculations and find the roots.
• Another simple way is by factorizing the quadratic equation.
• The obtained factors are the roots.

Graphical Method:

• Plot the graph for the random values of x.
• The points where the curve meets the x-axis are the roots.

Examples of Quadratic Equations

(i) 5x² + 3x + 2 = 0 is a quadratic equation.

(ii) x – (1/x) = 6 is a quadratic equation

On solving this, we get x * x – 1 = 6 * x

x² – 1 = 6x

x² – 6x – 1 = 0

(iii) x² + √2x + 7 = 0 is not a quadratic equation.

(iv) x² + 5 = 0 is a quadratic equation.

(v) x² = 0,√ x² + 2x + 1 are quadratic equations.

Solved Examples on Finding Roots of the Equation

Example 1.

Solve the roots of equation 3x² – 22x – 16 by factorization?

Solution:

Given the quadratic equation is 3x² – 22x – 16

= 3x² – 24x + 2x – 16

= 3x(x – 8) + 2(x – 8)

= (x – 8) (3x + 2)

x – 8 = 0 and 3x + 2 = 0

x = 8 and 3x = -2, x = -2/3

The roots of the equation are α = 8, β = -2/3.

Therefore, solution set (8, -2/3)

Example 2.

Solve 9x² + 25x + 10 = 0 by using the quadratic formula.

Solution:

Given that,

9x² + 25x + 10 = 0

Roots of equation formula is (α, β) = [-b ± √(b² – 4ac)] / 2a

In the equation, a = 9, b = 25, and c = 10

Substitute these values in the equation.

α = [-b + √(b² – 4ac)] / 2a

= [-9 + √(9² – 4 (9) (10)] / 2 (9)

= [-9 + √(81 – 360)] / 18

= [-9 + √(-279)] / 18

= [-9 + √(279i²)] / 18

= [-9 + 16.70i] / 18

β = [-b – √(b² – 4ac)] / 2a

= [-9 – √(9² – 4 (9) (10)] / 2 (9)

= [-9 – √(81 – 360)] / 18

= [-9 – √(-279)] / 18

= [-9 – √(279i²)] / 18

= [-9 – 16.70i] / 18

Roots are α = [-9 + 16.70i] / 18, β = [-9 – 16.70i] / 18

Therefore, solution set ([-9 + 16.70i] / 18, [-9 – 16.70i] / 18)

Example 3.

Find the roots of the equation 1/(x + 4) – 1/(x – 7) = 11/30.

Solution:

Given that,

1/(x + 4) – 1/(x – 7) = 11/30

[(x – 7) – (x + 4)] / (x + 4) (x – 7) = 11/30

[x – 7 -x – 4] / (x² + 4x – 7x – 28) = 11/30

(-11) / (x² – 3x – 28) = 11/30

-1/ (x² – 3x – 28) = 1/30

-30 = x² – 3x – 28

x² – 3x – 28 + 30 = 0

x² – 3x + 2 = 0

x² – 2x – x + 2 = 0

x(x – 2) -1(x – 2) = 0

(x – 1) (x – 2) = 0

(x – 1) = 0 and (x – 2) = 0

x = 1, x = 2

The roots of quadratic equation is α = 1, β = 2.

Therefore, the solution set is (1, 2)

Example 4.

Find the values of a for which the quadratic expression (x – a) (x – 10) + 1 = 0 has integral roots.

Solution:

Given that,

(x – a) (x – 10) + 1 = 0

x² – ax – 10x + 10a + 1 = 0

x² – (a + 10)x + 10a + 1 = 0

Discriminant = b² – 4ac = (a + 10)² – 4 . 1. (10a + 1)

= a² + 20a + 100 – 40a – 4

= a² – 20a + 100 – 4

= (a – 10)² – 4

The quadratic equation will have integral roots, if the value of discriminant > 0, D is a perfect square, a = 1 and b and c are integers.

(a – 10)² – Discriminant = 4

Since discriminant is a perfect square. Hence, the difference of two perfect square in R.H.S will be 4 only when D = 0 and (a – 10)² = 4.

(a – 10) = ± 2

a = 2 + 10 or a = -2 + 10

= 12 and 8

Therefore, values of a are 8, 12.

Frequently Asked Questions on Quadratic Equations

1. What are the uses of the quadratic equations?

The quadratic equations can be used in calculating the area of rooms, speed of an object, determining product profit in the business. It can also be used in athletics and sports.

2. Write the differences between quadratic expression and quadratic equation?

The quadratic equation is a polynomial of degree 2 or any equation where quadratic expression is in the form of ax² + bx + c for equation or expression a ≠ 0. The quadratic formula is used to solve the roots of the equation.

3. What is the simplest form of a quadratic equation?

The standard form of a quadratic equation is ax² + bx + c – 0, where a ≠ 0. The coefficients a, b, c are real numbers.

4. Describe the nature of the roots?

The discriminant b² – 4ac > 0, roots are real, unequal and distinct. In case the discriminant = 0, roots are real, equal, and coincident. If the discriminant < 0, then roots are imaginary and unequal.

Practice Test on Linear Inequation has different types of questions. Students can test their skills and knowledge on linear inequations problems by solving all the provided questions on this page. The questions are mainly related to inequalities and finding the solution to the given inequation and draw a graph for the obtained solution set. You can easily draw a graph on a numbered line.

1. Write the equality obtained?

(i) On subtracting 1 from each side 3 > 7

(ii) On adding 3 to each side 12 < 5

(iii) On multiplying (-2) to each side 11 < 4

(iv) On multiplying 4 to each side 15 > 2

Solution:

(i) 3 – 1 > 7 – 1

2 > 6

(ii) 12 + 3 < 5 + 3

15 < 8

(iii) 11 x (-2)  4 x (-2)

-22 < -8

22 > 8

(iv) 15 x 4 > 2 x 4

60 > 8

2. Write the word statement for the following?

(i) x ≥ 15

(ii) x < 2

(iii) x ≤ -5

(iv) x > 16

(v) x ≠ 6

Solution:

(i) The variable x is greater than equal to 15. The possible values of x are 15 and more than 15.

(ii) The variable x is less than 2. The possible values of x are less than 2.

(iii) The variable x is less than and equal to -5. The possible values of x are less than -5.

(iv) The variable x is greater than 16. The possible values of x are more than 16.

(v) The variable x is not equal to 6. The possible values of x are all real numbers other than 6.

3. Find the solution set for each of the following inequations. x ∈ N

(i) x + 5 < 12

(ii) x – 6 > 5

(iii) 5x + 10 ≥ 17

(iv) 2x + 3 ≤ 6

Solution:

(i) x + 5 < 12

Subtract 5 from both sides.

x + 5 – 5 < 12 – 5

x < 7

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set S = {1, 2, 3, 4, 5, 6}

(ii) x – 6 > 5

Add 6 to both sides.

x – 6 + 6 > 5 + 6

x > 11

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set S = {12, 13, 14, 15, . . . }

(iii) 5x + 10 ≥ 17

Subtract 10 from both sides.

5x + 10 – 10 ≥ 17 – 10

5x ≥ 7

Divide 5 by each side.

5x/5 ≥ 7/5

x ≥ 1.4

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set S = {2, 3, 4, 5 . . .}

(iv) 2x + 3 ≤ 6

Subtract 3 from both sides of the inequation

2x + 3 – 3 ≤ 6 – 3

2x ≤ 3

Both sides of the inequation divide by 2.

2x/2 ≤ 3/2

x ≤ 1.5

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set S = {1, 1.5}

4. Find the solution set for each of the following inequations and represent it on the number line.

(i) 3 < x < 10, x ∈ N

(ii) 3x + 2 ≥ 6, x ∈ N

(iii) 3x/2 < 5, x ∈ N

(iv) -4 < 2x/3 + 1 < – 2, x ∈ N

Solution:

(i) 3 < x < 10, x ∈ N

The two cases are 3 < x and x < 10

It can also represent as x > 3 and x < 10

Replacement set = {1, 2, 3, 4, 5 . .}

The solution set for x > 3 is 4, 5, 6, 7 . . . i.e P = {4, 5, 6, 7 . . .}

And the solution set for x < 10 is 1, 2, 3, 4, 5, 6, 7, 8, 9 i.e Q = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Therefore, solution set of the given inequation = P ∩ Q = {4, 5, 6, 7, 8, 9}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(ii) 3x + 2 ≥ 6, x ∈ N

Subtract 2 from both sides

3x + 2 – 2 ≥ 6 – 2

3x ≥ 4

Divide each side by 3

3x/3 ≥ 4/3

x ≥ 1.33

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set S = {2, 3, 4, 5, . . }

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(iii) 3x/2 < 5, x ∈ N

Multiply both sides by 2.

3x/2 x 2 < 5 x 2

3x < 10

divide both sides by 3

3x/3 < 10/3

x < 3.33

Replacement set = {1, 2, 3, 4, 5 . .}

Solution Set S = {1, 2}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(iv) -4 < 2x/3 + 1 < – 2, x ∈ N

The two cases are -4 < 2x/3 + 1 and 2x/3 + 1 < – 2

Case I: -4 < 2x/3 + 1

Subtract 1 from both sides

-4 – 1 < 2x/3 + 1 – 1

-5 < 2x/3

Multiply each side by 3

-5 x 3 < 2x/3 x 3

-15 < 2x

Divide each side by 2

-15/2 < 2x/2

-7.5 < x

x > 7.5

Replacement Set = {1, 2, 3, 4, 5 . .}

Solution Set P = {8, 9, 10, 11 . . . }

Case II: 2x/3 + 1 < – 2

Subtract 1 from both sides

2x/3 + 1 – 1 < – 2 – 1

2x/3 < -3

Multiply 3 to both sides

2x/3 x 3 < -3 x 3

2x < -9

Divide both sides by 2

2x/2 < -9/2

x < -4.5

4.5 > x

Replacement set = {1, 2, 3, 4, 5 . .}

Solution set Q = {1, 2, 3}

Therefore, required solution set S = P ∩ Q

S = Null

5. Find the solution set for each of the following and represent the solution set graphically?

(i) x – 6 < 4, x ∈ W

(ii) 6x + 2 ≤ 20, x ∈ W

(iii) 7x + 3 < 5x + 9, x ∈ W

(iv) 3x – 7 > 5x – 1, x ∈ I

Solution:

(i) x – 6 < 4, x ∈ W

Add 6 to both sides

x – 6 + 6 < 4 + 6

x < 10

Replacement set = {0, 1, 2, 3, 4, 5, 6, …}

Therefore, solution set S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(ii) 6x + 2 ≤ 20, x ∈ W

Subtract 2 from both sides

6x + 2 – 2 ≤ 20 – 2

6x ≤ 18

Divide each side by 6

6x/6 ≤ 18/6

x ≤ 3

Replacement set = {0, 1, 2, 3, 4, 5, 6, …}

Therefore, solution set S = {0, 1, 2, 3}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(iii) 7x + 3 < 5x + 9, x ∈ W

Move variables to one side and constants to other side of inequation

7x – 5x < 9 – 3

2x < 6

Divide each side by 2

2x/2 < 6/2

x < 3

Replacement set = {0, 1, 2, 3, 4, 5, 6, …}

Therefore, solution set S = {0, 1, 2}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

(iv) 3x – 7 > 5x – 1, x ∈ I

Move variables to one side and constants to another side of inequation

-7 + 1 > 5x – 3x

-6 > 2x

divide 2 by each side

-6/2 > 2x/2

-3 > x

Replacement set ={ . . . -4, -3, -2, -1, 0, 1, 2, 3, . . .}

Solution set = { -2, -1, 0, 1, 2, . . . }

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

Are you still looking for a simple procedure on how to represent the solution set of an inequation on a graph? If yes, then stay on this page. Here we are giving a detailed step by step explanation on finding Solution Set for Linear Inequations with the solved examples. So, refer to the following sections and solve the questions easily.

Graphical Representation of the Solution Set of an Inequation

We generally use a number line to represent the solution set of an inequation on a graph. Following are the steps to represent the solution set of a linear inequation on a graph.

• At first, solve the given linear inequation and find the solution set for it.
• Mark the solution set on a number line by putting dot.
• If the solution set is infinite, then put three more dots to indicate infiniteness.

Questions on Finding the Solution Set of an Inequation and their Representation

Example 1.

Solve the inequation 4x – 6 < 10, x ∈ N and represent the solution set graphically?

Solution:

Given linear inequation is 4x – 6 < 10

Add 6 to the both sides of inequation

= 4x – 6 + 6 < 10 + 6

= 4x < 16

Divide both sides of the inequation by 4

= 4x/4 < 16/4

= x < 4

So, the replacement set = {1, 2, 3, 4, . . }

Therefore, the solution set S = {1, 2, 3} or S = {x : x ∈ N, x < 4}

Let us mark the solution set graphically.

The solution set is marked on the number line by dots.

Example 2.

Solve the inequation 8x + 4 > 20, x ∈ W and represent the solution set graphically?

Solution:

Given linear inequation is 8x + 4 > 20

Subtract 4 from both sides

= 8x + 4 – 4 > 20 – 4

= 8x > 16

Divide both sides by 8

= 8x/8 > 16/8

= x > 2

Replacement set = {0, 1, 2, 3, 4, . . . }

Therefore, solution set = {3, 4, 5, . . } or S = {x : x ∈ W, x > 2}

Let us mark the solution set graphically.

The solution set is marked on the number line by dots. We put three more dots indicate the infiniteness of the solution set.

Example 3.

Solve -2 ≥ x ≥ 5, x ∈ I, and represent the solution set graphically?

Solution:

Given linear inequation is -2 ≥ x ≥ 5

This has two inequations,

-2 ≥ x and x ≥ 5

Replacement set = {. . . -3, -2, -1, 0, 1, 2, 3 . . .}

Solution set for the inequation -2 ≥ x is -2, -1, 0, 1, 2, 3, . . i.e S = {-2, -1, 0, 1, 2, 3 . . } = P

And the solution set for the inequation x ≥ 5 is 5, 6, 7, 8 . . i.e Q = {5, 6, 7, 8 . . .}

Therefore, solution set for the given inequation = P ∩ Q

= {5, 6, 7, 8, 9 . . . }

or S = {x : x ∈ I, -2 ≥ x ≥ 5}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

Example 4.

Solve 0 < 3x – 10 ≤ 12, x ∈ R and represent the solution set graphically.

Solution:

Given linear inequation is 0 < 3x – 10 ≤ 12

It has two cases.

Case I:

0 < 3x – 10

Add 10 to both sides

0 + 10 < 3x – 10 + 10

10 < 3x

Divide both sides by 3

10/3 < 3x/3

10/3 < x

Case II:

3x – 10 ≤ 12

Add 10 to both sides

3x – 10 + 10 ≤ 12 + 10

3x ≤ 22

Divide both sides by 3

3x/x ≤ 22/3

x ≤ 22/3

S ∩ S’ = {3.33 < x ≤ 7.33} x ∈ R

= {x : x ∈ R 3.33 < x ≤ 7.33}

Properties of Inequation or Inequalities page gives detailed information about six properties. Each Property is explained step by step by considering a few examples. Learn and understand the properties easily with the help of solved examples provided below. You can seek Complete Information about Linear Inequations and their definitions all on our page.

Properties of Inequation or Inequalities

The six different properties of linear inequalities or linear inequation are mentioned here. These properties describe how arithmetic operations show the effect on the linear inequations.

Property I

The inequation remains unchanged if the same quantity is added to both sides of it.

Example:

x – 3 < 4

Add 3 to the both sides

= x – 3 + 3 < 4 + 3

= x < 7

Property II:

The inequation remains unchanged if the same quantity is subtracted from the both sides of the inequation.

Example:

x + 3 < 4

Subtract 3 from both sides

x + 3 – 3 < 4 – 3

x < 1

Property III:

The inequation remains unchanged if the same positive number is multiplied to both sides of it.

Example:

x/3 > 5

multiply 3 to both sides

= x/3 x 3 > 5 x 3

= x > 15

Property IV:

The inequation changes if the same negative number is multiplied to both sides of it. Actually, it reverses.

Example:

x/5 < 6

multiply -5 to the both sides

= x/5 x (-5) < 6 x (-5)

= -x < -30

= x > 30

Property V:

The inequation remains unchanged if the same positive number is divided by both sides of the inequation.

Example:

5x > 20

dividing both sides by 5

= 5x/5 > 20/5

= x > 4

Property VI:

The inequation changes when the same negative number is divided by both sides. It reverses.

Example:

-3x < 12

Dividing both sides by -3

= -3x/-3 > 12/-3

= x > -4

Solved Examples on Properties of Linear Inequalities

Example 1.

Write the inequality obtained for each of the following statements.

(i) On subtracting 9 from both sides of 21 > 10.

(ii) On multiplying each side of 8 < 12 by -2.

Solution:

(i) We know that subtracting the same number from both sides of inequality does not change the inequality.

21 – 9 > 10 – 9

= 12 > 1

(ii) We know that multiplying each side of an inequality by the same negative number reverses that inequality.

= 8 x -2 < 12 x -2

= -16 < -24

= -16 > -24

Example 2.

Find the inequality obtained for the following statements.

(i) On adding 2 to both sides of 8 < 56

(ii) On dividing each side of 8 < 56 by -16.

Solution:

(i) We know that adding the same number to both sides of inequality does not change the inequality.

8 + 2 < 56 + 2

= 10 < 58

(ii) We know that dividing each side of an inequality by the same negative number reverses that.

8/-16 < 56/-16

= -1/2 > -7/2

Example 3.

Write the inequality obtained for the following statement.

On multiplying each side of 18 > 8 by 5.

Solution:

We know that multiplying each side of an inequality by the same positive number does not change the inequality.

18 x 5 > 8 x 5

= 90 > 40

On this page, we will learn about what is linear inequality and linear inequations and the steps to solve the linear inequalities problems. You will also get the properties of inequation or inequalities. Check out the representation of the solution set on the real line in the following sections. We have provided Solved Problems along with a detailed explanation so that you can better understand the concept.

What are Linear Inequation and Linear Inequalities?

Linear Inequation is a statement indicating the value of one quantity or algebraic expressions that are not exactly equal to one another.

Inequalities are nothing but the symbols enclosed between two or more algebraic expressions or quantities. The open sentence which involves <, ≤, >, ≥, and ≠ symbols are called the inequalities.

Some of the examples of Linear Inequation are listed below.

• x < 6
• y ≥ 25
• x + 3 > 40
• p ≠ 10

Linear Inequation

An inequation that contains one variable and that variable highest power is one then is known as the linear inequation in that variable. To make a linear equation into inequation, you have to replace the equal to symbol with the inequality sign. The statements of any of the forms ax + b < 0, ax + b > 0, ax + b≥ 0, ax + b ≤ 0 are the linear inequations invariable x, where a, b are real numbers and a is not equal to zero.

some of the examples of the linear inequation with variable y are included below:

• 3y + 6 ≥ 0
• 9 – y < 0
• 2y > 0
• 25 + 5y ≤ 0

Questions on Replacement Set or Domain of a Variable

Example 1.

Find the replacement set for the inequation x ≤ 9. The replacement set is a set of whole numbers?

Solution:

We know that whole numbers W = {0, 1, 2, 3 . . . }.

Replace x with some values of W. Some values of x from W satisfy the inequation and some don’t. Here, the values 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 satisfy the given inequation x ≤ 9 while the other values don’t.

Thus, the set of all those values of variables that satisfy the given inequation is called the solution set of the given inequation.

Therefore, the solution set for the inequation x ≤ 9 is S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} or S = {x : x ∈ W, x ≤ 9}

Example 2.

Find the replacement set for the inequation x > 2. Let the replacement set be the set of natural numbers?

Solution:

We know that natural numbers N = {1, 2, 3, 4, 5, 6}

Replace x with some values of N. Some values of x from N satisfy the inequation and some don’t. Here, the values 3, 4, 5, . . . satisfy the given inequation x > 2 while the other values don’t.

Thus, the set of all those values of variables that satisfy the given inequation is called the solution set of the given inequation.

Therefore, the solution set for the inequation x > 2 is S = {3 4, 5, 6, . . . }

Example 3.

Find the replacement set and the solution set for the inequation x ≥ -2 when the replacement set is an integer?

Solution:

Replacement set I = {. . . -3, -2, -1, 0, 1, 2, 3, . . . }

Solution set S = {-2, -1, 0, 1, 2, . . . } or S = { x : x ∈ I, x ≥ – 2}

Example 4.

Find the solution set for the following linear inequations.

(i) x < 5 where replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}

(ii) x ≥ 7 where replacement set is { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(iii) x ≠ 3 where replacement set is { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Solution:

(i) Solution Set S = {-4, -3, -2, -1, 0, 1, 2, 3, 4} S = {x : x ∈ I, -4 < x ≤ 4}

(ii) Solution set S = {7, 8, 9, 10} or S = {x : x ∈ N, 7 < x < 10}

(iii) Solution set S = {0, 1, 2, 4, 5, 6, 7, 8, 9, 10} or S = {x : x ∈ N, x ≠ 3}

Frequently Asked Questions on Linear Inequality

1. What is the difference between a linear equation and a linear inequality?

The linear equation is an equation that has one or two variables and those exponents are one. Linear inequation also has one variable whose exponent is one. Between two algebraic expressions, the = symbol is enclosed in a linear equation, linear inequality signs are enclosed in a linear inequation. The graph of inequalities is a dashed line but the equation is a solid line in any situation.

2. What is linear inequality?

Linear inequality contains any symbols of inequality. It represents the data that is not equal in graph form. It involves a linear function.

Linear inequations demonstrate the value of one quantity or algebraic expression which is not equal to another. These inequations are used to compare any two quantities. Get some solved examples on linear inequations and steps to draw graphs, the system of linear inequations in the following sections of this page.

• Linear Inequality and Linear Inequations
• Properties of Inequation or Inequalities
• Representation of the Solution Set of an Inequation
• Practice Test on Linear Inequations

What are Linear Inequations?

Linear inequalities are the expressions where any two values are compared by the inequality symbols. These are the equations that contain all mathematical symbols except equal to (=). The values can be numerical or algebraic or a combination of both.

The inequality symbols are < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to), ≠ (not equal to). The symbols < and > shows the strict inequalities and the symbols ≤ and ≥ represents the slack inequalities.

Example:

x < 3, x ≥ 5, y ≤ 8, p > 10, m ≠ 1.

Linear Inequalities Graphing

We can plot the graph for linear inequalities like an ordinary linear function. But, for a linear function, the graph represents a line and for inequalities, the graph shows the area of the coordinate plane that satisfies the inequality condition. The linear inequalities graph divides a coordinate plane into two parts. One part of the coordinate plane is called the borderline where it represents the solutions for inequality. The borderline represents the conditions <, >, ≤ and ≥.

Students who are willing to plot a graph for the linear inequations can refer to the following steps.

• Arrange the given linear inequation in such a way that, it should have one variable ‘y’ on the left-hand side of the symbol and the remaining equation on the right-hand side.
• Plot the graph for linear inequation by putting the random values of x.
• Draw a thicker and solid line for y≤ or y≥ and a dashed liner for y< or y> conditions.
• Now, draw shades as per the linear inequalities conditions.

System of Linear Inequalities

A system of linear inequalities in two variables includes at least two inequalities in the variables. By solving the linear inequality you will get an ordered pair. So basically, the solution to all linear inequalities and the graph of the linear inequality is the graph displaying all solutions of the system.

Questions on Linear Inequations

Example 1.

Solve the inequality x + 5 < 10?

Solution:

Given that,

x + 5 < 10

Move variable x to the one side of inequation.

= x < 10 – 5

= x < 5

Replacement set = {0, 1, 2, 3, 4, 5, 6, 7 . . . }

Solution set for the inequation x + 5 < 10 is 1, 2, 3, 4

Therefore, solution set s = {. . . 1, 2, 3, 4}

Let us mark the solution set graphically.

The solution set is marked on the number line by dots. We put three more dots indicate the infiniteness of the solution set.

Example 2.

Solve the inequation 3 < y ≤ 10?

Solution:

Given that,

3 < y ≤ 10

This has two inequations,

3 < y and y ≤ 10

Replacement set = {. . . -3, -2, -1, 0, 1, 2, 3, 4, , 6, 7, . . . }

Solution set for the inequation 3 < y is 4, 5, 6, 7, . . .  i.e Q = {4, 5, 6, . . . }

Solution set for the inequation y ≤ 10 is . . . . 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 i.e P = {. . . . 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Therefore, solution set of the given inequation = P ∩ Q = {4, 5, 6, 7, 8, 9}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

Example 3.

Solve the inequality 4 ( x + 2 ) − 1 > 5 − 7 ( 4 − x )?

Solution:

Given that,

4 ( x + 2 ) − 1 > 5 − 7 ( 4 − x )

4 x + 8 − 1 > 5 − 28 + 7 x

4 x + 7 > − 23 + 7 x

− 3 x > − 30

x < 10

Replacement Set = {1, 2, 3, 4, 5 . . . }

Solution set is 1, 2, 3, 4, 5, 6, 7, 8, 9 i.e S = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Let us represent the solution set graphically.

The solution set is marked on the number line by dots.

Frequently Asked Questions on Linear Inequations

1. What is meant by linear inequations?

Linear inequations involve linear expressions in two variables by using the relational symbols like <, >, ≤, ≥, and ≠.

2. How do you solve linear inequation?

Solving linear inequations having a single variable is very easy. All you have to do is simplify both the sides of the condition and get the variable term on one side and all other terms on the other side. And them either multiply/ divide the coefficient of the variable to obtain the solution.

3. What are the examples of linear inequations?

Some of the examples of linear inequality are x < 8, y > 15, z ≠ 7 + 9, p + 1 ≤ 2, 3 ≥ (z + 15).