# Simultaneous Linear Equations Definition | How to Solve Simultaneous Equations?

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.