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# Solving System of Equations – Methods & Examples

## How to Solve System of Equations?

By now, you have got the idea of how to solve linear equations containing a single variable. What if you were when presented with **multiple linear equations containing more than one variable**? A set of linear equations with two or more variables is known as a **system of equations.**

There are several methods of solving systems of linear equations.

This article will learn **how to solve linear equations using the commonly used methods**, namely substitution and elimination.

### Substitution method

Substitution is a method of solving linear equations in which a variable in one equation is isolated and then used in another equation to solve for the remaining variable.

*The general steps for substitution are:*

- Make the subject of the formula for a variable in one of the given equations.
- Substitute the value of this variable in the second equation.’
- Solve the equation to get the value of one of the variables.
- Substitute the obtained value in any of the equations to also get the value of the other variable.

*Let’s solve a couple of examples using the substitution method.*

*Example 1*

Solve the systems of equations below.

b = a + 2

a + b = 4.

__Solution__

Substitute the value of b into the second equation.

a + (a + 2) = 4

Now solve for a

a +a + 2 = 4

2a + 2 = 4

2a = 4 – 2

a = 2/2 = 1

Substitute the obtained value of a in the first equation.

b = a + 2

b = 1 + 2

b = 3

Hence, the solution for the two-equation is: a =1 and b=3.

*Example 2*

Solve the following equations using substitution.

7x – 3y = 31 ——— (i)

9x – 5y = 41 ——— (ii)

__Solution__

From equation (i),

7x – 3y = 31

Make y the subject of the formula in equation:

7x – 3y = 31

Subtract 7x from both sides of the equation 7x – 3y = 31 to get;

– 3y = 31 – 7x

3y = 7x – 31

3y/3 = (7x – 31)/3

Therefore, y = (7x – 31)/3

Now substitute the equation y = (7x – 31)/3 into the second equation:9x – 5y = 41

9x – 5 × (7x – 31)/3 = 41

Solving the equation gives;

27x – 35x + 155 = 41 × 3

–8x + 155 – 155 = 123 – 155

–8x = –32

8x/8 = 32/8

x = 4

By substituting the value of x in the equation y = (7x – 31)/3, we get;

y = (7 × 4 – 31)/3

y = (28 – 31)/3

y = –3/3

y = –1

Therefore, the solution to these systems of equation is x = 4 and y = –1

*Example 3*

Solve the following sets of equations:

2x + 3y = 9 and x – y = 3

__Solution__

Make x the subject of the formula in the second equation.

x = 3 + y.

Now, substitute this value of x in the first equation: 2x + 3y = 9.

⇒ 2(3 + y) + 3y = 9

⇒ 6 + 2y + 3y = 9

y = ⅗ = 0.6

Substitute the obtained value of y in the second equation – y =3.

⇒ x = 3 + 0.6

x = 3.6

Therefore, the solution is x = 3.6 and y = 0.6

### Elimination method

*The following steps are followed when solving systems of equations using the elimination method:*

- Equate the coefficients of the given equations by multiplying with a constant.
- Subtract the new equations common coefficients have same signs and add if the common coefficients have opposite signs,
- Solve the equation resulting from either addition or subtraction
- Substitute the obtained value in any of the equations to get the value of the other variable.

*Example 4*

4a + 5b = 12,

3a – 5b = 9

__Solution__

Since the coefficients b are the same in the two equations, we vertically add the terms.

4a+3a) +(5b – 5b) = 12 + 9

7a = 21

a = 21/ 7

a = 3

substitute the obtained value of a=3 in the equation the first equation

4(3) + 5b = 12,

12 + 5b = 12

5b = 12-12

5b =0

b = 0/5 = 0

Therefore, the solution is a =3 and b = 0.

*Example 5*

Solve using elimination method.

2x + 3y = 9 ———–(i)

x – y = 3 ———–(ii)

__Solution__

Multiply the two equations by 2 and perform subtraction.

2x + 3y = 9

(-)

2x – 2y = 6

-5y = -3

y = ⅗ = 0.6

Now substitute the obtained value of y in the second equation: x – y = 3

x – 0.6 = 3

x = 3.6

Therefore, the solution is: x = 3.6 and y= 0.6