In this chapter we will learn to solve linear equation with two variables using substitution method.
To understand the chapter, you should have fair knowledge of linear and simultaneous equations.
Substitution Method
This method works when we have two linear equations with two variables.
Using the method we can find common point that satisfies both the given equation.
Let we have two linear equations with x & y variables.
The equation can be solved by following below steps;
(a) Find value of x with respect to y or vice – versa using one of the given equation.
(b) Now put this value of x in the second equation.
On solving the equation, you will get exact value of y.
(c) Now put this value of y in any of the given equation and you will get exact value of x.
By following these three steps, you can solve given set of linear equations.
Let us see some solved examples for better clarity.
Example 01
Solve the below equations using substitution method.
5x + 2y = 16
x + 8y = 26
Solution
Take the second equation and find the value of x with respect to y.
x + 8y = 26
x = 26 – 8y
Now substitute this value of x in first equation.
5x + 2y = 16
5 (26 – 8y) + 2y = 16
130 – 40y + 2y = 16
-38y = 16- 130
-38y = -114
y = 114 / 38
y = 3
Here we got the exact value of y.
Now put this y value in any of the given equation & you will get value of x.
Taking first equation.
5x + 2y = 16
5x + 2(3) = 16
5x + 6 = 16
5x = 16 – 6
5x = 10
x = 10 / 5
x = 2
Hence, (2, 3) is the solution of given set of equations.
Example 02
Solve the below equations using substitution method.
2x + y = 1
4x – 2y = 6
Solution
Taking first equation and find value of y with respect of x.
2x + y = 1
y = 1 – 2x
Now substitute this y value in second equation.
4x – 2y = 6
4x – 2 (1 – 2x) = 6
4x – 2 + 4x = 6
8x = 6 + 2
8x = 8
x = 1
Here we have got the exact value of x.
Put the value of x in any of the given equation to find exact value of y.
Taking first equation
2x + y = 1
2(1) + y = 1
2 + y = 1
y = 1 – 2
y = -1
Hence, (1, -1) is the solution of given equations.
Example 03
Solve the below equations using substitution method.
3x – 2y = 4
-2x + 2y = -1
Solution
Take first equation and find value of x with respect of y.
3x – 2y = 4
3x = 4 + 2y
x = \mathtt{\frac{4+2y}{3}}
Now substitute this value of x in second equation.
-2x + 2y = -1
\mathtt{-2\left(\frac{4+2y}{3}\right) +2y=-1}\\\ \\ \mathtt{\frac{-8-4y}{3} +2y=-1}\\\ \\ \mathtt{\frac{-8-4y+6y}{3} =-1}\\\ \\ \mathtt{-8+2y=-3}\\\ \\ \mathtt{2y=\ -3+8}\\\ \\ \mathtt{y\ =\ \frac{5}{2}}
Here we have got exact value of y. Now put this value in any of the given equation to find exact value of x.
Taking first equation.
\mathtt{3x\ -\ 2y\ =\ 4}\\\ \\ \mathtt{3x\ -\ 2\left(\frac{5}{2}\right) =4}\\\ \\ \mathtt{3x\ -5\ =\ 4}\\\ \\ \mathtt{3x\ =\ 4\ +5}\\\ \\ \mathtt{3x\ =\ 9}\\\ \\ \mathtt{x\ =\ \frac{9}{3}}\\\ \\ \mathtt{x\ =\ 3}
Hence, ( 3, 5/2) is the solution.
Example 04
Solve the below equations using substitute method.
x – y = 2
y + 2x = 4
Solution
Take the first equation and find the value of x.
x – y = 2
x = y + 2
Now substitute this value of x in second equation.
y + 2x = 4
y + 2 (y + 2) = 4
y + 2y + 4 = 4
3y = 4 – 4
3y = 0
y = 0
We have got exact value of y. Put this value of y in any of the given equation to find value of x.
Taking first equation.
x – y = 2
x – 0 = 2
x = 2
Hence, (2, 0) is the solution of given expression.
Example 05
Solve the below equation using substitute method.
2x – 6y = -6
7x – 8y = 5
Solution
Take the first equation and find value of y with respect to x.
\mathtt{2x\ -\ 6y\ =\ -6}\\\ \\ \mathtt{2x\ =\ 6y\ -\ 6}\\\ \\ \mathtt{x\ =\ \frac{6y-6}{2}}\\\ \\ \mathtt{x\ =\ \frac{2( 3y-3)}{2}}\\\ \\ \mathtt{x\ =\ 3y-3}
Put this value of x in second equation.
7x – 8y = 5
7 (3y – 3) – 8y = 5
21y -21 – 8y = 5
13y= 5 + 21
13y = 26
y = 26 / 13
y = 2
Here we got exact value of y. Put this value of y in any of the given equation to get the exact values of x.
Taking first equation;
2x – 6y = -6
2x – 6(2) = -6
2x – 12 = -6
2x = -6 + 12
2x = 6
x = 3
Hence, (3, 2) is the solution of given linear equation.