## SUBSTITUTION METHOD EXAMPLES

In this page substitution method examples, we are going to see some example problems of solving linear equation using substitution method.

what is substitution method ?

Solving system of equation by substitution method, involves solving any one of the given equation for either 'x' or 'y' and plugging that in the other equation and solve that equation for another variable.

## Steps in substitution method

Step 1 :

Solve any one of the equations either x = or y =

Step 2 :

Substitute the value that we got from step 1 in the other equation.

Step 3 :

Now we have got the value of any one of the variables x or y.

Step 4 :

Apply this value  in step 1 in order to get the value of other variable.

Let us consider the following example problem to understand the substitution method.

## Substitution Method Example

Question 1 :

Solve the following equations by substitution method

5 x - 3 y - 8 = 0  and 2x - 3 y - 5  = 0

Solution :

5 x - 3 y - 8 = 0  ------(1)

2 x - 3 y - 5  = 0  ------(2)

x = -2 and y = 3

Solution is (-2,3)

Question 2 :

Solve the following equations using substitution method

5x - 3y - 8 = 0  and  2x - 3y - 5  = 0

Solution :

5x - 3y - 8 = 0  ------(1)

2x - 3y - 5  = 0  ------(2)

-3y  =  -5x + 8

3y  =  5x - 8

Apply the value of 3y in the second equation, we get

2x - (5x - 8)  - 5  =  0

2x - 5x + 8 - 5  =  0

-3x + 3  =  0

-3x  =  -3  ==>  x  =  -3/(-3)  ==> x  =  1

Substitute x  =  1 in the equation 3y  =  5x - 8, we get

3y  =  5(1) - 8

3y  =  5 - 8

3y  =  -3  ==>  y  =  (-3)/3  ==>  y  = -1

Hence x = 1 and y = -1 is the solution.

Question 3 :

Solve the following equations by substitution method

y  =  6x - 11 and -2x - 3y  =  -7

Solution :

y  =  6x - 11 ------(1)

-2x - 3y  =  -7 ------(2)

Substitute the value of y in the second equation, we get

-2x - 3(6x - 11)  =  -7

-2x - 18x + 33  =  -7

-20x + 33  =  -7

Subtract 33 on both sides

-20x + 33 - 33  =  -7 - 33

-20x  =  -40

Divide by -20 on both sides

-20x/(-20)  =  -40/(-20)

x  =  2

Substitute x  =  2 in the first equation, we get

y  =  6(2) - 11

y  =  12 - 11 ==>  1

Hence x = 2 and y = 1 is the solution

Question 4 :

Solve the following equations by substitution method

2x − 3y = −1 and y = x − 1

Solution :

2x − 3y = −1 -----(1)

y = x − 1  -----(2)

Substitute y = x - 1 in the first equation

2x - 3(x - 1)  =  -1

2x - 3x + 3  =  -1

-x + 3  = -1

Subtract by 3 on bot sides,

-x + 3 - 3  =  -1 - 3

-x  =  -4  ==> x  =  4

Apply x  =  4 in the equation y  =  x - 1

y  =  4 - 1  ==>  y  =  3

Hence x  =  4 and y  =  3 is the solution.

Let us see the next example of the topic "Substitution method examples".

Question 5 :

Solve the following equations by substitution method

y = −3x + 5 and 5x − 4y = −3

Solution :

y = −3x + 5 -------(1)

5x − 4y = −3 -------(2)

Substitute y  =  -3x + 5 in the second equation

5x - 4 (-3x + 5)  =  -3

5x + 12x - 20  =  -3

17x  =  -3 + 20

17x  =  17

Divide by 17 on both sides

17x/17  =  17/17  ==> x  =  1

Applying x  =  1 in the first equation, we get

y  =  -3(1) + 5

y  =  -3 + 5  ==>  y  =  2

Hence x = 1 and y = 2 is the solution.

Let us see the next example of the topic "Substitution method examples".

Question 6 :

Solve the following equations by substitution method

−3x − 3y = 3 and  y = −5x − 17

Solution :

−3x − 3y = 3 -------(1)

y = −5x − 17  -------(2)

Substitute y  =  -5x - 17 in the first equation

−3x − 3(-5x - 17) = 3

-3x + 15x + 51  =  3

12x + 51  =  3

Subtract by 51 on both sides

12x + 51 - 51  =  3 - 51

12x =  -48

Divide by 12 on both sides

12x/12  =  -48/12  ==>  x  =  -4

Applying x  =  -4 in the second equation

y  =  -5(-4)  - 17

y  =  20 - 17

y  =  3

Hence x = -4 and y = 3 is the solution.

Let us see the next example of the topic "Substitution method examples".

Question 7 :

Solve the following equations by substitution method

y = 5x − 7 and −3x − 2y = −12

Solution :

y = 5x − 7  -------(1)

−3x − 2y = −12  -------(2)

Substitute y  =  5x − 7 in the second equation

−3x − 2(5x - 7) = −12

−3x − 10x + 14 = −12

-13x + 14 =  -12

Subtract 14 on both sides

-13x + 14 - 14  =  -12 - 14

-13x  =  -26

Divide by -13 on both sides

-13x/(-13)  =  -26/(-13)

x  =  2

Apply x = 2 in the equation y  =  5x - 7

y = 5(2)  - 7

y  =  10 - 7 ==>  3

Hence the solution is x = 2 and y = 3.

Let us see the next example of the topic "Substitution method examples".

Question 8 :

Solve the following equations by substitution method

−4x + y = 6 and −5x − y = 21

Solution :

−4x + y = 6  -------(1)

−5x − y = 21  -------(2)

From the first equation,

y  =  6 + 4x

Substitute y  =  6 + 4x in the second equation

−5x − (6 + 4x) = 21

-5x - 6 - 4x  =  21

-9x - 6  =  21

-9x - 6 + 6  =  21 + 6

-9x  =  27

Divide by -9 on both sides

-9x/(-9)  =  27/(-9)

x  =  -3

Apply x = -3 in the equation y  =  6 + 4x

y = 6 + 4(-3)

y  =  6 - 12  =  -6

Hence the solution is x = -3 and y = -6.

Let us see the next example of the topic "Substitution method examples".

Question 9 :

Solve the following equations by substitution method

2x + y = 20 and 6x − 5y = 12

Solution :

2x + y = 20  -------(1)

6x − 5y = 12  -------(2)

From the first equation, we get

y  =  20 - 2x

Substitute y  =  20 - 2x in the second equation

6x − 5 (20 - 2x) = 12

6x - 100 + 10x  =  12

16x - 100  = 12

16x - 100 + 100  =  12 + 100

16x  =  112

Divide by 16 on both sides

16x/16  =  112/16

x  =  7

Apply x  =  7 in the equation y  =  20 - 2x

y  =  20 - 2(7)

y  =  20 - 14  ==>  6

Hence the solution is x = 7 and y = 6.

Let us see the next example of the topic "Substitution method examples".

Question 10 :

Solve the following equations by substitution method

y = −2 and 4x − 3y = 18

Solution :

y = −2 -------(1)

4x − 3y = 18  -------(2)

Apply y  =  -2 in the second equation

4x − 3(-2) = 18

4x + 6  =  18

Subtract by 6 on both sides

4x + 6 - 6  =  18 - 6

4x  =  12

Divide by 1 on both sides

4x/4  =  12/4

x  =  3

Hence the solution is x = 3 and y = -2.

After having gone through the stuff given above, we hope that the students would have understood "Substitution method examples".

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