# SOLVING SYSTEMS OF EQUATIONS BY SUBSTITUTION

## About "Solving systems of equations by substitution"

Solving systems of equations by substitution :

Here we are going to see some example problems to understand how to solve system of equations by substitution.

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.   Solving systems of equations by substitution

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.

After having gone through the stuff given above, we hope that the students would have understood "Solving systems of equations by substitution".

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