**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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