**Solving Equations Using Elimination and Substitution :**

In this section, we will learn, how to solve linear equations by substitution method.

We use the following steps to solve a linear equations in substitution method and elimination method.

**Step 1 :**

Solve one of the equations for one of its variables.

**Step 2 :**

Substitute the expression from step 1 into the other equation and solve for the other variable.

**Step 3 :**

Substitute the value from step 2 into either original equation and solve to find the value of the variable in step 1.

**Step 1 :**

By taking any one equations from the given two, first multiply by some suitable non-zero constant to make the co-efficient of one variable (either x or y) numerically equal.

**Step 2 :**

If both coefficients which are numerically equal of same sign, then we may eliminate them by subtracting those equations.

If they have different signs, then we may add both the equations and eliminate them.

**Step 3 :**

After eliminating one variable, we may get the value of one variable.

**Step 4 :**

The remaining variable is then found by substituting in any one of the given equations.

**Example 1 :**

Solve the following pairs of linear equations by the elimination method and the substitution method

3x – 5y – 4 = 0 and 9x = 2y + 7

**Solution :**

3x – 5y = 4 ---------(1)

9x - 2y = 7---------(2)

**Elimination method**

(1) ⋅ 2 ==> 9x - 15y = 12

9x - 2y = 7

(-) (+) (-)

--------------

-13y = 5 ==> y = -5/13

By applying the value of y in (1), we get

3x – 5 (-5/13) = 4

3x + (25/13) = 4

(39x + 25)/13 = 4

39x + 25 = 4(13)

39x + 25 = 52

39x = 52 – 25

39x = 27

x = 27/39

x = 9/13

So, the solution is x = 9/13 and y = -5/13

**Substitution method**

3 x – 5 y = 4 ---------(1)

9 x - 2 y = 7---------(2)

**Step 1 :**

Find the value of one variable in terms of another variable

-2y = 7 – 9 x

y = (9 x – 7)/2

**Step 2 :**

By applying the value of y in (1), we get

3x – 5(9 x – 7)/2 = 4

(6 x - 45x + 35)/2 = 4

-39 x + 35 = 4(2)

-39 x = 8 - 35

-39 x = - 27

x = -27/(-39)

x = 9/13

**Step 3 :**

Apply x = 9/13 in the equation y = (9 x – 7)/2

y = [9(9/13) – 7]/2

y = [(81/13) – 7]/2

y = [(81 –91)/26]

y = -10/26

y = -5/13

Therefore solution is x = 9/13 and y = -5/13

**Example 2 :**

Solve the following pairs of linear equations by the elimination method and the substitution method

(x/2) + (2y/3) = -1 and x – (y/3) = 3

**Solution :**

(3 x + 4 y)/6 = -1

3x + 4y = -6 ---------(1)

x – (y/3) = 3

(3 x – y)/3 = 3

3x – y = 9 ---------(2)

**Elimination method**

** 3x + 4 y = -6**

(2) ⋅ 4 ==> 12**x – 4y = 36**

** -----------------**

** 15x = 30**

** x = 30/15 ==> x**** = 2**

**By applying the value of x in (2), we get**

**3(2) - y = 9**

**6 - y = 9**

**y = 6 - 9**

**y = -3**

**Substitution method**

3x + 4y = -6 ---------(1)

3x – y = 9 ---------(2)

**Step 1 :**

Find the value of one variable in terms of another variable

-y = 9 – 3 x

y = 3x - 9

**Step 2 :**

Applying the value of y in (1), we get

3x + 4(3x-9) = -6

3x + 12x – 36 = -6

15x – 36 = -6

15 x = -6 + 36

15 x = 30

x = 30/15 ==> x = 2

**Step 3 :**

Apply x = 2 in the equation y = 3x - 9

y = 3(2) – 9

y = 6 - 9

y = -3

So, the solution is x = 2 and y = -3

After having gone through the stuff given above, we hope that the students would have understood, solving equations using elimination and substitution.

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