Question

Solve the following simultaneous equations by the substitution method.

4(x + 4) – 3(y + 2) = 5, 3(x – 1) + 2y + 1 = 24

Answer 1

Given: 4(x + 4) – 3(y + 2) = 5, 3(x – 1) + 2y + 1 = 24

Step 1: Simplify each equation

First equation:

4(x + 4) – 3(y + 2) = 5 

⇒ 4x + 16 – 3y – 6 = 5

⇒ 4x – 3y + 10 = 5

⇒ 4x – 3y = 5 – 10

⇒ 4x – 3y = –5

Second equation:

3(x – 1) + 2y + 1 = 24 

⇒ 3x – 3 + 2y + 1 = 24

⇒ 3x + 2y – 2 = 24

⇒ 3x + 2y = 24 + 2

⇒ 3x + 2y = 26

Step 2: Express (x) in terms of (y) from the first equation

4x – 3y = –5 

⇒ 4x = 3y – 5

⇒ `x = (3y - 5)/(4)`

Step 3: Substitute `(x = (3y - 5)/(4))` in the second equation

3x + 2y = 26 

⇒ `3((3y - 5)/(4)) + 2y = 26`

Multiply both sides by 4 to clear the denominator:

3(3y – 5) + 8y = 104 

⇒ 9y – 15 + 8y = 104

⇒ 17y – 15 = 104

⇒ 17y = 119

⇒ `y = 119/17`

⇒ y = 7

Step 4: Find (x) using (y = 7)

`x = (3y - 5)/4` 

`x = (3(7) - 5)/4` 

`x = (21 - 5)/4` 

`x = 16/4`

x = 4