Question

Solve the following system of equations by the elimination method:

3(x – 2) + 5(y + 1) = 20, 4(x – 1) + 3(y + 3) = 22

Answer 1

Given:

3(x – 2) + 5(y + 1) = 20

4(x – 1) + 3(y + 3) = 22

Step 1: Simplify each equation

Expand the expressions inside parentheses:

3x – 6 + 5y + 5 = 20 

⇒ 3x + 5y – 1 = 20

⇒ 3x + 5y = 21

4x – 4 + 3y + 9 = 22

⇒ 4x + 3y + 5 = 22

⇒ 4x + 3y = 17

Our system becomes:

3x + 5y = 21   ...(i)

4x + 3y = 17   ...(ii)

Step 2: Eliminate one variable

Multiply equation (i) by 4 and equation (ii) by 3, to align the coefficients of (x): 

4(3x + 5y) = 4 × 21 

⇒ 12x + 20y = 84   ...(iii) 

3(4x + 3y) = 3 × 17 

⇒ 12x + 9y = 51   ...(iv)

Now subtract (iv) from (iii) to eliminate (x):

(12x + 20y) – (12x + 9y) = 84 – 51 

11y = 33 

y = 3

Step 3: Substitute (y = 3) back into one of the original simplified equations

Using equation (i):

3x + 5(3) = 21

3x + 15 = 21 

3x = 6 

x = 2