Question

Solve the following system of equations by the elimination method:

`(x + 1)/3 + (y - 3)/4 = 3, (x - 2)/3 + (y + 1)/2 = 5`

Answer 1

Given the system of equations:

`(x + 1)/3 + (y - 3)/4 = 3` 

`(x - 2)/3 + (y + 1)/2 = 5`

Step 1: Eliminate denominators by multiplying both sides of each equation by the least common multiple of denominators

For the first equation, multiply both sides by 12 (LCM of 3 and 4): 

`12 xx ((x + 1)/3 + (y - 3)/4) = 12 xx 3`

4(x + 1) + 3(y – 3) = 36 

4x + 4 + 3y – 9 = 36 

4x + 3y – 5 = 36 

4x + 3y = 41   ...(Equation 1)

For the second equation, multiply both sides by 6 (LCM of 3 and 2): 

`6 xx ((x - 2)/3 + (y + 1)/2) = 6 xx 5`

2(x – 2) + 3(y + 1) = 30 

2x – 4 + 3y + 3 = 30 

2x + 3y – 1 = 30 

2x + 3y = 31   ...(Equation 2)

Step 2: Subtract Equation 2 from Equation 1 to eliminate (y):

(4x + 3y) – (2x + 3y) = 41 – 31 

4x – 2x + 3y – 3y = 10 

2x = 10

x = 5

Step 3: Substitute (x = 5) into Equation 2:

2(5) + 3y = 31 

10 + 3y = 31 

3y = 21

y = 7

The solution to the system is x = 5, y = 7.