Question

Solve the following equations:-
(x + 1)/2 + (y - 1)/3 = 8
(x - 1)/3 + (y + 1)/2 = 9

Answer 1

Given, (x + 1)/2 + (y - 1)/3 = 8
(x - 1)/3 + (y + 1)/2 = 9

first of all solve each equation separately,
(x + 1)/2 + (y - 1)/3 = 8

=> 3(x + 1) + 2(y - 1) = 6 × 8

=> 3x + 3 + 2y - 2 = 48

=> 3x + 2y = 47 ............(1)


(x - 1)/3 + (y + 1)/2 = 9

=> 2(x - 1) + 3(y + 1) = 9

=> 2x - 2 + 3y + 3 = 6 × 9 = 54

=> 2x + 3y = 53 ...........(2)

now, multiplying 3 with equation (1) and 2 with equation (2) then subtracting (2) from (1) ,

=> 3(3x + 2y) - 2(2x + 3y) = 47 × 3 - 2 × 53

=>9x - 4x = 141 - 106 = 35

=> 5x = 35

=> x = 7

put x = 7 in equation (1),

3x + 2y = 47

=> 3 × 7 + 2y = 47

=> 21 + 2y = 47

=> 2y = 47 - 21 = 26

=> y = 13

hence, x = 7 and y = 13

Answer 2

Dear Student,

Answer: x= 7, y =13

Solution:

$\frac{x + 1}{2} + \frac{y - 1}{3} = 8 \\ \\ \frac{3(x + 1) + 2(y - 1)}{6} = 8 \\ \\ 3x + 3 + 2y - 2 = 48 \\ \\ 3x + 2y = 47 \: \: \: \: eq1$
$\frac{x - 1}{3} + \frac{y + 1}{2} = 9 \\ \\ \frac{2(x - 1) + 3(y + 1)}{6} = 9 \\ \\ 2x - 2 + 3y + 3 = 54 \\ \\ 2x + 3y = 53 \: \: \: \: eq2$
add eq1 and eq 2

$5x + 5y = 100 \\ \\ x + y = 20 \: \: \: eq3$
subtract both eq 1 and eq2

$x - y = - 6 \: \: \: eq4$
add eq 3 and eq4

$2x = 14 \\ \\ x = 7 \\ \\ x + y = 20 \\ 7 + y = 20 \\ \\ y = 20 - 7 \\ \\ y = 13$
Hope it helps you.