Question

$\frac{x+1}{2}+\frac{y-1}{3}=8,\frac{x-1}{3}+\frac{y+1}{2}=9.$

Answer 1

The given equations are:
$\frac{x+1}{2}+\frac{y-1}{3}=8$ ..............(i)
$\frac{x-1}{3}+\frac{y+1}{2}=9$ ...............(ii)
On multiplying each equation by 6, we get:
3(x + 1) + 2(y − 1) = 48
⇒ 3x + 3 + 2y − 2 = 48
⇒ 3x + 2y = 47 .............(iii)
Also, 2(x − 1) + 3(y + 1) = 54
⇒ 2x − 2 + 3y + 3 = 54
⇒ 2x + 3y = 53 ..........(iv)
On multiplying (iii) by 3 and (iv) by 2, we get:
9x + 6y = 141 .........(v)
4x + 6y = 106 ..........(vi)
On subtracting (vi) from (v), we get:
5x = (141 − 106) = 35
⇒ x = 7
On substituting x = 7 in (iii), we get:
3 × 7 + 2y = 47
⇒ 2y = (47 − 21) = 26
⇒ y = 13
Hence, required solution is x = 7 and y = 13.