Correct Answer: 3$x$ + 5$y$ = 1

Solution :

Equation of straight lines are following
$7x + 11y = 3$..............................(i)
And $8x + y = 15$............................(ii)
On applying (i) × 1 – (ii) × 11, we get
$(7x + 11y) - (8x + y) × 11 = 3 - 15 × 11$
⇒ $(7x + 11y) - (88x + 11y) = 3 - 165$
⇒ $-81x = - 162$
⇒ $x = 2$
On putting $x = 2$ in equation (i), we get
$7 × 2 + 11y = 3$
⇒ $11y = 3 - 14 $
⇒ $11y = -11$
⇒ $y = -1$
So, the point of intersection is $(2, -1)$
Checking options one by one by putting $(2, -1)$, we get:
Option (i) $2x + y = 2$
LHS = $2 × 2 - 1 = 3 ≠ RHS$
Option (ii) $2x – y = 1$
LHS = $2 × 2 + 1 = 5 ≠ RHS$
Option (iii) $3x + 5y = 1$
LHS = $3 × 2 + 5 × - 1 = 1 = RHS$
Option (iv) $3x + 2y = 3$
LHS = $3 × 2 + 2 × -1 = 4 ≠ RHS$
Hence, the correct answer is 3$x$ + 5$y$ = 1.