If $2x + 3y-8=0$ and $3x-2y-12 = 0$, then $4x^2+y^2-4x$ is equal to:

48

The correct answer is Option (4) → 48

Given equations:

1. $2x + 3y - 8 = 0 \Rightarrow 2x + 3y = 8$
2. $3x - 2y - 12 = 0 \Rightarrow 3x - 2y = 12$

Step 1: Solve for x and y

Multiply (1) by 2 and (2) by 3:

• $4x + 6y = 16$
• $9x - 6y = 36$

Add them:

$13x = 52 \Rightarrow x = 4$

Substitute $x = 4$ into $2x + 3y = 8$:

$8 + 3y = 8 \Rightarrow y = 0$

Step 2: Evaluate the expression

$4x^2 + y^2 - 4x$

Substitute $x = 4, y = 0$:

$4(4^2) + 0 - 4(4) = 4(16) - 16 = 64 - 16 = 48$