If \(2x+3y-8=0\) and \(3x-2y-12=0\), then \(4x^2+y^2-4x\) is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 48
Two linear equations in \(x\) and \(y\) fix both values; they are found by elimination and then put into the expression.
\(2x+3y=8\) (1)
\(3x-2y=12\) (2)
Finding \(x\) and \(y\) (2 × (1) + 3 × (2), the \(y\) terms cancel):
\(4x+6y+9x-6y=16+36\)
\(13x=52\), so \(x=4\)
From (1): \(8+3y=8\), so \(y=0\)
The expression:
\(4x^2+y^2-4x=4\times16+0-16=48\)