Two lines \(3x + 4y = 12\) and \(2x + 3y = 16\) intersect at the point:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (−28, 24)
The point where two lines meet lies on both lines, so it is the solution of the two equations taken together. Making the \(y\) terms equal and subtracting removes \(y\).
First line: \(3x+4y=12\) (1)
Second line: \(2x+3y=16\) (2)
3 × (1): \(9x+12y=36\); 4 × (2): \(8x+12y=64\)
Subtracting, the \(y\) terms cancel: \(x=36-64=-28\)
Putting \(x=-28\) in (1): \(-84+4y=12\)
\(4y=96\), so \(y=24\) and the lines meet at (−28, 24)