The correct option is A (4, 2)
x+2y=8 ----- (1)
2x−3y=2 ----- (2)
First use formula for cross multiplication method:
xb1c2−b2c1=yc1a2−a1c2=−1a2b2−a2b1
So, from equation (1) and (2) we can write the value of a, b and c.
x2×2−(−3)×8=y8×2−1×2=−11×(−3)−2×2
x4+24=y16−2=−1−3−4
x28=y14=−1−7
x28=−1−7
x=−28−7=4
y14=−1−7
y=−14−7
y=2
The solution is (4, 2)