Question

Solve the following equations by substitution method.

$3a-2b=-10;2a+3b=2$

• A. $a=-2,b=2$
• B. $a=-1,b=2$
• C. $a=-4,b=2$
• D. $a=-7,b=2$

Answer 1

The correct option is A $a=-2,b=2$

$3a-2b=-10;2a+3b=2$

$a=\frac{2b-10}{3}\Rightarrow$ Put it in second equation.

$2\times \frac{2b-10}{3}+3b=2$

$4b-20+9b=6$
$13b=26$
$\therefore b=2$

$a=\frac{2b-10}{3}=\frac{2\left(2\right)-10}{3}=-2$

$a=-2$ and $b=2$