Question

Help with math problem for algebra 2

Answer 1

Answer:

$a=10$

Step-by-step explanation:

Equation 1:  $2a+3b=-1$

Equation 2: $4a+2b=26$

Subtract 3b from both sides in Equation 1:

$2a=-1-3b$

Divide both sides by 2:

$\frac{2a}{2}=-\frac{1-3b}{2}$

$a=\frac{-1-3b}{2}$

Substitute this into Equation 2:

$4* \frac{-1-3b}{2}+2b=26$

$2\left(-3b-1\right)+2b=26$

$-6b-2+2b=26$

$-4b-2=26$

Add 2 to both sides:

$-4b=28$

Divide both sides by -4:

$\frac{-4b}{-4}=\frac{28}{-4}$

$b=-7$

We know that $a=\frac{-1-3b}{2}$. Plug in b = -7 value:

$a=\frac{-1-3\left(-7\right)}{2}$

$a=\frac{-1+3\cdot \:7}{2}$

$a=\frac{20}{2}$

$a=10$