Question

Help with math problem for algebra 2

Answer 1

`a=10`

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:

`(2a)/(2)=-(1-3b)/(2)`

`a=(-1-3b)/(2)`

Substitute this into Equation 2:

`4* (-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:

`(-4b)/(-4)=(28)/(-4)`

`b=-7`

We know that
`a=(-1-3b)/(2)`. Plug in b = -7 value:

`a=(-1-3\left(-7\right))/(2)`

`a=(-1+3\cdot \:7)/(2)`

`a=(20)/(2)`

`a=10`