Question

What is the solution to this? I would like some help.

Answer 1

A)

`\begin{gathered} \angle3(x-8)\text{ and }\angle2(x+10) \\ \text{Are alternate angles, therefore:} \\ \angle3(x-8)=\angle2(x+10) \\ 3(x-8)=2(x+10) \\ 3x-24=2x+20 \\ \text{Solve for x:} \\ x=44 \end{gathered}`

B)

`\begin{gathered} \angle4x-12\text{ and }\angle120\text{ are supplementary angles, so:} \\ \angle4x-12+\angle120=180 \\ 4x-12+120=180 \\ \text{Solve for x:} \\ 4x=180-120+12 \\ 4x=72 \\ x=(72)/(4) \\ x=18 \end{gathered}`