\begin{aligned}
3x-5y&=44 \\
2x+8y&=-50
\end{aligned}
The solution to the given system is \( (x,y) \). What is the value of \(y\)?
This type of question used to be a huge pain. Here’s how someone doing this on the old paper SAT used to do this. To start, make the numbers in front of \( x \) or the numbers in front of \( y \) the same. One way to accomplish that is multiply both sides of the top equation by 2 and multiply both sides of the bottom equation by 3:
\begin{aligned}
3x-5y&=44 \\
2(3x-5y)&=2(44) \\
6x-10y&=88 \\
\ \\
2x+8y&=-50 \\
3(2x+8y) &= 3(-50) \\
6x+24y&=-150
\end{aligned}
Now the system is ready for elimination:
\begin{aligned}
6x-10y &= 88 \\
(-) \ \ 6x+ 24y &= -150 \\
\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } & \overline{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\
-34y &= 238 \\
y &= -7
\end{aligned}
We could stop there for the SAT. If this were math class, we’d probably need both values. To continue solving for \(x\), substitute \(-7\) for \(y\) in one of the original equations:
\begin{aligned}
3x-5y&=44 \\
3x-5(-7)&=44 \\
3x+35&=44 \\
3x&=9 \\
x&=3
\end{aligned}
The best way for everyone, whether you’re scoring 200 or 800 on math, is to have Desmos figure it out. This approach is definitely easier, way faster, and less prone to error. Use your mouse to hover over where the lines cross each other and you get the answer:
Now you try: