### Strategy #1:  Using the PIA (Plug In Answers) strategy:

• (A) Is not the right answer because when you plug in -80 for $x^{2}$  (which doesn’t actually make any sense since $x^{2}$ can’t equal a negative number), you get $-80+y^{^{2}}=160$.  After adding 80 to both sides, you have $y^{2}=240$.  So $y\approx&space;15.5$.  If you plug that back into the second equation, you get $15.5=-3x$ which gives you $x\approx&space;5.16$.  If $x^{^{2}}=-80$, as we originally presumed, x would not exist in the real number system.  Even if you mistakenly thought you could take the square root of a negative number, you might get $x\approx&space;-8.9$ when you tried to take the square root of both sides, which is not equal to $5.16$.
• (B) is not the right answer because when you plug $4$ in for $x^{2}$ in the first equation, you get $4+y^{2}=160$, which gives you $y^{2}=156$.  So after you take the square root of both sides, $y\approx&space;\pm&space;12.5$.  Looking at the second equation, if $x^{^{2}}=4$, then after you take the square root of both sides you get $x\doteq&space;\pm&space;2$.  Plugging $\pm&space;2$ in for x in the second equation gives you y=$\pm&space;6$. This does not match the $y$ value we got from the first equation.
• (C) is the correct answer!!! When you plug $16$ in for $x^{2}$ in the first equation, you end up with $16+y^{2}=160$.  After subtracting  from both sides, you are left with $y^{2}=144$.  After taking the square root of both sides, $y=\pm&space;12$.  Looking at the second equation, if $x^{2}=16$ then $x=\pm&space;4$.  So we get $y=-3(\pm&space;4)$ which means $y=\pm&space;12$.  This is the same thing we got when we plugged into the first equation!  Now we know (C) is the answer!
• (D) is not the right answer because when you plug $144$ in for $x^{2}$ in the first equation, you get $144+y^{2}=160$, which gives you $y^{2}=16$.  So after you take the square root of both sides, $y=\pm&space;4$.  Looking at the second equation, if $x^{2}=144$, then after you take the square root of both sides you get $x=\pm&space;12$.  Plugging $\pm&space;12$ in for x in the second equation gives you y=$\pm&space;36$. This does not match the  value we got from the first equation

### Strategy #2:  Using the substitution method

You need to find a solution that works for both equations.  Since the y is isolated in the second equation you can substitute  into  in the first equation.

• (A) is not the right answer. You might have gotten this answer if you forgot to square the $-3$ when you substituted $-3x$ in for $y$.  If this was your mistake, you may have gotten $x^{2}-3x^{2}=160$.  Upon combining like terms, you may have ended up with $-2x^{2}=160$.  After dividing by $-2$ you would get $x^{2}=-80$.
• (B) is not the right answer. You might have gotten this answer if you solved for $x$, instead of $x^{2}$.
• (D) is not the right answer. You might have gotten this answer if you solved for $y^{2}$, instead of $x^{2}$.
• (C) IS the correct answer!! If you substituted correctly by plugging in $-3x$ for $y$ like this: $x^{2}+(-3x)^2=160$ and then correctly simplified the left-hand side like this: $x^{2}+9x^{2}=160$ leading to $10x^{2}=160$, you could then divide both sides by $10$, leaving you with the answer: $x^{2}=16$.