Answer to Math #1

Answer: C

Difficulty: Medium

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 $y\approx 15.5$ . If you plug that back into the second equation, you get $15.5=-3x$ which gives you $x\approx 5.16$ . 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 $x\approx -8.9$ 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, $y\approx \pm 12.5$ . Looking at the second equation, if $x^{^{2}}=4$ , then after you take the square root of both sides you get $x\doteq \pm 2$ . Plugging $\pm 2$ in for x in the second equation gives you y= $\pm 6$ . 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, $y=\pm 12$ . Looking at the second equation, if $x^{2}=16$ then $x=\pm 4$ . So we get $y=-3(\pm 4)$ which means $y=\pm 12$ . 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, $y=\pm 4$ . Looking at the second equation, if $x^{2}=144$ , then after you take the square root of both sides you get $x=\pm 12$ . Plugging $\pm 12$ in for x in the second equation gives you y= $\pm 36$ . 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: $x^{2}=16$ .