SAT Prep

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.