Multiple choice

Solve the following equations: $9x + y - 8z = 0$, $4x - 8y + 7z = 0$, $yz + zx + xy = 47$.

  1. $x=\pm3; y=\pm5; z=\pm4$
  2. $x=\pm 2; y=\pm4; z=\pm1$
  3. $x=x=\pm4; y=3; z=\pm 5$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the linear system 9x+y-8z=0 and 4x-8y+7z=0 gives x:y:z = (7-64) : (-32-63) : (-72-4) = -57 : -95 : -76 = 3 : 5 : 4. Substituting x=3k, y=5k, z=4k into yz+zx+xy=47: 20k^2 + 12k^2 + 15k^2 = 47 => 47k^2 = 47 => k=1 or -1. Thus x=3, y=5, z=4 or x=-3, y=-5, z=-4.

Keep practicing — related questions