Multiple choice

Solve the following pair of equations: $\sqrt{2}x+3y=\sqrt{3}; \, \, \sqrt{3}x+3y=\sqrt{2}$

  1. $x=-1, \, y=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{3}$
  2. $x=1, \, y=\displaystyle \frac{\sqrt{5}-\sqrt{2}}{3}$
  3. $x=1, \, y=\displaystyle \frac{\sqrt{7}-\sqrt{13}}{3}$
  4. $x=1, \, y=\displaystyle \frac{\sqrt{10}-\sqrt{5}}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Subtracting the first equation from the second: (sqrt(3) - sqrt(2))x = sqrt(2) - sqrt(3). This implies x = -1. Substituting x = -1 into the first equation: -sqrt(2) + 3y = sqrt(3), so 3y = sqrt(3) + sqrt(2), which gives y = (sqrt(3) + sqrt(2)) / 3.

Keep practicing — related questions