Solve system of equations $ \sqrt{2}x+3y= a $ and $x-\sqrt{3}y=4$

Question

$$ \sqrt{2}x+3y= a $$ $$x-\sqrt{3}y=4$$ What is the easiest way to solve this system? I always seem to get square roots of 2 and 3s and don't know what to do.

Answer 1

By elimination we have

• $ \sqrt{2}x+3y= a$
• $x-\sqrt{3}y=4\implies \sqrt 2x-\sqrt{6}y=4\sqrt 2$

and subtracting

• $3y+\sqrt 6 y=a-4\sqrt 2 \implies y=\frac{a-4\sqrt 2}{3+\sqrt 6 }$

and then we can find $x$ from

• $x-\sqrt{3}y=4\implies x=4+\sqrt{3}y=4+\sqrt{3}\frac{a-4\sqrt 2}{3+\sqrt 6 }$

Answer 2

While it may not be the easiest way, here is a very systematic way:

It's a linear system that can be written as $$ \begin{pmatrix} \sqrt{2} & 3\\ 1 & -\sqrt{3} \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} a\\4 \end{pmatrix}, $$ so, due to $$ \det{ \begin{pmatrix} \sqrt{2} & 3\\ 1 & -\sqrt{3} \end{pmatrix} } = -\sqrt{6}-3\neq 0 $$ there exists a unique solution. We can compute this solution by left multiplying the above equation with the inverse of the matrix which yields $$ \begin{pmatrix} x\\ y \end{pmatrix}= \begin{pmatrix} \sqrt{2} & 3\\ 1 & -\sqrt{3} \end{pmatrix}^{-1} \begin{pmatrix} a\\4 \end{pmatrix}. $$ I guess you know how to compute this inverse.