Do the square root of 2, 3 and 4 form a right triangle?
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Treat the matrices on the right side of each equation like you would a constant.
Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.
Then you can eliminate <em>Y</em> by taking the sum
4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>
==> <em>X</em> = (4<em>A</em> + <em>B</em>)/11
Similarly, you can eliminate <em>X</em> by using
-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>
==> <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11
It follows that
Similarly, you would find
You can solve the second system in the same fashion. You would end up with