Answer
$x=3$ and $y=1$
Work Step by Step
Given the equations $3x+4y=13$ and $5x-9y=6$,
$3x+4y=13 \rightarrow 4y=13-3x \rightarrow y=\dfrac{13-3x}{4}$
$5x-9y=6 \rightarrow -9y=6-5x \rightarrow y=\dfrac{6-5x}{-9}.$
Matching equations, we have
$\dfrac{13-3x}{4} = \dfrac{6-5x}{-9} \rightarrow -9(13-3x) = 4(6-5x) \rightarrow -117+27x = 24-20x \rightarrow 27x+20x = 24+117 \rightarrow 47x = 141 \rightarrow x = \dfrac{141}{47} = 3$.
Substituting $x$, we have
$y=\dfrac{13-3x}{4} = \dfrac{13-3(3)}{4} = \dfrac{13-9}{4} = \dfrac{4}{4} = 1$.
Therefore $x=3$ and $y=1$.
