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Pelo método da substituição, temos
![\left \{ {{2x-3y=3}\quad(1) \atop {5x-y=1}\quad(2)} \right. \\\\
2x-3y=3 \Rightarrow 2x=3+3y \Rightarrow x=\frac{3+3y}{2}\\\\
substituindo \;em\;(2)\\\\
5(\frac{3+3y}{2})-y=1 \Rightarrow 5\left(3+3y\right)-2y=2\\\\
13y+15=2 \Rightarrow 13y=-13 \Rightarrow \framebox[1.5\width]{y = -1}\\\\
Substituindo \;y\; em\;\;x=\dfrac{3+3y}{2}\\\\
x=\frac{3+3\left(-1\right)}{2} \Rightarrow \framebox[1.5\width]{x = 0}](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B2x-3y%3D3%7D%5Cquad%281%29+%5Catop+%7B5x-y%3D1%7D%5Cquad%282%29%7D+%5Cright.+%5C%5C%5C%5C%0A2x-3y%3D3+%5CRightarrow+2x%3D3%2B3y+%5CRightarrow+x%3D%5Cfrac%7B3%2B3y%7D%7B2%7D%5C%5C%5C%5C%0Asubstituindo+%5C%3Bem%5C%3B%282%29%5C%5C%5C%5C%0A5%28%5Cfrac%7B3%2B3y%7D%7B2%7D%29-y%3D1+%5CRightarrow+5%5Cleft%283%2B3y%5Cright%29-2y%3D2%5C%5C%5C%5C%0A13y%2B15%3D2+%5CRightarrow+13y%3D-13+%5CRightarrow+%5Cframebox%5B1.5%5Cwidth%5D%7By+%3D+-1%7D%5C%5C%5C%5C%0A%0ASubstituindo+%5C%3By%5C%3B+em%5C%3B%5C%3Bx%3D%5Cdfrac%7B3%2B3y%7D%7B2%7D%5C%5C%5C%5C%0Ax%3D%5Cfrac%7B3%2B3%5Cleft%28-1%5Cright%29%7D%7B2%7D+%5CRightarrow++%5Cframebox%5B1.5%5Cwidth%5D%7Bx+%3D+0%7D "\left \{ {{2x-3y=3}\quad(1) \atop {5x-y=1}\quad(2)} \right. \\\\
2x-3y=3 \Rightarrow 2x=3+3y \Rightarrow x=\frac{3+3y}{2}\\\\
substituindo \;em\;(2)\\\\
5(\frac{3+3y}{2})-y=1 \Rightarrow 5\left(3+3y\right)-2y=2\\\\
13y+15=2 \Rightarrow 13y=-13 \Rightarrow \framebox[1.5\width]{y = -1}\\\\
Substituindo \;y\; em\;\;x=\dfrac{3+3y}{2}\\\\
x=\frac{3+3\left(-1\right)}{2} \Rightarrow \framebox[1.5\width]{x = 0}")
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