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Can someone help me with this substitution method for algebra Y=4*-3x 5x-y=22

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Konrad509idn
[tex]y=4-3x\\ 5x-y=22\\\\ 5x-(4-3x)=22\\ 5x-4+3x=22\\ 8x=26\\ x=\dfrac{26}{8}=\dfrac{13}{4}\\\\ y=4-3\cdot\dfrac{13}{4}\\ y=4-\dfrac{39}{4}\\ y=\dfrac{16}{4}-\dfrac{39}{4}\\ y=-\dfrac{23}{4}\\\\ \boxed{(x,y)=\left(\dfrac{13}{4},-\dfrac{23}{4}\right)}[/tex]
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