\(\mathrm{Subsititute\:}y=2+\frac{5}{6}\cdot \:x \)
 \(\begin{bmatrix}4x-3\left(2+\frac{5}{6}x\right)=3\end{bmatrix}\)
 \(\mathrm{Isolate}\:x\:\mathrm{for}\:4x-3\left(2+\frac{5}{6}x\right)=3:\quad x=6\)
 \(\mathrm{For\:}y=2+\frac{5}{6}\cdot \:x\)
 \(\mathrm{Subsititute\:}x=6\)
 \(2+\frac{5}{6}\cdot \:6=7\)
 \(y=7, \text{Solution is x=6, y=7}\)
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