[tex]Solution, y=\left(\frac{1}{3}\right)x\cdot \:2+2x-1,\:y-5=x\quad :\quad x=\frac{18}{5},\:y=\frac{43}{5}[/tex]
[tex]Steps:[/tex]
[tex]\mathrm{Subsititute\:}y=\left(\frac{1}{3}\right)x2+2x-1, \begin{bmatrix}\frac{1}{3}x\cdot \:2+2x-1-5=x\end{bmatrix}[/tex]
[tex]\mathrm{Isolate}\:x\:\mathrm{for}\:\frac{1}{3}x2+2x-1-5=x:\quad x=\frac{18}{5}, \mathrm{For\:}y=\left(\frac{1}{3}\right)x2+2x-1, \mathrm{Subsititute\:}x=\frac{18}{5}, y=\frac{1}{3}\cdot \frac{18}{5}\cdot \:2+2\cdot \frac{18}{5}-1[/tex]
[tex]\frac{1}{3}\cdot \frac{18}{5}\cdot \:2+2\cdot \frac{18}{5}-1=\frac{43}{5}, y=\frac{43}{5}[/tex]
[tex]\mathrm{The\:solutions\:to\:the\:system\:of\:equationts\:are:}, x=\frac{18}{5},\:y=\frac{43}{5}[/tex]
Hope This Helps!!!
