Answer
The value of \[{{x}^{2}}-\left( xy-y \right)\]is \[161\].
Work Step by Step
The equation\[\frac{x}{5}-2=\frac{x}{3}\].
Subtract \[\frac{x}{5}\] from both sides,
\[\begin{align}
& \frac{x}{5}-2-\frac{x}{5}=\frac{x}{3}-\frac{x}{5} \\
& -2=\frac{2x}{15} \\
\end{align}\]
Apply cross-products principle,
\[\begin{align}
& -2\times 15=2x \\
& -30=2x \\
& \frac{-30}{2}=x \\
& -15=x
\end{align}\]
So, \[x=-15\]
Next equation is\[-2y-10=5y+18\].
Add \[2y\]to both sides,
\[\begin{align}
& -2y-10+2y=5y+18+2y \\
& -10=7y+18 \\
\end{align}\]
Subtract \[18\] from both sides,
\[\begin{align}
& -10-18=7y-18-18 \\
& -28=7y \\
& -4=y \\
\end{align}\]
So, \[y=-4\]
Now put \[x=-15\] and \[y=-4\] in the expression\[{{x}^{2}}-\left( xy-y \right)\].
\[\begin{align}
& {{x}^{2}}-\left( xy-y \right)={{\left( -15 \right)}^{2}}-\left( -15\times -4-\left( -4 \right) \right) \\
& =225-\left( 60+4 \right) \\
& =225-64 \\
& =161
\end{align}\]
Therefore, value of \[{{x}^{2}}-\left( xy-y \right)\]is \[161\].
