Answer
The value of \[{{x}^{2}}-\left( xy-y \right)\]is\[-5\].
Work Step by Step
The equation\[\frac{3x}{2}+\frac{3x}{4}=\frac{x}{4}-4\].
Subtract \[\frac{x}{4}\] from both sides,
\[\begin{align}
& \frac{3x}{2}+\frac{3x}{4}-\frac{x}{4}=\frac{x}{4}-4-\frac{x}{4} \\
& \frac{6x}{4}+\frac{3x}{4}-\frac{x}{4}=-4 \\
& \frac{6x+3x-x}{4}=-4 \\
& \frac{8x}{4}=-4 \\
\end{align}\]
Apply cross-products principle,
\[\begin{align}
& 8x=-4\times 4 \\
& 8x=-16 \\
& x=-2 \\
\end{align}\]
So, \[x=-2\]
Next equation is \[5-y=7\left( y+4 \right)+1\].
Use distributive property,
\[\begin{align}
& 5-y=7y+28+1 \\
& 5-y=7y+29 \\
\end{align}\]
Add \[y\]to both sides,
\[\begin{align}
& 5-y+y=7y+29+y \\
& 5=8y+29 \\
\end{align}\]
Subtract \[29\] from both sides,
\[\begin{align}
& 5-29=8y+29-29 \\
& -24=8y \\
& y=-3 \\
\end{align}\]
So, \[y=-3\]
Now put \[x=-2\] and \[y=-3\] in the expression \[{{x}^{2}}-\left( xy-y \right)\].
\[\begin{align}
& {{x}^{2}}-\left( xy-y \right)={{\left( -2 \right)}^{2}}-\left( -2\times -3-\left( -3 \right) \right) \\
& =4-\left( 6+3 \right) \\
& =4-9 \\
& =-5
\end{align}\]
Therefore, value of \[{{x}^{2}}-\left( xy-y \right)\]is\[-5\].
