What is the solution to the system of equations?Use the substitution method to solve.6=−4x+y −5x−y=21Enter your answer in the boxes.( , ​​)

We have [tex]\begin{bmatrix}6=-4x+y\\ -5x-y=21\end{bmatrix}[/tex]

So we want to isolate x for 6.

[tex]6=-4x+y \ \textgreater \ Switch\: sides \ \textgreater \ -4x+y=6 [/tex]

[tex]\mathrm{Subtract\:}y\mathrm{\:from\:both\:sides} \ \textgreater \ -4x+y-y=6-y \ \textgreater \ -4x=6-y[/tex]

[tex]\mathrm{Divide\:both\:sides\:by\:}-4 \ \textgreater \ \frac{-4x}{-4}=\frac{6}{-4}-\frac{y}{-4} \ \textgreater \ Simplify[/tex]

[tex]\frac{-4x}{-4} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}:\ \frac{-a}{-b}=\frac{a}{b} \ \textgreater \ \frac{4x}{4} [/tex]

[tex]\mathrm{Divide\:the\:numbers:}\:\frac{4}{4}=1 \ \textgreater \ x[/tex]

[tex]\frac{6}{-4}-\frac{y}{-4} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{6-y}{-4} [/tex]

[tex]\mathrm{Apply\:the\:fraction\:rule}:\ \frac{a}{-b}=-\frac{a}{b} \ \textgreater \ -\frac{-y+6}{4} \ \textgreater \ x=-\frac{6-y}{4}[/tex]

[tex]\mathrm{Subsititute\:}x=-\frac{6-y}{4} \ \textgreater \ \begin{bmatrix}-5\left(-\frac{6-y}{4}\right)-y=21\end{bmatrix}[/tex]

Isolate y for [tex]-5\left(-\frac{6-y}{4}\right)-y=21[/tex]

[tex]\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ 5\cdot \frac{6-y}{4}-y=21 [/tex]

Multiply [tex]5\cdot \frac{6-y}{4} \ \textgreater \ \mathrm{Multiply\:fractions}:\ \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{\left(6-y\right)\cdot \:5}{4}[/tex]

[tex]\frac{5\left(6-y\right)}{4}-y=21 \ \textgreater \ \mathrm{Multiply\:both\:sides\:by\:}4 \ \textgreater \ \frac{5\left(6-y\right)}{4}\cdot \:4-y\cdot \:4=21\cdot \:4[/tex]

[tex]Refine\: it\ \textgreater \ 5\left(6-y\right)-4y=84 \ \textgreater \ Expand\: 5\left(6-y\right)[/tex]

[tex]\mathrm{Distribute\:parentheses\:using}:\:a\left(b+c\right)=ab+ac[/tex]

Where a=5, b=6, c=-y

[tex]5\cdot \:6+5\left(-y\right) \ \textgreater \ Apply\:minus-plus\:rules \ \textgreater \ +\left(-a\right)=-a [/tex]

[tex]5\cdot \:6-5y \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:5\cdot \:6=30 \ \textgreater \ 30-5y[/tex]

[tex]30-5y-4y=84 \ \textgreater \ \mathrm{Add\:similar\:elements:}\:-5y-4y=-9y [/tex]

[tex]30-9y=84 \ \textgreater \ \mathrm{Subtract\:}30\mathrm{\:from\:both\:sides} \ \textgreater \ 30-9y-30=84-30[/tex]

[tex]-9y=54 \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-9 \ \textgreater \ \frac{-9y}{-9}=\frac{54}{-9} \ \textgreater \ y=-6[/tex]

[tex]\mathrm{For\:}x=-\frac{6-y}{4}, \mathrm{Subsititute\:}y=-6 \ \textgreater \ x=-\frac{6-\left(-6\right)}{4}\quad \Rightarrow \quad x=-3[/tex]

Therefore...
[tex]The\:solutions\:to\:the\:system\:of\:equationts\:are \ \textgreater \ y=-6,\:x=-3[/tex]