Question

What is the solution to the system of equations?

Use the substitution method to solve.

6=−4x+y
−5x−y=21

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Answer 1

The first one is -6 and the second one is -3

Explanation:

Answer 2

We have
`\begin{bmatrix}6=-4x+y\\ -5x-y=21\end{bmatrix}`

So we want to isolate x for 6.


`6=-4x+y \ \textgreater \ Switch\: sides \ \textgreater \ -4x+y=6`


`\mathrm{Subtract\:}y\mathrm{\:from\:both\:sides} \ \textgreater \ -4x+y-y=6-y \ \textgreater \ -4x=6-y`


`\mathrm{Divide\:both\:sides\:by\:}-4 \ \textgreater \ (-4x)/(-4)=(6)/(-4)-(y)/(-4) \ \textgreater \ Simplify`


`(-4x)/(-4) \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}:\ (-a)/(-b)=(a)/(b) \ \textgreater \ (4x)/(4)`


`\mathrm{Divide\:the\:numbers:}\:(4)/(4)=1 \ \textgreater \ x`


`(6)/(-4)-(y)/(-4) \ \textgreater \ \mathrm{Apply\:rule}\:(a)/(c)\pm (b)/(c)=(a\pm \:b)/(c) \ \textgreater \ (6-y)/(-4)`


`\mathrm{Apply\:the\:fraction\:rule}:\ (a)/(-b)=-(a)/(b) \ \textgreater \ -(-y+6)/(4) \ \textgreater \ x=-(6-y)/(4)`


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

Isolate y for
`-5\left(-(6-y)/(4)\right)-y=21`


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

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


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


`Refine\: it\ \textgreater \ 5\left(6-y\right)-4y=84 \ \textgreater \ Expand\: 5\left(6-y\right)`


`\mathrm{Distribute\:parentheses\:using}:\:a\left(b+c\right)=ab+ac`

Where a=5, b=6, c=-y


`5\cdot \:6+5\left(-y\right) \ \textgreater \ Apply\:minus-plus\:rules \ \textgreater \ +\left(-a\right)=-a`


`5\cdot \:6-5y \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:5\cdot \:6=30 \ \textgreater \ 30-5y`


`30-5y-4y=84 \ \textgreater \ \mathrm{Add\:similar\:elements:}\:-5y-4y=-9y`


`30-9y=84 \ \textgreater \ \mathrm{Subtract\:}30\mathrm{\:from\:both\:sides} \ \textgreater \ 30-9y-30=84-30`


`-9y=54 \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-9 \ \textgreater \ (-9y)/(-9)=(54)/(-9) \ \textgreater \ y=-6`


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

Therefore...

`The\:solutions\:to\:the\:system\:of\:equationts\:are \ \textgreater \ y=-6,\:x=-3`