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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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2 Answers

3 votes

Answer:

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

Explanation:

User Dave Graves
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2 votes
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
User Channafow
by
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