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-6x + 6y = 6 -6x + 3y = -12 | Please solve using elimination method and PLEASE show work or walk through work

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\begin{array}{cccl} -6x+6y&=&6\\\\ -6x+3y&=&-12 \end{array} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{cccl} \text{\LARGE -1}(-6x+6y&=&6)\\\\ -6x+3y&=&-12 \end{array}\hspace{5em} \begin{array}{cccl} ~~ 6x-6y&=&-6\\\\ -6x+3y&=&-12\\\cline{1-3} ~~ 0 ~~ -3y&=&-18 \end{array} \\\\[-0.35em] ~\dotfill


-3y=-18\implies y=\cfrac{-18}{-3}\implies \boxed{y=6} \\\\\\ \stackrel{\textit{now substituting on the 2nd equation}}{-6x+3(6)=-12\implies }-6x+18=-12 \\\\\\ -6x=-30\implies x=\cfrac{-30}{-6}\implies \boxed{x=5} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (5~~,~~6)~\hfill

User Ejntaylor
by
8.1k points
5 votes

Answer:

x = 5 , y = 6

Explanation:

Given:

  • -6x + 6y = 6
  • -6x + 3y = -12

Solve:

Multiply the second equation by -1

-1 ( - 6x + 3y = -12)

⇒ 6x - 3y = 12

Then add/subtract: {-6x + 6x = 0 (aka nothing) { 6y - 3y = 3y} {6 + 12 = 18}

-6x + 6y = 6

6x - 3y = 12

⇒ 3y = 18

Divide both sides by 3:

3y/3 = 18/3

y = 6

Since now we know what y is, We can plug it into the original equation. You can either use the first one or the second one. I'll be using the first one.

-6x + 6y = 6

-6x + 6(6) = 6 ⇒ plug 6 in the equation since y = 6

-6x + 36 =6

-6x + 36 - 36 = 6 - 36 ⇒ Subtract both sides by 36

⇒ -6x = - 30

Divide both sides by -6:

-6x/-6 = -30/-6

x = 5

As a result, the answer is;

x = 5, y = 6

RevyBreeze

User Niza Siwale
by
8.4k points

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