Algebra 2 PLEASE HELP!!!!

Lil Kush · **Joined:** Mon, 23 Apr 2012 17:24:58 UTC **Posts:** 36

Solve each system by the addition method.

-2x + 3y = -7
3x + y = 16

Put your answer in ordered pair (x, y) form.

Shadow · **Posted:** Mon, 23 Apr 2012 18:09:14 UTC

Lil Kush wrote:

Solve each system by the addition method.

-2x + 3y = -7
3x + y = 16

Put your answer in ordered pair (x, y) form.

What have you tried? Are you familiar with examples you've seen in class where your teacher has used the addition method?

Lil Kush · **Joined:** Mon, 23 Apr 2012 17:24:58 UTC **Posts:** 36

I'm taking this online so I have no help. I did however ask one of the math teachers if the could help. He showed me how to plug it in the calculator but when i took my test and used his advice it failed me.

Soroban · **Posted:** Tue, 24 Apr 2012 05:35:58 UTC

Hello, Lil Kush!

You should have been shown the procedure before assigning this problem.

Quote:

Solve by the addition method: . $\begin{array}{cccc}\text{-}2x + 3y &=& \text{-}7 & [1] \\ 3x + y &=& 16 & [2] \end{array}$

With the "addition method", we hope to ADD the two equations and have one of the variables drop out.
. . Then we can solve for the remaining variable . . . and we're on our way!

If we add the equation now, we get: . $x + 4y \:=\:9$
. . We still have both an

and a

. . . that didn't work.

But here is something we can do . . .
. . We can multiply an equation by any number (except zero).

We can match the

-terms "match up" if we multiply equation [2] by -3.

$\begin{array}{cccccccccc}\text{Multiply [2] by -3:} & -9x - 3y &=& -48 \\ \text{Add [1]:} & -2x + 3y &=& -7 \end{array} \quad\text{ and the }y\text{-terms cancel out.}$

$\text{So we have: }\:-11x \:=\:-55 \quad\Rightarrow\quad x \:=\:5$

$\text{Substitute into [2]: }\:3(5) + y \:=\:16 \quad\Rightarrow\quad y \:=\:1$

$\text{Therefore: }\:(x,\,y) \:=\:(5,\,1)$

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Algebra 2 PLEASE HELP!!!!

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