elimination method; please help me solve

**runner940** · October 25th 2008, 09:38 AM

I keep getting different answers when I try to solve this problem, please help me!

Eq 1: 4/3x+6y=(-1)
Eq 2: 4x - 4y =(13/3)

**Soroban** · October 25th 2008, 10:17 AM

Hello, runner940!

$\begin{array}{cccc}\frac{4}{3}x+6y &=& \text{-}1 & {\color{blue}[1]}\\<br /> 4x - 4y &=& \frac{13}{3} & {\color{blue}[2]} \end{array}$

First eliminate the fractions . . .

$\begin{array}{ccccc}\text{Multiply {\color{blue}[1]} by 3:} & 4x + 18y &=&\text{-}3 & {\color{blue}[3]} \\ \text{Multiply {\color{blue}[2]} by 3:} & 12x - 12y &=& 13 & {\color{blue}[4]} \end{array}$

$\begin{array}{cccc}\text{Multiply {\color{blue}[3]} by 2:} & 8x + 36y &=& \text{-}6 \\ \text{Multiply {\color{blue}[4]} by 3:} & 36x - 36y &=& 39 \end{array}$

Add: . $44x \:=\:33 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{3}{4}}$

Substitute into [1]: . $4(\tfrac{3}{4}) + 18y \:=\:\text{-}3 \quad\Rightarrow\quad18y \:=\:\text{-}6 \quad\Rightarrow\quad\boxed{ y \:=\:-\frac{1}{3}}$

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