# System of Equations - Help

• Nov 3rd 2008, 08:54 AM
cnmath16
System of Equations - Help
Solve the following system of equations using the elimination method

3x-9y=10 (1)
4x+y=7 (2)

I have completed this question, but I would like someone to look over it to make sure that I did the right thing, because it does not look like the right answer to me. Any help or comments would be appreciated.
Thanks!

3x-9y=10 (1)
4x+y=7 (2)
By elimination:
Take equation (2) and multiply by 9
Thus, (2) x 9 gives you 36x-9y = 63 (3)
Which gives you 39x = 73 ... divide both sides by 39 and x= 73/39 or 1.87

Substitute x (1.87) into Equation #1 to find y:
3(73/39)-9y=10
..This is where I get confused.
• Nov 3rd 2008, 09:30 AM
Soroban
He4llo, cnmath16!

Quote:

Solve the following system of equations using the elimination method

. . $\displaystyle \begin{array}{cccc}3x-9y&=&10 & [1] \\ 4x+y&=&7 & [2] \end{array}$

$\displaystyle \begin{array}{cccccc}\text{Multiply [2] by 9:} & 36x + 9y &=& 63 \\ \text{Add [1]:} & 3x - 9y &=& 10 \end{array}$
Which gives you: .$\displaystyle 39x \:= \:73 \quad\Rightarrow\quad x \:=\:\tfrac{73}{39}$ . . . . Good!
Substitute into [2]: .$\displaystyle 4\left(\tfrac{73}{39}\right) + y \;=\;7 \quad\Rightarrow\quad y \:=\:7 - \tfrac{292}{39} \quad\Rightarrow\quad y \:=\:\text{-}\tfrac{19}{39}$
Therefore: .$\displaystyle \begin{Bmatrix}\; x &=&\dfrac{73}{39}\; \\ \\[-3mm] \; y &=& \text{-}\dfrac{19}{39}\; \end{Bmatrix}$