# System of linear equations.

• Aug 9th 2010, 08:22 PM
gabriel
System of linear equations.
how do i solve for this system of equations

11x-3y=-36
12y-9x=39
?
tried reordering the terms but what else?
• Aug 9th 2010, 08:51 PM
pickslides
Quote:

Originally Posted by gabriel
how do i solve for this system of equations

11x-3y=-36
12y-9x=39
?
tried reordering the terms but what else?

$11x-3y=-36$ ...(1)
$12y-9x=39$ ...(2)

From (1) $11x-3y=-36\implies -3y=-36-11x \implies y=12+\frac{11}{3}x$

Then (2) $12y-9x=39\implies 12\left(12+\frac{11}{3}x\right)-9x=39$ Can you finish it from here?
• Aug 9th 2010, 09:08 PM
gabriel
yeppp i got thnx good refreshment to the dusty mind
• Aug 9th 2010, 11:14 PM
Prove It
Quote:

Originally Posted by gabriel
how do i solve for this system of equations

11x-3y=-36
12y-9x=39
?
tried reordering the terms but what else?

The alternative method (the method used for more equations with more unknowns) is the elimination method...

$11x - 3y = -36$
$-9x + 12y = 39$.

Multiply the first equation by 4 to give

$44x-12y=-144$
$-9x+12y=39$

$(44x-12y)+(-9x+12y)=-144+39$

$35x=-105$

$x = -3$.

Substitute into one of the original equations...

$11(- 3) - 3y = -36$

$-33- 3y = -36$

$-3y = -3$

$y = 1$

So the solution is $(x,y) = (-3,1)$.
• Aug 9th 2010, 11:39 PM
earboth
Quote:

Originally Posted by Prove It
...

$(44x-12y)+(-9x+12y)=-144+39$

$35x=105$ <=== unfortunately you made a small sign mistake here

$x = 3$.

Substitute into one of the original equations...

$11\cdot 3 - 3y = -36$

$33- 3y = -36$

$-3y = -69$

$y = 23$.

So the solution is $(x,y) = (3,23)$.

$(44x-12y)+(-9x+12y)=-144+39$

$35x=-105$

$x = -3$.
• Aug 10th 2010, 12:03 AM
Prove It
Yes I did, edited :)