# Thread: System of linear equations.

1. ## 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?

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

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

Then (2) $\displaystyle 12y-9x=39\implies 12\left(12+\frac{11}{3}x\right)-9x=39$ Can you finish it from here?

3. yeppp i got thnx good refreshment to the dusty mind

4. 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...

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

Multiply the first equation by 4 to give

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

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

$\displaystyle 35x=-105$

$\displaystyle x = -3$.

Substitute into one of the original equations...

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

$\displaystyle -33- 3y = -36$

$\displaystyle -3y = -3$

$\displaystyle y = 1$

So the solution is $\displaystyle (x,y) = (-3,1)$.

5. Originally Posted by Prove It
...

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

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

$\displaystyle x = 3$.

Substitute into one of the original equations...

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

$\displaystyle 33- 3y = -36$

$\displaystyle -3y = -69$

$\displaystyle y = 23$.

So the solution is $\displaystyle (x,y) = (3,23)$.
$\displaystyle (44x-12y)+(-9x+12y)=-144+39$

$\displaystyle 35x=-105$

$\displaystyle x = -3$.

6. Yes I did, edited