# Thread: Linear Equation by Elimination

1. ## Linear Equation by Elimination

Solve by elimination.
x+2/6 - 3(y+2)/2 = 1

x-2/2 + y-1/3 = 0

For the second one, this is what i did:

multiply by -3 (x-2/2 + y-1/3 = 0)

y = 3x - 4

For the first one:

multiply by 2 (x+2/6 - 3(y+2)/2 = 1)

1/3x + 2/3 -3y -6 = 2

2. ## Re: Linear Equation by Elimination

Do you mean:
$x+2/6 - 3(y+2)/2 = 1$

$x-2/2 + y-1/3 = 0$
(this is how you wrote it)
or

$\frac{x+2}{6} - \frac{3(y+2)}{2} = 1$
$\frac{x-2}{2} + \frac{y-1}{3} = 0$

3. ## Re: Linear Equation by Elimination

Some brackets used where they're needed would be nice...

4. ## Re: Linear Equation by Elimination

Originally Posted by Scopur

$\frac{x+2}{6} - \frac{3(y+2)}{2} = 1$
$\frac{x-2}{2} + \frac{y-1}{3} = 0$
yea thanks, this. how would you simplify it, so that its easier to find x,y?

Anyone?

6. ## Re: Linear Equation by Elimination

Take the 2nd equation and isolate for $y$.

$\frac{y-1}{3} = - \frac{x-2}{2}$

$y-1 = - \frac{3}{2}(x-2)$

$y-1 = -\frac{3}{2}x + 3$

$y = -\frac{3}{2}x + 4$

Now take this and plug it into your first equation for y.