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

Am i doing this right so far? i'm really confused at this one please help

Re: Linear Equation by Elimination

Do you mean:

$\displaystyle x+2/6 - 3(y+2)/2 = 1 $

$\displaystyle x-2/2 + y-1/3 = 0 $

(this is how you wrote it)

or

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

$\displaystyle \frac{x-2}{2} + \frac{y-1}{3} = 0 $

Re: Linear Equation by Elimination

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

Re: Linear Equation by Elimination

Quote:

Originally Posted by

**Scopur**

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

$\displaystyle \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?

Re: Linear Equation by Elimination

Re: Linear Equation by Elimination

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

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

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

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

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

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