Linear Equation by Elimination

• Dec 2nd 2012, 05:16 PM
Vince604
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

• Dec 2nd 2012, 05:59 PM
Scopur
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

$\displaystyle \frac{x+2}{6} - \frac{3(y+2)}{2} = 1$
$\displaystyle \frac{x-2}{2} + \frac{y-1}{3} = 0$
• Dec 2nd 2012, 06:04 PM
Prove It
Re: Linear Equation by Elimination
Some brackets used where they're needed would be nice...
• Dec 2nd 2012, 06:34 PM
Vince604
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?
• Dec 3rd 2012, 09:24 AM
Vince604
Re: Linear Equation by Elimination
Anyone?
• Dec 3rd 2012, 09:38 AM
Scopur
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.