Linear Equation by Elimination

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December 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

Am i doing this right so far? i'm really confused at this one please help
December 2nd 2012, 05:59 PM
Scopur

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

$\frac{x+2}{6} - \frac{3(y+2)}{2} = 1$
$\frac{x-2}{2} + \frac{y-1}{3} = 0$
December 2nd 2012, 06:04 PM
Prove It

Re: Linear Equation by Elimination

Some brackets used where they're needed would be nice...
December 2nd 2012, 06:34 PM
Vince604

Re: Linear Equation by Elimination

Quote:

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?
December 3rd 2012, 09:24 AM
Vince604

Re: Linear Equation by Elimination

Anyone?
December 3rd 2012, 09:38 AM
Scopur

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.

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