# Thread: Help Solving Linear Equations with fractions

1. ## Help Solving Linear Equations with fractions

Okay, I am stuck on linear equations again. I am not understanding what to do with the fractions? I have read my text and our online lecture but obviously something isn't sinking in. The two equations that I am stuck on is as follows:

1. -3x + 6y = 2

2x + (2/3) y = 1

2. (x/3) + (y/4) = 7
(x/4) + (y/3) = 7

Looking at #2, one can see that the answer is 12 but I am not clear on how it should be written.

For # 1 what do I do with the 2/3?

Thanks in advance for the help

2. Hey!

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

Multiply second equation by -9.

$-3x + 6y = 2$ (a)
$-18x - 6y = -9$ (b)

$(-3x + -18x) + (6y - 6y) = 2 - 9$
$-21x = -7$
$x = \frac{1}{3}$.

Can you do the rest?

3. Are you trying to solve these equations simultaneously?

$-3x + 6y = 2$
$\phantom{-}2x + \frac{2}{3}y = 1$.

Multiply the first equation by 2 and the second equation by 3

$-6x + 12y = 4$
$\phantom{-}6x + 2y = 3$

$(-6x + 12y) + (6x + 2y) = 4 + 3$

$14y = 7$

$y = \frac{1}{2}$.

Substitute this back into one of the original equations - the first will do.

$-3x + 6y = 2$

$-3x + 6\left(\frac{1}{2}\right) = 2$

$-3x + 3 = 2$

$-3x = -1$

$x = \frac{1}{3}$.

So the solution is $(x, y) = \left(\frac{1}{3}, \frac{1}{2}\right)$.

4. Hi, thank you so much, you have been a tremendous help to me. I think I can handle it from here :-)

5. Wonderful, thank you very much. That is exactly what I needed to see, how it was broken down. How about the other one:

(x/3) + (y/4) = 7
(x/4) + (y/3) = 7

I know the answer is 12 but I'm not sure how to write it out in the proper format.

6. Sure thing! We're going to solve it basically the same way

$\frac{x}{3} + \frac{y}{4} = 7$ (a)
$\frac{x}{4} + \frac{y}{3} = 7$ (b)

Multiply (a) by -3 and (b) by 4

$-x + \frac{-3y}{4} = -21$ (a)
$x + \frac{4y}{3} = 28$ (b)

$(-x + x) + \frac{-3y}{4} + \frac{4y}{3} = -21 + 28 = 7$