The Purplemath Forums

by **collograph** on Sat Nov 14, 2009 5:33 am

I am having some trouble with the elimination method - was able to do all the other problems that were assigned- but was unable to figure out the last two problems - can someone help me? .....thank you.

1) 6x+2y-4=0 and 10x+7y=-8

2) 2/3x+y=-5/3 and x-1/3y=-13/3

**Sponsor**

by **stapel_eliz** on Sat Nov 14, 2009 2:34 pm

What have you tried? Where are you stuck? For instance, after you moved the 4 to the right-hand side of the first equation in the first exercise, multiplied the first equation by 7 and the second equation by -2, and added down, what did you get?

Please be complete. Thank you! :wink:

by **collograph** on Sat Nov 14, 2009 6:19 pm

I was able to figure the first one out- but for the second problem....

2/3x+y=-5/3 and x-1/3y=-13/3

....would I just multiply everythng in both equations by 1/3 (or just 1/3 for the first one and -1/3 for the second one)? Once I get past that- am confident about the rest of it.

by **stapel_eliz** on Sat Nov 14, 2009 6:46 pm

As posted, the equations for the second system are as follows:

. . . . . $\mbox{(1) }\, \frac{2}{3x}\, +\, y\, =\, -\frac{5}{3}$

. . . . . $\mbox{(2) }\, x\, -\, \frac{1}{3y}\, =\, -\frac{13}{3}$

From your comments, however, I think you mean the following:

. . . . . $\mbox{(1) }\, \frac{2}{3}x\, +\, y\, =\, -\frac{5}{3}$

. . . . . $\mbox{(2) }\, x\, -\, \frac{1}{3}y\, =\, -\frac{13}{3}$

If you mean the latter, then yes, multiplying each equation by "3" would be a good way to start, especially if you're aiming to avoid fractions as much as possible. Then multiply the second equation by 3 again, and add down. :wink:

by **collograph** on Sat Nov 14, 2009 7:35 pm

The second one....... (2/3x+y=-5/3.....).......

I wasn't sure about multiplyng it by just 3......or 3/1......or 1/3 (I think that is a major part of why I feel confused). And do I multiply them both (or just one)by The negative number (3....or 3/1...etc.). :oops:

by **stapel_eliz** on Sun Nov 15, 2009 1:56 pm

Since $3\, =\, \frac{3}{1},$ multiplying by either is the same as multiplying by the other. Use whichever form works better for you. :wink:

And since both equations have the same denominator that you want to clear, multiply each by the same value.

by **collograph** on Sun Nov 15, 2009 11:02 pm

I have been trying for over an hour - I still can't get the problem to come out right :( . Yuck- I am to the point where I want to quit but at the same time- got no choice in the matter.

by **stapel_eliz** on Sun Nov 15, 2009 11:45 pm

What did you get after you followed the steps provided? You multiplied both equations by 3 to clear the denominators:

. . . . . $\mbox{(1) }\, 2x\, +\, 3y\, =\, -5$

. . . . . $\mbox{(2) }\, 3x\, -\, y\, =\, -13$

Then you multiplied the second equation by 3 again:

. . . . . $\mbox{(1) }\, 2x\, +\, 3y\, =\, -5$

. . . . . $\mbox{(2) }\, 9x\, -\, 3y\, =\, -39$

You added down to get:

. . . . . $11x\, =\, -44$

Where are you stuck after this point?

Please be complete. Thank you! :wink:

The Purplemath Forums

Elimination Method: 6x+2y-4=0 and 10x+7y=-8

Elimination Method: 6x+2y-4=0 and 10x+7y=-8

Re: Elimination Method: 6x+2y-4=0 and 10x+7y=-8

Re: Elimination Method: 6x+2y-4=0 and 10x+7y=-8

Re: Elimination Method: 6x+2y-4=0 and 10x+7y=-8