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At least try this yourself, then I'll check your working and answer.
Do you know how to use the linear combination method?
no, I suck at everything that has to do with graph's or geometry
Here, try reading through this: http://www.mathnstuff.com/math/algebra/asystem.htm It shows thorough step-by-step on how to use linear combination
do you know how to use it?
This is actually the dumbest subject, because how many times are we going to use graphs? I haven't used a graph for as long as I lived
I'm not a big fan of math either haha. The link I gave you shows step-by-step with an example equation. If you plug your own equation in as you read, you'll be able to solve it and future questions regarding the linear combination method.
This isn't to do with graphs or geometry. @kxttxn posted an excellent link on it so give it a go and we can check your working.
Please just help me by teaching me or something, I am failing algebra by 1%
That link tells you exactly what to do, just read it. When you've read it, you'll be able to do this question and we can check what you've done is correct. If there's anything specific you get stuck on then feel free to ask.
please help me on this, I only have 15 minutes to do it!
Have you looked at the site at all?
Alright, have you tried working the equation out how the site shows you to do it?
yea but I can't figure it out for the life of me
Well what work have you got so far?
I think D, but i could be VERY wrong
Let me check the answer real quick and I'll get back with you
Hurry please, my time is running down quickly
Ok just solved it and you got the correct answer (D), great job! :)
wow.. shot in the dark.. haha
Ah I was hoping you had solved through it. Definitely refer back to that website in the future though, it's helped me a lot myself!
Just gonna post a solution here so you know what to do in the future: \[2x + 3y = - 17\]\[5x + 2y = -4\]We need to have the same number of x's or y's so we can cancel them by subtraction. We could also do it by addition if one was the negative of the other. I'm going to try and get the y's equal because the numbers are lower, and we can see that the lowest common multiple of 3 and 2 is 6 so we need to multiply the first equation by 2 and the second by 3: \[4x+6y=-34\]\[15x+6y=-12\]Now we have an equal number of y's, we can subtract one equation from the other. I will subtract the first from the second because that will leave a positive number of x's:\[(15x+6y)-(4x+6y)=-12-(-34)\]\[11x=22\]\[x=2\]Now we substitute this into an earlier equation, say \( 2x+3y=17\) to get \(4+3y=-17\) so \(y=-7\).