Please help me solve this problem using elimination method, Thank you

2x+3y=34
25x+30y=350

By inspection, 3 and 30 and the least common multiple is 30, so you can multiply the first equation by 10 and subtract the two equations.

Any help?

We aren't here to do your homework for you, so I'll show you an example.




The idea is to get a positive and negative of the same number on one of the variables. It doesn't matter if it's the x's or the y's. Use the one that is easiest to do. You need a common multiple. (It's like finding a common denominator.)

If I wanted the x's to be the same number, that would be 12 (multiple of both 3 and 4), which can be done but is more work. But it would be easy to get the y's as the same number because I can use 2 and that's easy to do. Since the 2 in the top equation is positive, I want the 2 in the bottom equation to be -2. (Don't worry -- if you keep going it should make sense why.)

I get it to be -2 by multiplying the y in the bottom equation by -2. But we can't just randomly multiply one side by something like that, so I have to multiply both sides through by that -2:


Distribute:


Now I've got an equation that looks different, but is still equivalent, and has that -2 on the y that I need. So I will now list both equations together again, with the bottom one changed as shown above. Hopefully at this point you can see a little more clearly what we were trying to accomplish. Note that I have a positive and negative version of the same number on those y's.




I then add them together.



Notice the point is that the 2 and -2 wipe each other out, so that I have nothing on that y. So I can simply eliminate that y. (I assume that's why they call it elimination. :D)



Divide both sides by 11 and you have:



Then plug that answer back into either one of the equations to solve for the other variable. Doesn't hurt to plug it into both and make sure you're getting the same answer, since that's the point. (And yes, I did do that to make sure I didn't screw up.)

Your numbers are bigger, but your equations also already have one variable that has a coefficient that's a nice, neat multiple of the other. And yours doesn't end up in fractions.