SOLUTION: I have tried and tried to get this right and when I check the answer and am not even near been correct: Problem: 4x + 3y = 59 2x + 3y = 43 added together you get: 6x + 6y = 102

Algebra ->  Systems-of-equations -> SOLUTION: I have tried and tried to get this right and when I check the answer and am not even near been correct: Problem: 4x + 3y = 59 2x + 3y = 43 added together you get: 6x + 6y = 102      Log On


   



Question 7687: I have tried and tried to get this right and when I check the answer and am not even near been correct:
Problem:
4x + 3y = 59
2x + 3y = 43 added together you get:
6x + 6y = 102
6x/6 = 102/6
x = 17
I know this is not correct because when substitute of x I am not correct? See
4(17) + 3y = 59
98 + 3y =59
HELP! I know that it is simple but, I keep doing it wrong. I had 75 problems, and the problems without negative numbers seem to mess me up.
Thanks

Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
So we start with
4x + 3y = 59
2x + 3y = 43
What we're going to do is eliminate one of the variables. By the looks of it, you've got a 3y on both equations. That's very nice of them because it makes it easier.

We're actually going to use a trick here. We're going to multiply the second equation by -1. BTW, doing this does NOT alter the equation. It just changes its appearance. So now you've got:

4x + 3y = 59
-2x - 3y = -43

Now calculate downwards like you did. You should get
2x = 16 <------ The 3y's cancel to 0.
x = 8.

Now to get the y, you can plug in the x = 8 in any of the two equations. You WILL get the same y-value. Let's just pick the first one:

4(8) + 3y = 59 ---> 32 + 3y = 59 ---> 3y = 27 ---> y = 9.

Let's try to do the second one just to see:

2(8) + 3y = 43 ---> 16 + 3y = 43 ---> 3y = 27 ---> y = 9. AHA! Same answer!
----------------------------------------------------------------------
What happens, then, if all of the coefficients are different? What do you do? Stay tuned.