SOLUTION: solve by the elimination method 7r-4s=33 4r+7s=56 I have tried setting it up like this 7r-4s=33 4r+7s=56 -------- 3r 3s=89 I don't know if I am supposed to put a + or

Question 259755: solve by the elimination method 7r-4s=33 4r+7s=56
I have tried setting it up like this
7r-4s=33
4r+7s=56
--------
3r 3s=89
I don't know if I am supposed to put a + or - between the 2 and to eliminate the variables I divide by 3 and get r,s = 29.6
this does not look right. Any help on how to solve would be appreciated. Thanks
Found 2 solutions by richwmiller, unlockmath:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
You are right it doesn't look right!
r=7 s=4
7r-4s=33
4r+7s=56
--------
3r 3s=89
your addition needs work.
7+4=11 not 3
but worse your approach is wrong
It is too early to add or subtract.
you have to get the same coefficient for x or y so that when you add or subtract either the x or the y disappear or are eliminated .
multiply the first by 7 and the second by 4
then we will add them to get rid of y's
7*(7r-4s=33)
4*(4r+7s=56)
49r-28s=7*33
16r+28s=4*56
add
65r=7*33+4*56
65r=455
r=7

Answer by unlockmath(1688) (Show Source): You can put this solution on YOUR website!
Hello,
First off we want to see if we can eliminate one of the variables.
7r-4s=33
4r+7s=56
Let's go for getting rid of "s". To do this we need to multiply the first equation by 7 and the second equation by 4 so it would turn out like this:
49r-28s=231
16r+28s=224
Now we can add both of these to be:
65r=455 Divide each side by 65 to result in:
r=7
Now we can plug 7 into either original equation to find "s".
49-4s=33 Subtract 49 from both sides to be:
-4s=-16 Now divide -4 into both sides to give us:
s=4
You can check to see if this is right by plugging s=4 and r=7 into both equations.
Make sense?
RJ
Check out a new book I wrote at:
www.math-unlock.com
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