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To solve this system using the addition method, you would need to multiply the first equation by what number in order for the y's to add out? 3x - y = 3 -2x + 2y = 6

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well you need the same coefficient for y if you want them to cancel. in the first one the coefficient of y is -1. and in the second it's 2. so -1 x ? = 2
-1 x -1 = 2 ?
You would the first equation by 2, getting 6x -2y = 6. when added to second equation the y's would "add out" or their sum is 0.

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I left out the word multiply, which goes in the first sentence between " would * the"
so what would you multiply the eqaution by?
@hero. Don't just post the answer next time unless you absolutely need to...
I thought I was checking someone's answer. @Falkqwer misled me
Yes he did :)
lol I did
As I have previously posted, if you want to eliminate the y's, multiply the first equation by 2. this gives you: 6x -2y = 6 Now add the 2nd equation -2x+2y =6 --------------- 4x = 12 x=3
Now you can solve for y.
There can only be one right answer
@radar is right cause your supposed to Solve it using the addition method.
Substituting for x in the 2nd equation getting -2(3) + 2y = 6 -6 +2y = 6 2y= 12 y=6 x=3, y=6 To be sure check them out
I found my mistake. Wha ta shame
Do we need to verify? Did you get (3,6) @Hero
@radar, what is it about "I found my MISTAKE" did you not understand?
It never hurts to verify substituting 3 for x, 6 for y 3x-y=3 or 3(3)-6=3 9-6=3 3=3 first one checks! -2x+2y=6 -2(3)+2(6)=6 -6 +12 = 6 6=6 checks
No need to persist. Be happy that you provided the right solution and move on.
@Hero, just wanted to verify my answer, so I was simply asking if you got the same (after your error discovery.
When I said I found my mistake, that implies that I acknowledge your correct answer. Furthermore I gave you a medal. As I said, no need to persist beyond that.
Thanks @Hero

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