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anonymous
 one year ago
3x6y=12
2x+6y=12
Solve by using elimination
anonymous
 one year ago
3x6y=12 2x+6y=12 Solve by using elimination

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what are your thoughts?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have no idea im so confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i forgot how to do it

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.0Pandabear78693  so you were taught how to do these?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1elimination involves adding the equations together so that one of the terms zeros out ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i was taught to do this but i forgot how to do it @wolf1728

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1do the equations have a similar term to them that will cancel when added up?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03x6y=12 + 2x+6y=12 _________ 5x=0 x=0 (Since the y's already canceled each other, I didn't have to multiply anything.) Now plug the xvalue in both equation and see if it works: 3(0)6y=12 6y=12 y=2 2(0)+6y=12 6y=12 y=2 So the answer is (0,2) I think Hope this helps!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i will add 3x and 2x

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1you will add all like terms, the idea is to get one of the terms to cancel itself out i the process.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1recall that aa = 0; known as the inverse of addition property or some such

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for helping @amistre64 and @Peaches15 btw ur welcome

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you thanks for explaining it @amistre64
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