What is the value of y in the solution to the following system of equations?
5x – 3y = –3
2x – 6y = –6
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OpenStudy (anonymous):
Multiply first equation with 2 and second equation with 5 and then subtract them if you really want the value of y..
OpenStudy (anonymous):
Do you multiply everything ?
OpenStudy (anonymous):
@waterineyes
OpenStudy (anonymous):
See first one I solve for you:
I said multiply first equation by 2,
2(5x - 3y = 3)
10x - 6y = 6
Now do the same for second equation but multiply it with 5..
OpenStudy (anonymous):
Sorry gotta go...
@lgbasallote please help @Caramel_823 I have to go now...
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OpenStudy (lgbasallote):
so what's left here?
OpenStudy (anonymous):
5(2x-6y)=-6
10x-30y=-30
OpenStudy (lgbasallote):
wait...the system is
5x - 3y = -3
2x - 6y = -6
right?
OpenStudy (anonymous):
Yes so after she multiplied everything how come her last number stayed positive ?
OpenStudy (lgbasallote):
can we start from beginning?
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OpenStudy (lgbasallote):
which wouldyou like to cancel? x or y?
OpenStudy (anonymous):
x because I'm trying to find y
OpenStudy (lgbasallote):
so that was why you multiplied by 5...okay let's do that
5x - 3y = -3
10x - 30y = -30
now what would you multiply the first equation so you can cancel out x?
OpenStudy (anonymous):
I think 2 would cancel it out
OpenStudy (lgbasallote):
okay let's try
10x - 6y = -6
10x -30y = -30
so how can we cancel x? what operation do we use?
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OpenStudy (anonymous):
multiplication
OpenStudy (lgbasallote):
hmm if you multiply the two equations you'll get a VERY BIG number haha
OpenStudy (lgbasallote):
and the choices are addition or subtraction
OpenStudy (anonymous):
subtraction
OpenStudy (lgbasallote):
you can only use those operation in elimination method remember?
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