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Algebra 7 Online
OpenStudy (destinyyyy):

Can someone tell me what I did wrong? If possible, solve the system of equations. Use any method.

OpenStudy (destinyyyy):

3x+5y=6 (1) 5x+7y=6 (2) -5(3x+5y=6) -15x-25y=-30 3(5x+7y=6) 15x+21y=30 -4y=0 y=0

pooja195 (pooja195):

3(5x+7y=6) 15x+21y=30

pooja195 (pooja195):

Theres something wrong with that.Keep in mind OPPOSITES CANCEL.

OpenStudy (destinyyyy):

What? O.o

pooja195 (pooja195):

15+15= 30 we want to cancel the x value

pooja195 (pooja195):

-5(3x+5y=6) -15x-25y=-30 same here -15+-15=-30

OpenStudy (destinyyyy):

-15x-25y=-30 15x+21y=30 4y=0

pooja195 (pooja195):

what did u do?

OpenStudy (destinyyyy):

I have no clue what you doing. I am doing substitution method. The x is not suppose to just disappear like you have.

pooja195 (pooja195):

-5(3x+5y=6)<---wrong -15x-25y=-30 5(3x+5y=6) 15+25y=30 15+25y=30 -15x-25y=-30

pooja195 (pooja195):

Get it?

OpenStudy (anonymous):

the mistake is in the second eq: \[3(5x+7y=6)\] you get: \[15x+21y=18\] because 3*6=18, not 30 !!

pooja195 (pooja195):

They made a mistake on both.

pooja195 (pooja195):

There shouldnt even be 2 sets of equations

pooja195 (pooja195):

@Destinyyyy 15+25y=30 -15x-25y=-30 solve this

OpenStudy (destinyyyy):

@Greg_D I see that now thank you.. Cannot believe I did that -.-

pooja195 (pooja195):

Im gonna let them finish up.

pooja195 (pooja195):

But there should not be 2 sets.

OpenStudy (anonymous):

it was a system of 2 eqs from the begining of the problem... the only mistake i can see is the one i pointed out earlier 3*6 is not 30

OpenStudy (destinyyyy):

I have no clue what your talking about @pooja195

OpenStudy (destinyyyy):

I know the final answer is (-3,3) ... But im not sure how they get -3

OpenStudy (anonymous):

@pooja195 there are a lot of ways to solve the system, the one @Destinyyyy is using is fine, there was just that little mistake

OpenStudy (anonymous):

once you get y=3, just replace that in any of the original equations...

OpenStudy (destinyyyy):

Im only allowed to use elimination, graph, substitution.

OpenStudy (destinyyyy):

Thank you @Greg_D

OpenStudy (anonymous):

so you arent allow to use "any" method

pooja195 (pooja195):

-5 (3x+5y=6) 3(5x+7y=6 ) -15x-25y=-30 15x+21y=18 -4y=-12 divde both sides by -4 y=3 15x+21(3)=18 15x+63=18 subtract 63 both sides 15x=-45 divde both sides by 15 x=-3

OpenStudy (destinyyyy):

Its any method of the three. I should of said that

pooja195 (pooja195):

notice how there is not 2 sets of equations in no way shape or for should u have 2 sets of equations even when doing elimination

OpenStudy (destinyyyy):

Um the 3 is suppose to go into one of the original equation.

OpenStudy (destinyyyy):

I know how to solve it. I just didnt see where I messed up. @Greg_D showed me my mistake.

OpenStudy (anonymous):

look if: \[y=3\] and: \[3x+5y=6\] then: \[3x+15=6\] you can get x from there...

OpenStudy (anonymous):

the same is done in the last part of @pooja195 last response :)

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