Destinyyyy
  • Destinyyyy
Can someone tell me what I did wrong? If possible, solve the system of equations. Use any method.
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Destinyyyy
  • 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.

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More answers

Destinyyyy
  • 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
Destinyyyy
  • Destinyyyy
-15x-25y=-30 15x+21y=30 4y=0
pooja195
  • pooja195
what did u do?
Destinyyyy
  • 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?
anonymous
  • 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
Destinyyyy
  • 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.
anonymous
  • 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
Destinyyyy
  • Destinyyyy
I have no clue what your talking about @pooja195
Destinyyyy
  • Destinyyyy
I know the final answer is (-3,3) ... But im not sure how they get -3
anonymous
  • 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
anonymous
  • anonymous
once you get y=3, just replace that in any of the original equations...
Destinyyyy
  • Destinyyyy
Im only allowed to use elimination, graph, substitution.
Destinyyyy
  • Destinyyyy
Thank you @Greg_D
anonymous
  • 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
Destinyyyy
  • 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
Destinyyyy
  • Destinyyyy
Um the 3 is suppose to go into one of the original equation.
Destinyyyy
  • Destinyyyy
I know how to solve it. I just didnt see where I messed up. @Greg_D showed me my mistake.
anonymous
  • anonymous
look if: \[y=3\] and: \[3x+5y=6\] then: \[3x+15=6\] you can get x from there...
anonymous
  • anonymous
the same is done in the last part of @pooja195 last response :)

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