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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
Theres something wrong with that.Keep in mind OPPOSITES CANCEL.
15+15= 30 we want to cancel the x value
-5(3x+5y=6) -15x-25y=-30 same here -15+-15=-30
-15x-25y=-30 15x+21y=30 4y=0
what did u do?
I have no clue what you doing. I am doing substitution method. The x is not suppose to just disappear like you have.
-5(3x+5y=6)<---wrong -15x-25y=-30 5(3x+5y=6) 15+25y=30 15+25y=30 -15x-25y=-30
the mistake is in the second eq: \[3(5x+7y=6)\] you get: \[15x+21y=18\] because 3*6=18, not 30 !!
They made a mistake on both.
There shouldnt even be 2 sets of equations
@Destinyyyy 15+25y=30 -15x-25y=-30 solve this
@Greg_D I see that now thank you.. Cannot believe I did that -.-
Im gonna let them finish up.
But there should not be 2 sets.
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
I have no clue what your talking about @pooja195
I know the final answer is (-3,3) ... But im not sure how they get -3
@pooja195 there are a lot of ways to solve the system, the one @Destinyyyy is using is fine, there was just that little mistake
once you get y=3, just replace that in any of the original equations...
Im only allowed to use elimination, graph, substitution.
Thank you @Greg_D
so you arent allow to use "any" method
-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
Its any method of the three. I should of said that
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
Um the 3 is suppose to go into one of the original equation.
I know how to solve it. I just didnt see where I messed up. @Greg_D showed me my mistake.
look if: \[y=3\] and: \[3x+5y=6\] then: \[3x+15=6\] you can get x from there...
the same is done in the last part of @pooja195 last response :)