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5x+4y=-30 and 3x-9y=-18 using solving systems of elimination

Let 5x+4y=-30 be ( 1 ) and 3x-9y=-18 be ( 2 ).

Solution:

Multiply ( 1 ) by 3 to have ( 3 ) and multiply ( 2 ) by 5 to have ( 4 ). Then substract ( 3 ) by ( 4 )  to eliminate x, that is:

3 ( 5x+4y=-30 ) ( 1 )    = >  15x + 12y = -90 ( 3 )

5 ( 3x-9y=-18 ) ( 2 )     = >  15x - 45y = -90  ( 4 )

57y = 0

y = 0, x = - 6

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5x+4y=-30 3x-9y=-18
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