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