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Solve the system by substitution

2x-4y=8

-3x=9y=-18

Thanks for your help

Since we have 2x - 4y = 8, we can add 4y to both sides. This gives 2x = 8 + 4y.

Dividing the entire equation by 2 gives x = 4 + 2y.

I believe the 2nd equation is suppose to really be -3x - 9y = -18.

Multiplying the entire equation by -1 gives 3x + 9y = 18.

Since x = 4 + 2y, we can multiply by 3 and get 3x = 12 + 6y.

Putting 3x = 12 + 6y into 3x + 9y = 18 gives 12 + 6y + 9y = 18.

Since 6+9 is 15, we have a 15y on the left side.

Subtracting 12 from both sides leaves a 6 on the right.

This gives 15y = 6.

Dividing both sides by 15 gives y = 6/15 = 2/5.

Putting that in the 1st equation gives 2x - 4(2/5) = 8.

Since 4(2/5) = 8/5, and 8 = 40/5, we can add 8/5 to both sides.

This gives 2x = 48/5.

Dividing by 2 on both sides gives x = 24/5.

Putting x = 24/5 and y = 2/5 into -3x - 9y gives -72/5 - 18/5 = -90/5.

Dividing -90 by 5 gives -18, which is what's there, so that has been done correctly.

Algebra

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