Do I have the Right solution?

[SCSmith]

4/5(10x + 15) - 20 = 2x + 3/4(8x + 4)
40/5x + 12 - 20 = 2x + 6x + 3
8x + 12 - 20 = 8x + 3
8x - 8 = 8x + 3
x = 11

When I substitute 11 for x I get the wrong solution. Am I doing something wrong?

 

[soroban]

Heloo, SCSmith!

(4/5)(10x + 15) - 20 = 2x + (3/4)(8x + 4)
8x + 12 - 20 = 2x + 6x + 3
8x - 8 = 8x + 3
x = 11 . <--- here!

You had: .8x - 8 .= .8x + 3

Subtract 8x from both sides: .-8 .= .3 . ?

We have a false statement. .The equation has no solution.

 

[Denis]

4/5(10x + 15) - 20 = 2x + 3/4(8x + 4)
40/5x + 12 - 20 = 2x + 6x + 3
8x + 12 - 20 = 8x + 3
8x - 8 = 8x + 3
x = 11

Perhaps the 2x on the right is 2 (typo) ?