Math Question!!!?

1/2 (x - 3) + (3/2 - x) = 5x
(x/2 - 3/2) + (3/2 - x) = 5x
(x/2 - x) + (3/2 - 3/2) = 5x
-x/2 = 5x

The only way for this to be true is for x=0.

Answer: x = 0
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½(x - 3) + (3/2 - x) = 5x . . . First subtract 5x

½(x - 3) + (3/2 - x) -5x = 0 . . .expand ½(x - 3)

x/2 - 3/2 + 3/2 - x - 5x = 0 . . .Combine like terms

x/2 - 6x = 0 . . .multiply term '6x' by 2/2 for a common denominator

x/2 - 12x/2 = 0 . . .Combine

11x/2 = 0

X must equal 0 in order for this equation to be true.
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Check:

½(x - 3) + (3/2 - x) = 5x
½(0 - 3) + (3/2 - 0) = 5(0)
-3/2 + 3/2 = 0

x=0 no other way because -3/2 and +3/2 cancelled each other

(1/2)(x - 3) + (3/2 - x) = 5x
(x - 3)/2 + 3/2 - x = 5x
2[(x - 3)/2 + 3/2 - x] = 2(5x)
x - 3 + 3 - 2x = 10x
x - 2x - 10x = 3 - 3
-11x = 0
x = 0/-11
x = 0

½(x - 3) + (3/2 - x) = 5x
½x - 3/2 + 3/2 - x = 5x
½x - x = 5x
½x = 6x
x = 12x
0 = 11x
x = 0
unless you've made an error with your parentheses.

1/2 (x - 3) + (3/2 - x) = 5x
-11/2x=0
x=0

This one isn't really solvable per se. I came out with 1/2 x = 5 x, in which case the only solution is zero. Your numbers cancel, so there is only a solution of Zero.

pinksicl, is it (1/2)(x-3) or 1[2(x-3)]?