Solve for x
2(3x-4)=4(1/2x-8)

x = -6

Step-by-step explanation:

Solve for x:

2 (3 x - 4) = 4 (x/2 - 8)

Put each term in x/2 - 8 over the common denominator 2: x/2 - 8 = x/2 - 16/2:

2 (3 x - 4) = 4 x/2 - 16/2

x/2 - 16/2 = (x - 16)/2:

2 (3 x - 4) = 4 (x - 16)/2

4/2 = (2×2)/2 = 2:

2 (3 x - 4) = 2 (x - 16)

Divide both sides by 2:

3 x - 4 = x - 16

Subtract x from both sides:

(3 x - x) - 4 = (x - x) - 16

3 x - x = 2 x:

2 x - 4 = (x - x) - 16

x - x = 0:

2 x - 4 = -16

Add 4 to both sides:

2 x + (4 - 4) = 4 - 16

4 - 4 = 0:

2 x = 4 - 16

4 - 16 = -12:

2 x = -12

Divide both sides of 2 x = -12 by 2:

(2 x)/2 = (-12)/2

2/2 = 1:

x = (-12)/2

The gcd of -12 and 2 is 2, so (-12)/2 = (2 (-6))/(2×1) = 2/2×-6 = -6:

x = -6

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.