Question 628345
problem: how do i solve a problem with a number outside the absolute value lines?
<pre>

{{{1/2}}}|6 - 2x|= 3x + 1

Clear the fraction by multiplying both sides by 2

  |6 - 2x| = 6x + 2

There are two cases to consider:

Case 1:  When the value of 6 - 2x has the same value as the right side.

  6 - 2x = 6x + 2
     -8x = -4
       x = {{{1/2}}}

Case 2:  When the value of 6 - 2x has -1 times the value of the right side.

  6 - 2x = -(6x + 2)
  6 - 2x = -6x - 2
      4x = -8
       x = -2

There appear to be two solutions {{{1/2}}} and -2. 

However we must check them to see that they satisfy the
original equation.

Checking {{{1/2}}}

{{{1/2}}}|6 - 2x|= 3x + 1

{{{1/2}}}|6 - 2({{{1/2}}})|= 3({{{1/2}}}) + 1

{{{1/2}}}|6 - 1|= 3({{{1/2}}}) + 1

{{{1/2}}}|5| = ({{{3/2}}}) + 1

{{{1/2}}}(5) = {{{3/2}}} + {{{2/2}}}

{{{5/2}}} = {{{5/2}}}

That checks, so {{{1/2}}} is a solution.

Checking -2

{{{1/2}}}|6 - 2x|= 3x + 1

{{{1/2}}}|6 - 2(-2)|= 3(-2) + 1

{{{1/2}}}|6 + 4|= -6 + 1

{{{1/2}}}|10| = -5

{{{1/2}}}(10) = -5

   5 = -5

So -2 is not a solutionh and we discard it.

The only solution is {{{1/2}}}

Edwin</pre>