Verified answer

You need to define ranges for x in which it would be solved.

The ranges would be based on values of x where (5-2x) = 0 and (x-4) = 0

a) for x < 2.5,

(x-4) <0 which means | x-4 | = -(x-4) = 4-x

(5-2x) >0 which means| 5-2x | = (5-2x)

Thus, in this range we solve:

4-x = 5-2x

=> x = 1

b) for 2.5 < x < 4,

(x-4) < 0 which means | x-4 | = -(x-4) = 4-x

(5-2x) < 0 which means | 5-2x | = -(5-2x) = 2x -5

Thus, in this range we solve:

4-x = 2x - 5

=> 9 = 3x

=> x = 3

c) for x > 4

(x-4) > 0 which means | x-4 | = (x-4)

(5-2x) < 0 which means | 5-2x | = -(5-2x) = 2x -5

Thus, in this range we solve:

x-4 = 2x - 5

=> x = 1

But since, the range is for x >4, this solution is invalid.

So, the answers are: x = 1 and x = 3

x=5 -4

5-4=1

2x=2*2=4

5-4=1

Here are a couple of good math web sites

math forum. Ask Dr. Math

Math.com Math Practice

since in absolute value , a # inside the | | cannot be negitive you need to take the absolute value out and make everything positive

so your new equation would be : x+4=5+2x

then you would subtract the four from both sides of the equation

x=1+2x

then subtract the 2x

-x=1

divide by -1 to get

x=1

hope it helped

just solve it all four ways

x - 4 = 5 - 2x

x - 4 = - 5 + 2x

-x + 4 = 5 - 2x

-x + 4 = - 5 + 2x

-3x - 2x^2 - 20

if it isnt that im gonna mess up my GCSE's sooo badly