7 l 3x-4 l + 8 = 15

First we ill subtract 8 from both sides:

==> 7 l 3x-4 l = 7

No divide by 7:

== l 3x -4 l = 1

Now we need to determine x, such that the absolute value is 1.

Then, we have two possible cases:

3x-4 = 1 OR -(3x-4) = 1

Let us solve both:

3x-4 = 1

==> 3x= 5

==> x1= 5/3

-(3x-4) = 1

==> -3x + 4 = 1

==> -3x = -3

==: x2= 1

Then the soluion is:

**x= { 1, 5/3}**

To solve 7|x-4| +8 = 15.

We subtract 8 from both sides of the equation:

7|x-4| = 15-8 = 7

7|x-4| = 7.

We divide both sides of the eqution by 7:

|x-4) = 1

There are 2 cases .

i)When x-4 > 0,

|x-4| = x-4 = 1.

x = 1+4 = 5.

ii) when x -4 < 0,

|x-4| = 4-X = 1

4 = 1+x

4-1 = x

3 = x.

Therefore the solutions are x = 5 or x = 3.

We'll subtract 8 both sides:

7 l 3x-4 l = 15 - 8

7 l 3x-4 l = 7

We'll divide by 7:

l 3x-4 l = 1

We'll have 2 cases:

1) 3x-4 for 3x-4 >=0

We'll add 4 both sides:

3x>=4

We'll divide by 3:

x>=4/3

The interval of admissible values for x is [4/3 , +infinite)

We'll solve the equation:

3x-4 = 1

We'll add 4 both sides:

3x = 5

x = 5/3

Since x = 5/3 belongs to the interval [4/3 , +infinite), the equation has the solution x = 5/3.

2) -3x+4 for 3x-4 <0

The solution of the equation has to belong to the interval

(-infinite, 4/3)

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