Manuela solved the equation 3−2|0.5x+1.5|=2 for one solution. Her work is shown below. 3−2|0.5x+1.5|=2 −2|0.5x+1.5|=−1 |0.5x+1.5|=0.5 0.5x+1.5=0.5 0.5x=−1 x=−2 What is the other solution to the equation? x=−6 x=−4 x=2 x=4

We want to find the other solution of the given absolute value equation.

The other solution is x = -4

We know that:

3 - 2*|0.5*x + 1.5| = 2

It has one solution given by:

- 2*|0.5*x + 1.5| = 2 - 3 = -1

|0.5*x + 1.5| = 0.5

0.5*x + 1.5 = 0.5

0.5*x = 0.5 - 1.5 = -1

0.5 = -1/x

Then we have x = -2

To get the other solution we need to remember that an absolute value equation can be written as:

|x - a| = b

or:

(x - a) = b

(x - a) = -b

Then the other solution to our equation comes from:

|0.5*x + 1.5| = 0.5

(0.5*x + 1.5) = -0.5

0.5*x = -0.5 - 1.5 = -2

x = -2/0.5 = -4

The other solution is x = -4

If you want to learn more, you can read:

link

Step-by-step explanation:

We'll just work on solving both so you can see what's involved in solving an absolute value equation. Because an absolute value is a distance, we can have that distance being both to the right on the number line of the number in question or to the left. For example, from 2 on the number line, the numbers that are 5 units away are 7 and -3. Using that logic, we will simplify the equation down so we can set up the 2 basic equations needed to solve for x.

If   then

 What you need to remember here is that you cannot distribute into a set of absolute values like you would a set of parenthesis. The -2 needs to be divided away:

Now we can set up the 2 main equations for this which are

.5x + 1.5 = .5  and .5x + 1.5 = -.5

Knowing that an absolute value will never equal a negative number (because absolute values are distances and distances will NEVER be negative), once we remove the absolute value signs we can in fact state that the expression on the left can be equal to a negative number on the right, like in the second equation above.

Solving the first one:

.5x = -1 so

x = -2

Solving the second one:

.5x = -2 so

x = -4

no this is a nonlinear negative