Intermediate Algebra/Solving Absolute Value Equations

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Absolute Values[edit]

Absolute Values represented using two |'s are common in Algebra. They are meant to signify the number's distance from 0 on a number line. So, if the number is negative, it becomes positive. And if the number was positive, it remains positive:

|4| = 4 \,
|-4| = 4 \,

For a formal definition:

If x \ge 0, then |x| = x
If x < 0, then |x| = -x \,

Please note that the opposite (-) of a negative number is positive.


Practice Problems[edit]

For all of these problems, a = -2 and b = 3. Evaluate the following expressions.

1. |a|

2. |b|

3. |b + a|

Answers[edit]

1. 2

2. 3

3. 1

Absolute Value Equations[edit]

Now, let's say that we're given the equation |k| = 8 and we are asked to solve for k. Well, what number would work if you plugged it in for k? 8 would work, but wouldn't -8 also work? That's why there can be two solutions to one equation (and later, even more solutions). So, let's look at this equation:

|2k + 6| = 8

Knowing that what is in the absolute value bars must equal 8 or -8, we can make two separate equations for this problem:

2k + 6 = 8 OR 2k + 6 = -8

Using our knowledge of solving equations, we would discover that k = 1, -7. Now, let's say we have 3|2k + 6| = 12. First of all, we'd have to divide both sides by 3 to get the absolute value by itself. Then, we'd set up the two different equations, 2k + 6 = 4 OR 2k + 6 = -4. Then, we'd solve,by subtracting the 6 from both sides and dividing both sides by 2 to get the k by itself, resulting in k = -1, -5.

What if the equation was -4|2k + 6| = 8? Well, we'd first divide like the previous problem, so the equation would look like this: |2k + 6| = -2. Do you notice anything strange? When you evaluate an absolute value, you always get a positive number, so this is No Solution.


Practice Problems[edit]

1. |k + 6| = 2k

2. |7 + 3a| = 11 - a

3. |2k + 6| + 6 = 0

Answers[edit]

1. \begin{matrix}|k + 6| & = & 2k \\ \ k + 6 & = & 2k \qquad k + 6 & = & -2k \\ \ 6 & = & k \qquad 6 & = & -3k \\ \ 6 & = & k \qquad -2 & = & k\end{matrix}

However, if we plug in -2 we get a negative number on the right side, which is impossible, so -2 does not work. However, 6 does. So, k = 6.


2. \begin{matrix}|7 + 3a| & = & 11 - a \\ \ 7 + 3a & = & 11 - a \qquad 7 + 3a & = & -11 + a \\ \ 7 + 4a & = & 11 \qquad 7 + 2a & = & -11 \\ \ 4a & = & 4 \qquad 2a & = & -18 \\ \ a & = & 1 \qquad a & = & -9\end{matrix}


3. \begin{matrix}|2k + 6| + 6 & = & 0 \\ \ |2k + 6| & = & -6 \\ \ No Solution\end{matrix} Notice that we get a negative number on the right side, which is impossible it becomes 5 because the negative becomes a positive so 3+2=5.

Lesson Review[edit]

An absolute value (represented with |'s) stands for the number inside's distance from 0 on the number line. Basically, it makes a negative positive and a positive remain positive. To solve an equation involving absolute values, you must get the absolute value by itself on one side and set it equal to the positive and negative version of the other side, because those are the two solutions the absolute value can output. However, check the solutions you get in the end; some might produce negative numbers on the right side, which are impossible because all outputs of an absolute value symbol are positive!

Lesson Quiz[edit]

Evaluate.

1. |-4|

2. |6-8|

Solve for a.

1. |3a - 4| = 5

2. 5|2a + 3| = 15

3. 3|4a - 2| - 12 = -3


Quiz Answers*!

1. 4

2. 2

1. a={3,-1/3}

2. a={0,-3}

3. a={5/4,-1/4}