2.4 – Equations with Absolute Values

Equations with Absolute Values

The absolute value of a number is its distance from zero. The absolute value of a number is always a positive number. We notate absolute value by using two vertical bars around the number (or expression). For example, the absolute value of 5 can be written $\left|5\right|$ .

The absolute value of 7, written $\left|7\right|$ is 7, since the number 7 is 7 units away from zero.

The absolute value of $-7$ written $\left|-7\right|$ is also 7, since the number $-7$ is also 7 units away from zero. The absolute value of a number disregards the direction it is from zero, and reports only the distance.

Find the absolute value of 10
$\left\|10\right\|=10$ Because 10 is 10 units away from zero.

Find the absolute value of negative 10
$\left\|-10\right\|=10$ Because -10 is 10 units away from zero.

The preceding examples may seem a bit silly, but absolute values have greater consequence once an unknown value, or an expression is placed inside:

Find the values of x that make the equation true:
$\left\|x\right\|=10$ We saw above that $\left\|10\right\|=10$ But we also saw above that $\left\|-10\right\|=10$ So then the equation has two solutions, $x$ can be either $10,$ or $-10$

Find the values of x that make the equation true:

$\left|x\right|=10$

We saw above that

$\left|10\right|=10$
But we also saw above that
$\left|-10\right|=10$
So then the equation has two solutions, $x$ can be either $10,$ or $-10$

We see from the above that when there is a variable in the absolute value, there are two possible answers that might make it true. Whenever there is an expression inside an absolute value equation, there are potentially two values that will make the equation true. We will use this in the next problem involving an expression inside the absolute value.

Solve for x:
$\|x+3\|=4$ Note from above, that whatever is inside the absolute values can either be equal to 4, or to -4. We then set up two equations, noting the two possible scenarios: $x+3=4 \hspace{1cm} or \hspace{1cm} x+3=-4$ Solving each equation separately gives $x=1 \hspace{1cm} or \hspace{1cm} x=-7$

Solve for x:

$|x+3|=4$

Note from above, that whatever is inside the absolute values can either be equal to 4, or to -4. We then set up two equations, noting the two possible scenarios:

$x+3=4 \hspace{1cm} or \hspace{1cm} x+3=-4$
Solving each equation separately gives
$x=1 \hspace{1cm} or \hspace{1cm} x=-7$

When there are other terms on the same side as the absolute value, it is important to eliminate those terms and isolated the absolute value before creating the two equations.

Solve for $x$ :
$3+\|x+4\| = 10$ To isolate the absolute value, we will subtract 3 from each side: $\|x+4\|=7$ Now we can set the expression inside the absolute value equal to the positive and negative values: $x+4 = 7 \hspace{1cm} or \hspace{1cm} x+4=-7$ and solve each: $x=3 \hspace{1cm} or \hspace{1cm} x=-11$ So then $x=\{-11,3\}$

Solve for $x$ :

$3+|x+4| = 10$
To isolate the absolute value, we will subtract 3 from each side:
$|x+4|=7$
Now we can set the expression inside the absolute value equal to the positive and negative values:
$x+4 = 7 \hspace{1cm} or \hspace{1cm} x+4=-7$
and solve each:
$x=3 \hspace{1cm} or \hspace{1cm} x=-11$

So then $x=\{-11,3\}$

Solve for $x$ :
$-6\|x-2\|=-24$ In this problem, the $-6$ is not part of the absolute value, it is simply multiplying it. We can eliminate the $-6$ by dividing both sides by it: $\frac{-6\|x-2\|}{\color{red}-6}=\frac{-24}{\color{red}-6}$ $\|x-2\|=4$ And can now break the expression out of the absolute value by setting it equal to 4 or $-4$ : $x-2=4 \hspace{1cm} or \hspace{1cm} x-2=-4$ And solving: $x=6 \hspace{1cm} or \hspace{1cm} x=-2$ Then $x=\{-2,6\}$

Solve for $x$ :

$-6|x-2|=-24$

In this problem, the $-6$ is not part of the absolute value, it is simply multiplying it. We can eliminate the $-6$ by dividing both sides by it:

$\frac{-6|x-2|}{\color{red}-6}=\frac{-24}{\color{red}-6}$

$|x-2|=4$
And can now break the expression out of the absolute value by setting it equal to 4 or $-4$ :
$x-2=4 \hspace{1cm} or \hspace{1cm} x-2=-4$
And solving:
$x=6 \hspace{1cm} or \hspace{1cm} x=-2$

Then $x=\{-2,6\}$

Not all absolute value equations have solutions. For example:

Solve for $x$
$\|x\|=-2$ Because the absolute value of a number is always positive, it is not possible to put any value in for $x$ to make a true statement. There is no solution to this equation. We can indicated this by using the emptyset symbol: $\varnothing$

Solve for $x$

$|x|=-2$

Because the absolute value of a number is always positive, it is not possible to put any value in for $x$ to make a true statement. There is no solution to this equation. We can indicated this by using the emptyset symbol: $\varnothing$

Solve for $x$
$\|x-1\|+6=2$ when we isolate the absolute value by subtracting 6 from each side, we have $\|x-1\|=-4$ We can stop here, because we have the absolute value set equal to a negative number. We need not go on. There is no solution to this equation.

Solve for $x$
$\|3x-18\|+21=4$ Isolating the absolute value: $\|3x-18\|=-17$ And we stop, as we note that there are no solutions

Solve for $x$

$|3x-18|+21=4$
Isolating the absolute value:
$|3x-18|=-17$

And we stop, as we note that there are no solutions

Solve for $x$
$2-4\|2x+3\|=-18$ It may be tempting to say that since there is a negative on the right of the equation that there are no solutions, but we must not make that decision until we have isolated the absolute value: And divide the negative 4 from each side: $\dfrac{-4\|2x+3\|}{\color{red}-4}=\dfrac{-20}{\color{red}-4}$ $\|2x+3\|=5$ And we can see that the absolute value is now set equal to a positive number. So we create our two equations and solve: $2x+3=5 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 2x+3=-5$ $2x=2 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 2x=-8$ $x=1 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ x=-4$

Solve for $x$

$2-4|2x+3|=-18$

It may be tempting to say that since there is a negative on the right of the equation that there are no solutions, but we must not make that decision until we have isolated the absolute value:

Rendered by QuickLaTeX.com

And divide the negative 4 from each side:

$\dfrac{-4|2x+3|}{\color{red}-4}=\dfrac{-20}{\color{red}-4}$

$|2x+3|=5$
And we can see that the absolute value is now set equal to a positive number. So we create our two equations and solve:
$2x+3=5 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 2x+3=-5$
$2x=2 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 2x=-8$
$x=1 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ x=-4$

We might find ourselves tasked with an equation involving two absolute values. This can get quite complicated, so we will deal with only one case, when one absolute value expression is set equal to another. In this case, we will simply note that the expressions in the absolute values are either the same, or opposite of each other. This again creates two scenarios. To solve, simply write that the expression in the first set of absolute values is either the same as, or the opposite of the second expression:

Solve the equation
$\|3x+4\|=\|5x-10\|$ First we will set the expressions inside the absolute values equal to each other: $3x+4=5x-10$ Our second equation involves the two being opposite of each other. We can write this simply by stating that the first expression is equal to the negative of the second: $3x+4=-(5x-10)$ $3x+4=-5x+10$ Note that EVERY term in the right hand side changed signs in the simplified version. This is important to ensure happens! We can again solve for both equations in similar fashion to previous exercises: $3x+4=5x-10 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 3x+4=-5x+10$ $14=2x \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 8x=6$ $7=x \ \ \ \ \ \ \ or \ \ \ \ \ \ \ x=\frac{3}{4}$

Solve the equation

$|3x+4|=|5x-10|$
First we will set the expressions inside the absolute values equal to each other:
$3x+4=5x-10$

Our second equation involves the two being opposite of each other. We can write this simply by stating that the first expression is equal to the negative of the second:

$3x+4=-(5x-10)$
$3x+4=-5x+10$
Note that EVERY term in the right hand side changed signs in the simplified version. This is important to ensure happens!

We can again solve for both equations in similar fashion to previous exercises:

$3x+4=5x-10 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 3x+4=-5x+10$
$14=2x \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 8x=6$
$7=x \ \ \ \ \ \ \ or \ \ \ \ \ \ \ x=\frac{3}{4}$

Solve for $x$
$\|2x-7\|=\|7x+8\|$ We set our two equations and solve: $2x-7=7x+8 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 2x-7={\color{red}-}(7x+8)$ $2x-7=7x+8 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 2x-7=-7x-8$ $-15=5x \ \ \ \ \ \ \ or \ \ \ \ \ \ \ 9x=-1$ $-3=x \ \ \ \ \ \ \ or \ \ \ \ \ \ \ x=\frac{-1}{9}$

PreCalculus I

2.4 – Equations with Absolute Values

Equations with Absolute Values

Modal title