1. Functions

Virginia SOL Algebra 2 - 2020 Edition

1.10 Absolute value equations

Lesson

Like other functions, a function that has the absolute value symbols in its definition becomes an equation when its function value is specified. The equation may then be solved. We may get $2$2, $1$1 or $0$0 solutions. Consider under when we will have the different number of solutions using the graph below.

Definition of absolute value

For any real numbers $a$`a` and $b$`b`, where $b\ge0$`b`≥0, if $\left|a\right|=b$|`a`|=`b`, then $a=b$`a`=`b` or $a=-b$`a`=−`b`.

Solve the equation $2|x-1|+3=21$2|`x`−1|+3=21.

**Think: **We can start solving this like a regular linear equation until we get to the absolute value sign. To be sure of finding all the solutions to such an equation, we must look at both sides of the absolute value function.

**Do:**

$2\left|x-1\right|+3$2|x−1|+3 |
$=$= | $21$21 | (State the original equation) | |||

$2\left|x-1\right|$2|x−1| |
$=$= | $18$18 | (Subtract $3$3 from both sides) | |||

$\left|x-1\right|$|x−1| |
$=$= | $9$9 | (Divide both sides by$2$2) | |||

$x-1$x−1 |
$=$= | $9$9 | Split into the two cases | $x-1$x−1 |
$=$= | $-9$−9 |

$x$x |
$=$= | $10$10 | (Add $1$1to both sides) | $x$x |
$=$= | $-8$−8 |

The existence of these solutions can be seen in the following graphical representation.

Solve the equation $9\left|x\right|=4$9|`x`|=4.

Write both solutions as equations on the same line separated by a comma.

Consider the equation $\left|5x\right|=15$|5`x`|=15.

On the same set of axes, graph the functions $y=\left|5x\right|$

`y`=|5`x`| and $y=15$`y`=15.Loading Graph...Hence determine the solutions to the equation $\left|5x\right|=15$|5

`x`|=15. Write the solutions on the same line, separated by a comma.

Solve $\left|4x-8\right|+1=13$|4`x`−8|+1=13.

Write both solutions as equations on the same line separated by a comma.

To say $x\le4$`x`≤4 is to say that $x$`x` can be any number positive or negative that is not greater than four.

On a number line we draw $x\le4$`x`≤4 as such:

If the notation is changed slightly to $|x|\le4$|`x`|≤4, then we are restricting $x$`x` to be any number between $-4$−4 and $4$4, inclusive.

On a number line we draw $|x|\le4$|`x`|≤4 as such:

You could check that if $x$`x` is a number less than $-4$−4, then the absolute value of the number is greater than $4$4. It follows that a statement like $|x|\le4$|`x`|≤4 is equivalent to a pair of inequalities that can be written $-4\le x\le4$−4≤`x`≤4.

See the different representations below:

Absolute Value | Inequality | Interval Notation | Number Line |
---|---|---|---|

$\left|x\right|\le4$|x|≤4 |
$-4\le x\le4$−4≤x≤4 |
$\left[-4,4\right]$[−4,4] | |

$\left|x\right|<2$|x|<2 |
$-2x<2 |
$\left(-2,2\right)$(−2,2) |

To solve an inequality (sometimes called an inequation) means to find out the range or ranges of values of the variable for which the statement is true.

Above we looked at $|x|\le4$|`x`|≤4, and said this was the same as restricting $x$`x` to be any number between $-4$−4 and $4$4, inclusive.

What about something like $\left|x\right|\ge2$|`x`|≥2? Well, any number greater than or equal to $2$2 will satisfy this equation, but also any values less than or equal to $-2$−2. This means that this single expression is represented by two disjoint intervals as show below.

See the different representations below:

Absolute Value | Inequality | Interval Notation | Number Line |
---|---|---|---|

$\left|x\right|\ge2$|x|≥2 |
$x\le-2$x≤−2 or $x\ge2$x≥2 |
$\left(-\infty,-2\right]\cup\left[2,\infty\right)$(−∞,−2]∪[2,∞) | |

$\left|x\right|>2$|x|>2 |
$x<-2$x<−2 or $x>2$x>2 |
$\left(-\infty,-2\right)\cup\left(2,\infty\right)$(−∞,−2)∪(2,∞) |

Solve $\left|x+2\right|>5$|`x`+2|>5.

**Think:** It is common practice to consider what happens to the statement in $2$2 parts. The positive side and the negative side.

**Do:**

Positive side means to consider and solve $+(x+2)>5$+(`x`+2)>5

$+(x+2)$+(x+2) |
$>$> | $5$5 | (original positive inequality) |

$x+2$x+2 |
$>$> | $5$5 | (remove unnecessary parentheses) |

$x$x |
$>$> | $3$3 | (subtract $2$2 from both sides) |

Negative side means to consider and solve $-(x+2)>5$−(`x`+2)>5

$-(x+2)$−(x+2) |
$>$> | $5$5 | (original positive inequality) |

$x+2$x+2 |
$<$< | $-5$−5 | (divide both sides by $-1$−1, remember this flips the sign) |

$x$x |
$<$< | $-7$−7 | (subtract $2$2 from both sides) |

This means that for values of $x$`x`, that are either greater then $3$3 or less than $-7$−7, then the inequality $|x+2|>5$|`x`+2|>5 will hold true.

This graph shows the solution set. Spot check values inside or outside the range to check.

Another way to think about the previous example is via translations.

Consider first the solution on a number line for $|x|>5$|`x`|>5. We would know that this means all values of $x$`x` on the line that are a distance of greater than $5$5 from zero. It would look like this.

The effect of the $+2$+2, is a horizontal translation of $2$2 units to the left. Resulting in our final solution:

Rewrite the inequality $\left|x\right|<2$|`x`|<2 without using absolute values.

Rewrite the inequality $\left|x\right|\ge5$|`x`|≥5 without using absolute values.

Solve $\left|4x-8\right|+1=13$|4`x`−8|+1=13.

Write both solutions as equations on the same line separated by a comma.

Solve absolute value linear equations and inequalities