Math Study Guide for the SAT Exam

Absolute Value

The absolute value of a number or expression is its distance from \(0\) on a number line. Absolute values are always positive because they represent a magnitude. They are represented by an expression surrounded by two vertical lines (e.g., \(\vert-3\vert\)).

As you can see, the absolute value of both \(-3\) and \(3\) is \(3\) because they are both three spaces away from \(0\) on the number line.

Absolute Values in Expressions and Equations

Expressions and equations containing absolute values can be manipulated in much the same way as expressions and equations lacking absolute values can be manipulated.

Consider the following:

\[-3 |2x + 3| = -12\]

The goal, in this case, is to isolate the absolute value; this is accomplished by dividing both sides by \(-3\):

\[|2x + 3| = 4\]

Now, to solve this isolated absolute value, you need to rewrite the problem as two equations, one equal to positive \(4\) and one equal to \(-4\). Then you can solve each equation for \(x\). First, \(\vert{2x+3}\vert = 4\) is rewritten as:

\[2x + 3 = 4\]

and

\[2x + 3 = -4\]

So, with a little basic algebra, we find:

\[x = \frac{1}{2}\] \[x = \frac{-7}{2}\]

These answers can be verified by substituting them back into \(x\) and evaluating.

Absolute Values in Inequalities

Inequalities can also contain an absolute value. In the case of inequalities containing a less than (\(<\)) or less than or equal to (\(\le\)) sign, the original inequality must be rewritten without the absolute value to indicate that the solution set includes all values between (and sometimes including) the positive and negative integer expressions in the original inequality. For example:

\[|6x -8| \le 52\]

is rewritten as:

\[-52 \le 6x - 8 \le 52\]

This can be broken apart to form two inequalities that can be solved individually:

\[-52 \le 6x - 8\]

and

\[6x - 8 \le 52\]

This gives us two answers:

\[\frac{-44}{6} \le x\]

and

\[x \le 10\]

This solution set would be graphed on a number line as two closed (shaded) circles at \(\frac{-44}{6}\) and \(10\), with the values between them being shaded as well. The closed circles indicate that the values at the beginning and end of the shaded line are also part of the solution.

Greater than (\(>\)) or greater than or equal to (\(\ge\)) inequalities containing an absolute value are solved similarly. For example, say we have this inequality:

\[2|3x-2| \ge 18\]

In this case, division by \(2\) results in:

\[|3x - 2| \ge 9\]

Which can now be rewritten as:

\[3x - 2 \ge 9\]

and

\[3x - 2 \le -9\]

So, finding the values of \(x\), we get:

\[x \ge \frac{11}{3}\]

and

\[x \le \frac{-7}{3}\]

This solution set would be graphed on a number line with a single closed circle at \(\frac{11}{3}\) and an arrow pointing to the right, indicating that all values greater than \(\frac{11}{3}\) are also true. Another closed circle would be at \(\frac{-7}{3}\) with an arrow pointing to the left of this value, indicating that all numbers smaller than \(\frac{-7}{3}\) are also true.