Number line, absolute value, inequalities

Number line

A number line is a line marked with number values. Each number tells you how far a point is from zero and in which direction.

Zero is the center of the number line. Positive numbers are to the right of zero, and negative numbers are to the left. Most number line problems ask about the distance between two points or a point’s distance from zero.

Absolute value

The absolute value of a number is its distance from zero on the number line. For example, $-3$ is $3$ units away from $0$, so $|-3|=3$. The absolute value of any positive number is the number itself. Because distance is never negative, absolute values are always positive.

In algebra, absolute value is written using two vertical bars. For example, “the absolute value of $-3$ is $3$” is written as $|-3|=3$.

The steps to solve an algebraic equation with an absolute value are simpler than they look:

1. Isolate the expression within the absolute value bars
2. Delete the absolute value bars
3. Write and solve the problem twice:
   a. Once normally, as it is without changes;
   b. Again with the opposite side of the equation multiplied by $-1$

Here’s an example that shows the full process:

$$2|3+x|=8$$

First, divide both sides by $2$ to isolate the absolute value expression:

$$|3+x|=4$$

Now, delete the bars and solve normally to get the first solution:

$$\begin{array}{rl}|3+x|& =4\\ 3+x& =4\\ x& =1\end{array}$$

Finally, write the problem again. After deleting the bars, multiply the right side by $-1$:

$$\begin{array}{rl}|3+x|& =4\\ 3+x& =4\ast (-1)\\ 3+x& =-4\\ x& =-7\end{array}$$

So, $x$ could be either $1$ or $-7$.

While practicing (and if you have time during the real exam), it’s a good idea to check your work by plugging your answers into the original equation.

$$\begin{array}{rl}2|3+x|& =8\\ 2|3+(1)|& =8\\ 2|4|& =8\\ 2\ast 4& =8\\ 8& =8✓\end{array}$$

$$\begin{array}{rl}2|3+x|& =8\\ 2|3+(-7)|& =8\\ 2|-4|& =8\\ 2\ast 4& =8\\ 8& =8✓\end{array}$$

Inequalities

You solve inequalities much like equations. If $3+x=5$, you subtract $3$ from both sides to get $x=2$. If the problem is an inequality, like $3+x>5$, you still subtract $3$ from both sides and get $x>2$.

The key difference is this: if you multiply or divide both sides by a negative number, you must flip the inequality sign. For example, $-2x>4$ becomes $x<-2$. You divide both sides by $-2$ to isolate $x$, and flipping the sign keeps the statement true.

Here’s a chart if you need a refresher on the inequality symbols:

Let’s walk through the steps to solve an absolute value inequality.

Solve for $x$:

$$(2/3)|x-5|\ge 2$$

First, isolate the absolute value by multiplying both sides by the reciprocal of the fraction. The fraction is $2/3$, so the reciprocal is $3/2$.

$$\begin{array}{rl}(2/3)|x-5|& \ge 2\\ (3/2)\times (2/3)|x-5|& \ge (3/2)\times 2\\ |x-5|& \ge 3\end{array}$$

Now, delete the bars and solve normally to get the first part of the solution:

$$\begin{array}{rl}|x-5|& \ge 3\\ x-5& \ge 3\\ x& \ge 8\end{array}$$

Finally, write the problem again. Multiply the right side by $-1$ and flip the inequality:

$$\begin{array}{rl}|x-5|& \ge 3\\ x-5& \le 3\times (-1)\\ x-5& \le -3\\ x& \le 2\end{array}$$

So the solution is $x\le 2$ or $x\ge 8$. In other words, $x$ cannot be between $2$ and $8$.

$$x\le 2or8\le x$$