# Linear Inequalities in the Number Line

An **inequality** is similar to an equation, but instead of saying two quantities are equal, it says that one quantity is greater than or less than another. For example, since $1$ is greater than $0$, we write $1>0$. Likewise, since $0$ is less than $1$, we write $0<1$. If we write $x>0$, then we mean that $x$ can be $1$, $2$, $\pi$, $0.000001$, or any other positive number. If we write $x<0$, then we mean that $x$ can be $-1$, $-2$, $-\pi$, $-0.000001$, or any other negative number.

We can also write $x \geq 0$ to mean that $x$ is greater than or equal to $0$. In $x>0$, the number $0$ is not a valid solution for $x$ because $0$ is not greater than $0$, but in $x \geq 0$, the number $0$ is a valid solution because $0$ is greater than *or equal to* $0$. Likewise, we can write $x \leq 0$ to mean that $x$ is less than or equal to $0$.

Inequalities can be solved much like equations: we can perform algebraic manipulations to both sides of the equation until we isolate the variable.

If we substitute any number that is greater than $5$, it will satisfy the original inequality. For example, if we substitute $x=6$, then the original inequality becomes $22>20$, which is true. Likewise, if we substitute $x=5.001$, then the original inequality becomes $18.004>18.002$, which is true. On the other hand, if we substitute any number that is $5$ or less, it will not satisfy the original inequality. For example, if we substitute $x=5$, then the original inequality becomes $18>18$, which is not true. Likewise, if we substitute $x=4$, then the original inequality becomes $14>16$, which is not true.

In manipulating inequalities, there is just one catch: **whenever we multiply or divide by a negative number, we have to flip the inequality.**

To understand why we need to flip the inequality whenever we multiply or divide by a negative sign, consider the example $1<2$. If we multiply or divide by $-1$, we reach $-1<-2$, which is not true. In order to keep the inequality true, we have to flip the inequality sign: $-1>-2$.

To visualize inequalities, we can plot them on a number line. An open (unfilled) circle around a point means that the point itself is NOT a solution, while a closed (filled) circle around a point means that the point itself is a solution.

The number line can help us understand why we have to flip the inequality sign whenever we multiply or divide by a negative number. Starting with $5<10$, we know that $10$ is the bigger number which is further from $0$. When we multiply or divide, $10$ is still going to be further from $0$ than $5$ is – but if we multiply or divide by a negative number, then $10$ will be further from $0$ in the negative direction, which means it will actually be the lesser number.

The number line is a great intuitive aid, but it takes a while to draw. To simultaneously leverage the benefit of number line intuition and avoid the headache of drawing actual number lines, it is common to use **interval notation**, which represents number line segments using parentheses for open circles and brackets for closed circles.

To indicate that a segment continues forever, we imagine it having an open circle at positive or negative infinity.

*Practice Problems*

Solve the following inequalities, writing the solutions in interval notation. (You can view the solution by clicking on the problem.)

## $1) \hspace{.5cm} 4x-1 \leq 3x+1$

*Solution:*

$x \leq 2 \hspace{.5cm} (-\infty, 2]$

## $2) \hspace{.5cm} 14 > 4-2x$

*Solution:*

$x>5 \hspace{.5cm} (-5,\infty)$

## $3) \hspace{.5cm} -2x < 6x+24$

*Solution:*

$x>-3 \hspace{.5cm} (-3,\infty)$

## $4) \hspace{.5cm} 10-5x>3-4x$

*Solution:*

$x<7 \hspace{.5cm} (-\infty,7)$

## $5) \hspace{.5cm} 10-7x \leq 3x-5$

*Solution:*

$x \geq \frac{3}{2} \hspace{.5cm} \left[ \frac{3}{2}, \infty \right)$

## $6) \hspace{.5cm} 22x+2 \leq 16x$

*Solution:*

$x \leq -\frac{1}{3} \hspace{.5cm} \left( -\infty, -\frac{1}{3} \right]$

## $7) \hspace{.5cm} -4x-7 < 2-5x$

*Solution:*

$x<9 \hspace{.5cm} (-\infty,9)$

## $8) \hspace{.5cm} -3 < 4(x+3)$

*Solution:*

$x > -\frac{12}{7} \hspace{.5cm} \left( -\frac{12}{7}, \infty \right)$

## $9) \hspace{.5cm} 9x \geq 11(x-2)-5x$

*Solution:*

$x \geq -\frac{22}{3} \hspace{.5cm} \left[ -\frac{22}{3}, \infty \right)$

## $10) \hspace{.5cm} 3-4x > 4(1-2x)+9x-5$

*Solution:*

$x < \frac{4}{5} \hspace{.5cm} \left( -\infty, \frac{4}{5} \right]$

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