Linear Inequalities and Absolute Value Inequalities

Use a bracket on the left of $\mathrm{-2}$ and parentheses after infinity: $\left[\mathrm{-2},\infty \right).$ The bracket indicates that $\mathrm{-2}$ is included in the set with all real numbers greater than $\mathrm{-2}$ to infinity.

We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at $-\infty$ and ends at $\mathrm{-1},$ which is written as $\left(-\infty ,\mathrm{-1}\right].$

The second interval must show all real numbers greater than or equal to $1,$ which is written as $\left[1,\infty \right).$ However, we want to combine these two sets. We accomplish this by inserting the union symbol, $\cup ,$ between the two intervals.

The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.

1. ⓐ
   $\begin{array}{ll}x-15<4\hfill & \hfill \\ x-15+15<4+15\hfill & \text{Add 15 to both sides}.\hfill \\ x<19\hfill & \hfill \end{array}$
2. ⓑ
   $\begin{array}{ll}6\ge x-1\hfill & \hfill \\ 6+1\ge x-1+1\hfill & \text{Add 1 to both sides}.\hfill \\ 7\ge x\hfill & \hfill \end{array}$
3. ⓒ
   $\begin{array}{ll}x+7>9\hfill & \hfill \\ x+7-7>9-7\hfill & \text{Subtract 7 from both sides}.\hfill \\ x>2\hfill & \hfill \end{array}$

1. ⓐ
   $\begin{array}{l}3x<6\hfill \\ \frac{1}{3}(3x)<(6)\frac{1}{3}\hfill \\ x<2\hfill \end{array}$
2. ⓑ
   $\begin{array}{ll}-2x-1\ge 5\hfill & \hfill \\ -2x\ge 6\hfill & \hfill \\ \left(-\frac{1}{2}\right)(-2x)\ge (6)\left(-\frac{1}{2}\right)\hfill & \text{Multiply by}-\frac{1}{2}.\hfill \\ x\le -3\hfill & \text{Reverse the inequality}.\hfill \end{array}$
3. ⓒ
   $\begin{array}{ll}5-x>10\hfill & \hfill \\ -x>5\hfill & \hfill \\ (-1)(-x)>(5)(-1)\hfill & \text{Multiply by}-1.\hfill \\ x<-5\hfill & \text{Reverse the inequality}.\hfill \end{array}$

Solving this inequality is similar to solving an equation up until the last step.

The solution set is given by the interval $\left(-\infty ,1\right],$ or all real numbers less than and including 1.

We begin solving in the same way we do when solving an equation.

The solution set is the interval $\left(-\infty ,\frac{15}{34}\right].$

The first method is to write two separate inequalities: $3\le 2x+2$ and $2x+2<6.$ We solve them independently.

Then, we can rewrite the solution as a compound inequality, the same way the problem began.

In interval notation, the solution is written as $\left[\frac{1}{2},2\right).$

The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time.

We get the same solution: $\left[\frac{1}{2},2\right).$

Let's try the first method. Write two inequalities:

The solution set is $\mathrm{-4}<x<\frac{5}{6}$ or in interval notation Notice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right, as they appear on a number line. See Figure 3.

Figure 3

We want the distance between $x$ and 5 to be less than or equal to 4. We can draw a number line, such as in Figure 4, to represent the condition to be satisfied.

Figure 4

The distance from $x$ to 5 can be represented using an absolute value symbol, $\left|x-5\right|.$ Write the values of $x$ that satisfy the condition as an absolute value inequality.

We need to write two inequalities as there are always two solutions to an absolute value equation.

If the solution set is $x\le 9$ and $x\ge 1,$ then the solution set is an interval including all real numbers between and including 1 and 9.

So $\left|x-5\right|\le 4$ is equivalent to $\left[1,9\right]$ in interval notation.

We are trying to determine where $y<0,$ which is when $-\frac{1}{2}|4x-5|+3<0.$ We begin by isolating the absolute value.

Next, we solve for the equality $|4x-5|=6.$

Now, we can examine the graph to observe where the y-values are negative. We observe where the branches are below the x-axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at $x=-\frac{1}{4}$ and $x=\frac{11}{4},$ and that the graph opens downward. See Figure 5.

Figure 5