Inequalities

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We discuss inequalities.

Devyn asked what score was needed on the third exam to have an average of 82. This gave us the equation $\displaystyle \dfrac {74 + 84 + x}{3} = 82$ . The solutions only tell us what will give an average exactly 82. It may have been more advantageous to ask what score was needed to have an average of at least 82, yielding not an equation, but an inequality, $\displaystyle \dfrac {74+84+x}{3} \geq 82$ .

As with equations, there are various types of inequalities.

A linear inequality in $x$ is an inequality which is equivalent to $ax + b >0$ , $ax + b \geq 0$ , $ax+b < 0$ , or $ax + b \leq 0$ . We solve the inequality in much the same manner as with a linear equation. The main differences come from changing the direction of the inequality when multiplying/dividing by a negative quantity and expressing our answers in interval notation.

Solve the inequality $\displaystyle \dfrac {2x}{3}-4 < 3\left ( 2-x \right )$ .

As with equations, we’ll start by simplifying and combining like terms. $\begin{align*} \dfrac{2x}{3} -4 &< 3\left( 2-x \right)\\ \dfrac{2x}{3} - 4 &< 6 - 3x\\ \dfrac{2x}{3}+3x &< 6 + 4\\ \dfrac{11}{3} x &< 10\\ x &< \dfrac{30}{11} \end{align*}$

All $x$ less than $\frac {30}{11}$ satisfy the inequality.

The solution is $\left ( -\infty , \dfrac {30}{11} \right )$ .

Nonlinear inequalities are more complicated. To solve them, we will use a tool called a sign chart. The process requires us to move all the nonzero terms of the inequality to one side, and factor.

Solve the inequality $\dfrac {x^3 \left (2x-5\right )^2}{x+1} > 0.$

Since the right-hand side is zero, we are really asking where the left-hand side will be positive. The only way a nice formula like the left-hand side can switch from positive to negative, or vice-versa, is if it is either zero or undefined at the point where it changes.

Notice where the left-hand side would be either zero or undefined. Those occur at $x = -1, 0, \frac {5}{2}$ . We’ll use those points to split the number line into four regions. Inside each of those regions, our factors are either always positive or always negative.

For $x < -1$ , we find $x^3 < 0$ , $(2x-5)^2 > 0$ , and $x+1 < 0$ . Together, that means the left-hand side of our inequality has the form $\displaystyle \dfrac { \textrm {negative} \cdot \textrm {positive}}{\textrm {negative}}$ . It is, therefore, positive in that region, and makes up part of our solution. We cannot include the endpoint $x=-1$ in the solution, as the fraction is undefined there.

In the same way, we see that in the intervals $\left ( 0, \frac {5}{2} \right )$ and $\left ( \frac {5}{2}, 0\right )$ , the left-hand side of the inequality is also positive. Notice that the inequality is NOT satisfied at $x=\frac {5}{2}$ , where the left-hand side equals zero.

The solution is $\left ( -\infty , -1 \right ) \cup \left ( 0, \frac {5}{2} \right ) \cup \left ( \frac {5}{2}, \infty \right )$ .

Solve the inequality $\dfrac {2x}{x-1} \leq \dfrac {5-x}{x-1}.$

Let’s start by moving all the terms to one side of the inequality. $\begin{align*} \dfrac{2x}{x-1} &\leq \dfrac{5-x}{x-1}\\ \dfrac{2x}{x-1} - \dfrac{5-x}{x-1} &\leq 0\\ \dfrac{3x - 5}{x-1} &\leq 0 \end{align*}$

The solution is $\displaystyle \left ( 1, \dfrac {5}{3} \right ]$ .

Find the solution of the inequality $\dfrac {x}{x-4} \geq \dfrac {2x-1}{x-4}.$

$\left [1, 4\right )$ $\left (-\infty , 1\right ] \bigcup \left (4, \infty \right )$ $\left (1, 4\right )$ $\left (-\infty , 1\right ) \bigcup \left (4, \infty \right )$ None of the above