## Linear Inequalities

Linear inequalities are often easy to solve. We just need to isolate the variable on one side of the inequality sign.

• Recall that we can
• add (or subtract) the same quantity (+ or $-$) from both sides of the inequality, or
• multiply (or divide) both sides of the inequality by a positive quantity, and the inequality will still hold.
• If we multiply or divide both sides of an inequality by a negative quantity, the direction of the inequality will be reversed (See).

Never multiply or divide both sides of an inequality by a quantity whose sign is unknown!

Example 1
Solve each inequality
(a) $7x-5\geq4x+4$
(b) $-4<\dfrac{7-5x}{2}\leq1$
Solution
(a)
\begin{align*}
\begin{array}{rlr}
7x-5 & \geq4x+4 & {\small\text{ (given inequality)}}\\
7x-4x-5 & \ge4x-4x+4 &{\small\text{ (subtract $4x$ from both sides)}}\\
3x-5 & \ge4 &{\small \text{(simplify)}}\\
3x & \geq9 &{\small\text{ (add 5 to both sides)}}\\
x & \geq3 &{\small \text{(divide both sides by 3)}}
\end{array}
\end{align*}
(b)
$-4<\frac{7-5x}{2}\leq1$

Multiply each term by 2:
$-8<7-5x\leq2$

Subtract 7 from each term:
$-15<-5x\leq-5$ Divide each term by $-5$:
$3>x\geq 1$ which can alternatively be rewritten as $1\leq x<3$. For the last step, recall that when we divide both sides of an inequality by a negative number, the direction of the inequality changes (see Section 1.4).

The graph of $y=-\frac{5}{2}x+\frac{7}{2}$ shows that for $1\leq x<3$, we have $-4<y\le1$.

• For part (b) of the above example, note that
$-4<\dfrac{7-5x}{2}\leq 1$ means
$-4<\dfrac{7-5x}{2}\qquad\text{and}\qquad\dfrac{7-5x}{2}\leq1.$ In fact, we have to solve two inequalities.

## Nonlinear Inequalities

To solve nonlinear inequalities, we follow these steps:

1. Move all terms to one side of the inequality sign and express the inequality in the form
$P>0,\qquad\text{or}\qquad P<0$ or
$P\geq0,\qquad\text{or}\qquad P\leq 0$ where $P$ is an expression in the variable (usually $x$) [to indicate its dependence on $x$, we may write it as $P(x)$ if we wish]
2.  Factor $P$
$P=Q_{1}Q_{2}\cdots Q_{n}$ where $Q_{1},…,Q_{n}$ are expressions in $x$.
3. Determine the zeros of each factor of $P$ (find the values for which each $Q_{i}$ is zero). These values divide the real line into intervals.
4. Make a sign table. Determine the sign of each factor in each interval.
5. Determine the sign of $P$ in each interval using the sign table. Recall that a product (or a quotient) that involves an even number of negative factors is positive and one that involves an odd number of negative factors is negative. If the inequality sign is $\geq$ or $\leq$, pay attention to the endpoints of the intervals and check if they satisfy the inequality.
• For fractions; that is, when
$P=\frac{Q_{1}}{Q_{2}}$ we follow the same steps.
• To determine the sign of each factor, we can choose an arbitrary number (called a test value) in that interval and find the sign of the factor. Alternatively, you can use the following facts:

• If $ax^{2}+bx+c$ does not have any real roots, then its sign is always the same as the sign of $a$
[because in this case the graph of $y=ax^{2}+bx+c$ is completely above or completely below the $x$-axis, and never intersects it]

• If $ax^{2}+bx+c$ has two real roots $r_{1}$ and $r_{2}$, then when $x$ is between $r_{1}$ and $r_{2}$ , its sign is the opposite of the sign of $a$ and outside that interval, its sign is the same as the sign of $a$.
[Because when $ax^{2}+bx+c$ has two real roots, then
$ax^{2}+bx+c=a(x-r_{1})(x-r_{2})$
Example 2
Solve $x^{2}+6>5x$
Solution
$x^{2}+6>5x$

Subtract $5x$ from each term:
$x^{2}-5x+6>0$

Factor:
$(x-2)(x-3)>0$

Because the sign of a product depends on the signs of its factors, we need to determine where each factor is positive or negative. We identify that the zeros of these factors are $x=2$ and $x=3$ and using these values we construct the following sign table. Because the product is positive where both factors are positive or both are negative, the solutions are the real numbers in the union $(-\infty,2)\cup(3,\infty)$
or
$\{x|\ x<2\ \text{or}\ x>3\},$ as illustrated in the last row of the following sign table.

Alternatively we can say because the multiple of $x^{2}$ ($a=1$) is positive, as discussed above we know that when $2<x<3$, the sign of the polynomial is opposite to the sign of $a$ (here negative) and outside of that interval, the sign of the polynomial is the same as the sign of $a$ (here positive).

The graph of $y=x^{2}-5x+6$ shows that when $x<2$ or $x>3$, we have $y>0$ (see the following figure)

Example 3
Solve $x^{3}+2x^{2}\le x+2$
Solution

Given inequality:
$x^{3}+2x^{2} \le x+2$

Move all terms to one side:
$x^{3}+2x^{2}-x-2 \le0$ Now we need to factor $x^{3}+2x^{2}-x-2$. We can do that by grouping terms as follows
\begin{align*}
x^{3}+2x^{2}-x-2 & =(x^{3}+2x^{2})-(x+2)\\
& =x^{2}(x+2)-(x+2)
\end{align*}

Now factor out $(x+2)$ and then use the Difference of Squares formula in Section 1.13

\begin{align*}
x^{3}+2x^{2}-x-2 &=(x+2)(x^{2}-1)\\
& =(x+2)(x-1)(x+1)
\end{align*}
So we need to find $x$ such that
$(x+2)(x-1)(x+1)\le0$ To determine the sign, first we identify the zeros that are $x=-2$, $x=-1$, and $x=1$, and then construct the following sign table

The last row of the sign table is obtained from the fact that the polynomial is the product of the three factors. We read from this table that $(x+2)(x-1)(x+1)$ or $x^{3}+2x^{2}-x-2$ is negative or zero on the interval $(-\infty,-2]$ and $[-1,1]$. We have included $x=-2$ and $x=\pm1$ because at these points, the polynomial is zero and the inequality is satisfied. The solution set can be expressed
as
$\{x|\ x\leq-2\quad\text{or}\quad-1\le x\leq1\}=(-\infty,2]\cup[-1,1].$ The graph of $y=x^{3}+2x-x-2$ shows that when $x\le2$ or $-1\leq x\leq1$, we have $y\le0$ (see the following figure).

Example 4
Solve $\dfrac{x}{x+1}\leq2$.
Solution
Recall that if we multiply both sides by a positive quantity, the direction of the inequality will be preserved but if we multiply both sides by a negative quantity, the direction of the inequality will be reversed. We do not know the sign of $x+1$, so if we wanted to multiply both sides by $x+1$, we would not be sure about the direction of the inequality.

Given inequality:
$\frac{x}{x+1} \leq 2$

Subtract 2 from both sides
$\frac{x}{x+1}-2 \leq0$

The common denominator is $x+1$, so we get
$\frac{x}{x+1}-\frac{2(x+1)}{x+1} \leq0$

Combine fractions:
$\frac{x-2x-2}{x+1} \le 0$

Simplify:
$\frac{-x-2}{x+1} \leq 0$ The sign of $(-2-x)/(x+1)$ is solely determined by the signs of $(-x-2)$ and $(x+1)$. To construct the sign table, we need to identify the zeros of the numerator and denominator: $x=-1$ and $x=-2$. These are the only values at which the fraction may change sign. Now we construct the following sign table.

For $x=-1$, the denominator becomes zero, and because division by zero is not defined, we must eliminate $x=-1$.

From the sign table, we read that $(-x-2)/(x+1)\leq0$ when
$x\le-2\qquad\text{or}\qquad x>-1$ and the solution set is
$(-\infty,-2]\cup(-1,\infty)$ We have included $x=-2$ because for $x=-2$ the numerator and hence the fraction are zero and the inequality is satisfied.

The graph of $y=x/(x+1)$ is plotted in the following figure. The solutions are the $x$-values for which the corresponding $y$-values lie on or below the horizontal line $y=2$.

Example 5
Solve $\dfrac{x+1}{x^{2}+1}>1$.
Solution

Given equation

$\frac{x+1}{x^{2}+1} >1$

Move all terms to one side
$\frac{x+1}{x^{2}+1}-1 >0$

The common denominator is $x^2+1$, so:
$\frac{x+1}{x^{2}+1}-\frac{x^{2}+1}{x^{2}+1} >0$

Combine the fractions:
$\frac{x-x^{2}}{x^{2}+1} >0$

Factor:
$\frac{x(1-x)}{x^{2}+1} >0$

Because $x^{2}+1>0$, the sign of the fraction is the same as the sign of the numerator $x(1-x)$. The zeros of the numerator are $x=0$ and $x=1$ that divide the real line into three intervals. Then we construct the sign table as follows.

The sign table shows that $x(1-x)/(x^{2}+1)>0$ when $0<x<1$. In other words, the solution set is
$\{x|\ 0<x<1\}=(0,1).$ Because the denominator is never zero, we do not have to exclude any value of $x$.

The graph of $y=(x+1)/(x^{2}+1)$, illustrated in the following figure, shows that $y>1$ only when $0<x<1$.