Solve Simple and Compound Linear Inequalities

Properties of Inequalities

Addition Property

If $a < b$, then $a+c < b+c$.

Multiplication Property

If $a < b$ and $c > 0$, then $a c < b c$.

If $a < b$ and $c < 0$, then $a c > b c$.

These properties also apply to $a \leq b, a > b$, and $a \geq b$.

Example 3

Demonstrating the Addition Property

Illustrate the addition property for inequalities by solving each of the following:

1. $x-15 < 4$
2. $6 \geq x-1$
3. $x+7 > 9$

Solution

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.

(a)

$\begin{array}{cl}x-15 < 4 \\x-15+15 < 4+15 &\quad \text { Add } 15 \text { to both sides. } \\x < 19\end{array}$

(b)

$\begin{array}{cl}6 \geq x-1 \\6+1 \geq x-1+1 & \qquad \text{Add 1 to both sides.}\\7 \geq x\end{array}$

(c)

$\begin{array}{cl}x+7 > 9 \\x+7-7 > 9-7 & \qquad \text{Subtract 7 from both sides.}\\x > 2\end{array}$

Try It #3

Solve: $3 x-2 < 1$.

Example 4

Demonstrating the Multiplication Property

Illustrate the multiplication property for inequalities by solving each of the following:

1. $3 x < 6$
2. $-2 x-1 \geq 5$
3. $5-x > 10$

Solution

(a)

$\begin{align}\begin{gathered}3 x < 6 \\\frac{1}{3}(3 x) < (6) \frac{1}{3} \\x < 2\end{gathered}\end{align}$

(b)

$\begin{array}{cl}-2 x-1 \geq 5 & \\ -2 x \geq 6 & \text { Multiply by }-\frac{1}{2} \\ \left(-\frac{1}{2}\right)(-2 x) \geq(6)\left(-\frac{1}{2}\right) & \text { Reverse the inequality. } \\ x \leq-3 & \end{array}$

(c)

$\begin{aligned} 5-x & > 10 & & \\-x & > 5 & & \\(-1)(-x) & > (5)(-1) & & \text { Multiply by }-1 \\ x & < -5 & & \text { Reverse the inequality } \end{aligned}$

Try It #4

Solve: $4 x+7 \geq 2 x-3$