When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equations. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number, we must reverse the inequality symbol.
properties of inequalities
[latex]\begin{array}{ll}\text{Addition Property}\hfill& \text{If }a< b,\text{ then }a+c< b+c.\hfill \\ \hfill & \hfill \\ \text{Multiplication Property}\hfill & \text{If }a< b\text{ and }c> 0,\text{ then }ac< bc.\hfill \\ \hfill & \text{If }a< b\text{ and }c< 0,\text{ then }ac> bc.\hfill \end{array}[/latex]
These properties also apply to [latex]a\le b[/latex], [latex]a>b[/latex], and [latex]a\ge b[/latex].
Illustrate the addition property for inequalities by solving each of the following:
[latex]x - 15<4[/latex]
[latex]6\ge x - 1[/latex]
[latex]x+7>9[/latex]
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.
[latex]\begin{array}{ll} \text{Given inequality:} & x - 15 < 4 \\ \text{Add 15 to both sides:} & x - 15 + 15 < 4 + 15 \\ \text{Simplify:} & x < 19 \end{array}[/latex]
In interval notation, the solution is expressed as: [latex](-\infty, 19)[/latex].
[latex]\begin{array}{ll} \text{Given inequality:} & 6 \geq x - 1 \\ \text{Add 1 to both sides:} & 6 + 1 \geq x - 1 + 1 \\ \text{Simplify:} & 7 \geq x \\ \text{or equivalently:} & x \leq 7 \end{array}[/latex]
In interval notation, the solution is expressed as: [latex](-\infty, 7][/latex]
[latex]\begin{array}{ll} \text{Given inequality:} & x + 7 > 9 \\ \text{Subtract 7 from both sides:} & x + 7 - 7 > 9 - 7 \\ \text{Simplify:} & x > 2 \end{array}[/latex]
In interval notation, the solution is expressed as: [latex](2, \infty)[/latex]
Illustrate the multiplication property for inequalities by solving each of the following:
[latex]3x<6[/latex]
[latex]-2x - 1\ge 5[/latex]
[latex]5-x>10[/latex]
[latex]\begin{array}{ll} \text{Given inequality:} && 3x &< 6 \\ \text{Multiply both sides by} \frac{1}{3} \text{or divide by 3:} && x &< \frac{6}{3} \\ \text{Simplify:} && x &< 2 \end{array}[/latex]
In interval notation, the solution is expressed as: [latex](-\infty, 2)[/latex]
[latex]\begin{array}{ll} \text{Given inequality:} && -2x - 1 &\geq 5 \\ \text{Add 1 to both sides:} && -2x - 1 + 1 &\geq 5 + 1 \\ \text{Simplify:} && -2x &\geq 6 \\ \text{Divide both sides by -2 (reverse inequality):} && x &\leq -3 \end{array}[/latex]
In interval notation, the solution is expressed as: [latex](-\infty, -3][/latex]
[latex]\begin{array}{ll} \text{Given inequality:} & 5 - x > 10 \\ \text{Subtract 5 from both sides:} & 5 - x - 5 > 10 - 5 \\ \text{Simplify:} & -x > 5 \\ \text{Multiply both sides by -1 (reverse inequality):} & x < -5 \end{array}[/latex]
In interval notation, the solution is: [latex](-\infty, -5)[/latex]
We can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.
Solve the following inequality:
[latex]13 - 7x\ge 10x - 4[/latex]
Solving this inequality is similar to solving an equation up until the last step.
[latex]\begin{array}{ll}13 - 7x\ge 10x - 4\hfill & \hfill \\ 13 - 17x\ge -4\hfill & \text{Move variable terms to one side of the inequality}.\hfill \\ -17x\ge -17\hfill & \text{Isolate the variable term}.\hfill \\ x\le 1\hfill & \text{Dividing both sides by }-17\text{ reverses the inequality}.\hfill \end{array}[/latex]
The solution set is given by the interval [latex]\left(-\infty ,1\right][/latex], or all real numbers less than and including [latex]1[/latex].
Solve the following inequality and write the answer in interval notation: