# 5.7: Linear Inequalities in One Variable

- Page ID
- 49372

## Inequalities

**Relationships of ****Inequality**

We have discovered that an equation is a mathematical way of expressing the relationship of equality between quantities. Not all relationships need be relationships of equality, however. Certainly the number of human beings on earth is greater than 20. Also, the average American consumes less than 10 grams of vitamin C every day. These types of relationships are not relationships of equality, but rather, relationships of **inequality**.

## Linear Inequalities

**Linear ****Inequality**

A **linear inequality** is a mathematical statement that one linear expression is greater than or less than another linear expression.

The following notation is used to express relationships of inequality:

\(>\) Strictly Greater Than

\(<\) Strictly Less Than

\(\ge\) Greater than or equal to

\(\le\) Less than or equal to

Note that the expression \(x > 12\) has infinitely many solutions. Any number strictly greater than 12 will satisfy the statement. Some solutions are \(13, 15, 90, 12.1, 16.3\) and \(102.51\).

## Sample Set A

The following **are** linear inequalities in one variable.

1. \(x \leq 12\)

2. \(x+7>4\)

3. \(y+3 \geq 2 y-7\)

4. \(P+26<10(4 P-6)\)

5. \(\dfrac{2 r-9}{5}>15\)

The following **are not** linear inequalities in one variable.

1. \(x^{2}<4\)

The term \(x^{2}\) is quadratic, not linear.

2. \(x \leq 5 y+3\)

There are two variables. This is a linear inequality in two variables.

3. \(y+1 \neq 5\)

Although the symbol \(\neq\) certainly expresses an inequality, it is customary to use only the symbols \(<,>, \leq, \geq\).

A linear equation, we know, may have exactly one solution, infinitely many solutions, or no solution. Speculate on the number of solutions of a linear inequality. (**Hint:** Consider the inequalities \(x<x−6\) and \(x\ge9\).)

A linear inequality may have infinitely many solutions, or no solutions.

## The Algebra of Linear Inequalities

Inequalities can be solved by basically the same methods as linear equations. There is one important exception that we will discuss in item 3 of the algebra of linear inequalities.

Let \(a, b\), and \(c\) represent real numbers and assume that:

\(a < b\) (or \(a > b\))

Then, if \(a < b\):

- \(a+c < b+c\) and \(a-c < b-c\).

If any real number is added to or subtracted from both sides of an inequality, the sense of the inequality remains unchanged. - If \(c\) is a
**positive**real number, then if \(a < b\),

\(ac < bc\) and \(\dfrac{a}{c} < \dfrac{b}{c}\).

If both sides of an inequality are multiplied or divided by the same positive number the sense of the inequality remains unchanged. - If \(c\) is a
**negative**real number, then if \(a < b\),

\(ac > bc\) and \(\dfrac{a}{c} > \dfrac{b}{c}\)

If both sides of an inequality are multiplied or divided by the same**negative**number,**the inequality sign must be reversed**(change direction) in order for the resulting inequality to be equivalent to the original inequality. (See problem 4 in the next set of examples.)

For example, consider the inequality \(3 < 7\).

For \(3<7,\) if 8 is added to both sides, we get

\(3+8<7+8\)

\(

11<15

\)

True

For \(3 < 7\), if 8 is subtracted from both sides, we get:

\(3-8 < 7-8\)

\(-5 < -1\)

True

For \(3 < 7\), if both sides are multiplied by 8 (a positive number), we get:

\(8(3) > 8(7)\)

\(24 < 56\)

True

For \(3<7\), if both sides are multiplied by −8 (a negative number), we get

\((−8)3>(−8)7\)

Notice the change in direction of the inequality sign.

\(−24>−56\)

True

If we had forgotten to reverse the direction of the inequality sign we would have obtained the incorrect statement \(−24<−56\).

For \(3<7\), if both sides are divided by 8 (a positive number), we get

\(\dfrac{3}{8} < \dfrac{7}{8}\)

True

For \(3 < 7\), if both sides are divided by -8 (a negative number), we get:

\(\dfrac{3}{-8} > frac{7}{-8}\)

True, since \(-.375 > .875\)

## Sample Set B

Solve the following linear inequalities. Draw a number line and place a point at each solution.

\(3x > 15\) Divide both sides by 3. The 3 is a positive number, so we need not reverse the sense of the inequality.

\(x > 5\)

Thus, all numbers strictly greater than 5 are solutions to the inequality \(3x > 15\)

\(2y-1 \le 16\) Add 1 to both sides.

\(2y \le 17\) Divide both sides 2.

\(y \le \dfrac{17}{2}\)

\(-8x + 5 < 14\) Subtract 5 both from both sides.

\(-8x < 9\) Divide both sides by -8. We must reverse the sense of the inequality since we are divide by a neagtive number.

\(x > -\dfrac{9}{8}\)

\(5-3(y+2) < 6y - 10\)

\(5-3y-6 < 6y-10\)

\(-3y-1 < 6y-10\)

\(-9y < -9\)

\(y > 1\)

\(\dfrac{2z+7}{-4} \ge -6\) Multiply by -4

\(2x+7 \le 24\) Notice the change in the sense of the inequality.

\(2z \le 17\)

\(z \le \dfrac{17}{2}\)

## Practice Set B

Solve the following linear inequalities.

\(y−6≤5\)

**Answer**-
\(y≤11\)

\(x+4>9\)

**Answer**-
\(x>5\)

\(4x−1≥15\)

**Answer**-
\(x≥4\)

\(−5y+16≤7\)

**Answer**-
\(y \ge \dfrac{9}{5}\)

\(7(4s−3)<2s+8\)

**Answer**-
\(s < \dfrac{29}{2}\)

\(5(1−4h)+4<(1−h)2+6\)

**Answer**-
\(h > \dfrac{1}{18}\)

\(18≥4(2x−3)−9x\)

**Answer**-
\(x≥−30\)

\(-\dfrac{3b}{16} \le 4\)

**Answer**-
\(b \ge \dfrac{-64}{3}\)

\(\dfrac{-7z+10}{-12} < -1\)

**Answer**-
\(z < -\dfrac{2}{7}\)

\(-x -\dfrac{2}{3} \le \dfrac{5}{6}\)

**Answer**-
\(x \ge \dfrac{-3}{2}

## Compound Inequalities

**Compound ****Inequality**

Another type of inequality is the **compound inequality**. A compound inequality is of the form:

\(a < x < b\)

There are actually two statements here. The first statement is \(a<x\). The next statement is \(x<b\). When we read this statement we say "\(a\) is less than \(x\)," then continue saying "and \(x\) is less than \(b\)."

Just by looking at the inequality we can see that the number \(x\) is between the numbers \(a\) and \(b\). The compound inequality \(a<x<b\) indicates "betweenness." Without changing the meaning, the statement \(a<x\) can be read \(x>a\). (Surely, if the number \(a\) is less than the number \(x\), the number \(x\) must be greater than the number \(a\).) Thus, we can read \(a<x<b\) as "\(x\) is greater than a and at the same time is less than \(b\)." For example:

\(4 < x < 9\).

The letter \(x\) is some number strictly between \(4\) and \(9\). Hence, \(x\) is greater than \(4\) and, at the same time, less than \(9\). The numbers \(4\) and \(9\) are not included so we use open circles at these points.

\(-2 < z < 0\).

The \(z\) stands for some number between \(-2\) and \(0\). Hence, \(z\) is greater than \(-2\) but also less than \(0\).

\(1 < x + 6 < 8\).

The expression \(x + 6\) represents some number strictly between \(1\) and \(8\). Hence, \(x + 6\) represents some number strictly greater than \(1\), but less than \(8\).

\(\dfrac{1}{4} \le \dfrac{5x-2}{6} \le \dfrac{7}{9}\).

The term \(\dfrac{5x-2}{6}\) represents some number between and including \(\dfrac{1}{4}\) and \(\dfrac{7}{9}\). Hence, \(\dfrac{5x-2}{6}\) represents some number greater than or equal to \(\dfrac{1}{4}\) but less than or equal to \(\dfrac{7}{9}\).

Consider problem 3 above, \(1<x+6<8\). The statement says that the quantity \(x+6\) is between \(1\) and \(8\). This statement will be true for only certain values of \(x\). For example, if \(x=1\), the statement is true since \(1<1+6<8\). However, if \(x=4.9\), the statement is false since \(1<4.9+6<8\) is clearly not true. The first of the inequalities is satisfied since \(1\) is less than \(10.9\), but the second inequality is not satisfied since \(10.9\) is not less than \(8\).

We would like to know for exactly which values of \(x\) the statement \(1<x+6<8\) is true. We proceed by using the properties discussed earlier in this section, but now we must apply the rules to all **three** parts rather than just the two parts in a regular inequality.

## Sample Set C

Solve \(1 < x + 6 < 8\).

\(1−6<x+6−6<8−6\) Subtract \(6\) from all three parts.

\(-5 < x < 2\)

Thus, if \(x\) is any number strictly between \(-5\) and \(2\), the statement \(1 < x+6 < 8\) will be true.

Solve \(-3 < \dfrac{-2x-7}{5} < 8\)

\(-3 < \dfrac{-2x-7}{5}(5) < 8(5)\) Multiply each part by \(5\).

\(-15 < -2x-7 < 40\). Add \(7\) to all three parts.

\(-8 < -2x < 47\) Divide all three parts by \(-2\).

\(4 > x > -\dfrac{47}{2}\) Remember to reverse the direction of the inequality signs.

\(-\dfrac{47}{2} < x < 4\). It is customary (but not necessary) to write the inequality so that inequality arrows point to the left.

Thus, if \(x\) is any number between \(-\dfrac{47}{2}\) and \(4\), the original inequality will be satisfied.

## Practice Set C

Find the values of x that satisfy the given continued inequality.

\(4<x−5<12\)

**Answer**-
\(9<x<17\)

\(−3<7y+1<18\)

**Answer**-
\(-\dfrac{4}{7} < y < \dfrac{17}{7}\)

\(0≤1−6x≤7\)

**Answer**-
\(-1 \le x \le \dfrac{1}{6}\)

\(-5 \le \dfrac{2x+1}{3} \le 10\)

**Answer**-
\(-8 \le x \le \dfrac{29}{2}\)

\(9 < \dfrac{-4x+5}{-2} < 14\)

**Answer**-
\(\dfrac{23}{4} < x < \dfrac{33}{4}\)

Does \(4<x<−1\) have a solution?

**Answer**-
No

## Exercises

For the following problems, solve the inequalities.

\(x+7<12\)

**Answer**-
\(x<5\)

\(y−5≤8\)

\(y+19≥2\)

**Answer**-
\(y≥−17\)

\(x−5>16\)

\(3x−7≤8\)

**Answer**-
\(x≤3\)

\(9y−12≤6\)

\(2z+8<7\)

**Answer**-
\(z < -\dfrac{1}{2}\)

\(4x−14>21\)

\(−5x≤20\)

**Answer**-
\(x≥−4\)

\(−8x<40\)

\(−7z<77\)

**Answer**-
\(z>−11\)

\(−3y>39\)

\(\dfrac{x}{4} \ge 12\)

**Answer**-
\(x≥48\)

\(\dfrac{y}{7} > 3\)

\(\dfrac{2x}{9} \ge 4\)

**Answer**-
\(x≥18\)

\(\dfrac{5y}{2} \ge 15\)

\(\dfrac{10x}{3} \le 4\)

**Answer**-
\(x \le \dfrac{6}{5}\)

\(-\dfrac{5y}{4} < 8\)

\(\dfrac{-12b}{5} < 24\)

**Answer**-
\(b>−10\)

\(\dfrac{-6a}{7} \le -24\)

\(\dfrac{8x}{-5} > 6\)

**Answer**-
\(x < -\dfrac{15}{4}\)

\(\dfrac{14y}{-3} \ge -18\)

\(\dfrac{21y}{-8} < -2\)

**Answer**-
\(y < \dfrac{16}{21}\)

\(−3x+7≤−5\)

\(−7y+10≤−4\)

**Answer**-
\(y≥2\)

\(6x−11<31\)

\(3x−15≤30\)

**Answer**-
\(x≤15\)

\(-2y + \dfrac{4}{3} \le -\dfrac{2}{3}\)

\(5(2x−5)≥15\)

**Answer**-
\(x≥4\)

\(4(x+1)>−12\)

\(6(3x−7)≥48\)

**Answer**-
\(x≥5\)

\(3(−x+3)>−27\)

\(−4(y+3)>0\)

**Answer**-
\(y<−3\)

\(−7(x−77)≤0\)

\(2x−1<x+5\)

**Answer**-
\(x<6\)

\(6y+12≤5y−1\)

\(3x+2≤2x−5\)

**Answer**-
\(x≤−7\)

\(4x+5>5x−11\)

\(3x−12≥7x+4\)

**Answer**-
\(x≤−4\)

\(−2x−7>5x\)

\(−x−4>−3x+12\)

**Answer**-
\(x>8\)

\(3−x≥4\)

\(5−y≤14\)

**Answer**-
\(y≥−9\)

\(2−4x≤−3+x\)

\(3[4+5(x+1)]<−3\)

**Answer**-
\(x<−2\)

\(2[6+2(3x−7)]≥4\)

\(7[−3−4(x−1)]≤91\)

**Answer**-
\(x≥−3\)

\(−2(4x−1)<3(5x+8)\)

\(−5(3x−2)>−3(−x−15)+1\)

**Answer**-
\(x<−2\)

\(−.0091x≥2.885x−12.014\)

What numbers satisfy the condition: twice a number plus one is greater than negative three?

**Answer**-
\(x>−2\)

What numbers satisfy the condition: eight more than three times a number is less than or equal to fourteen?

One number is five times larger than another number. The difference between these two numbers is less than twenty-four. What are the largest possible values for the two numbers? Is there a smallest possible value for either number?

**Answer**-
First number: any number strictly smaller that 6.

Second number: any number strictly smaller than 30.

No smallest possible value for either number.

No largest possible value for either number.

The area of a rectangle is found by multiplying the length of the rectangle by the width of the rectangle. If the length of a rectangle is 8 feet, what is the largest possible measure for the width if it must be an integer (positive whole number) and the area must be less than 48 square feet?

## Exercises for Review

Simplify\((x^2y^3z^2)^5\)

**Answer**-
\(x^{10}y^{15}z^{10}\)

Simplify \(−[−(−|−8|)]\).

Find the product. \((2x−7) (x+4)\).

**Answer**-
\(2x^2 + x - 28\)

Twenty-five percent of a number is \(12.32\). What is the number?

The perimeter of a triangle is 40 inches. If the length of each of the two legs is exactly twice the length of the base, how long is each leg?

**Answer**-
16 inches