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Algebra 1 : Linear Inequalities

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Example Question #1 : Solving Inequalities

Solve this inequality.

$\left | \frac{4x+3}{x+6}\right |\eq>1$

Possible Answers:

$x\eq>-\frac{9}{5}, x \eq<1$

$x\eq>-\frac{9}{5}, x\eq>1$

$x\eq<\frac{9}{5}, x\eq>-1$

$x \eq< -\frac{9}{5}, \hspace x \eq>1$

$x<-\frac{9}{5}, x\eq<1$

Correct answer:

$x \eq< -\frac{9}{5}, \hspace x \eq>1$

Explanation:

Split the inequality into two possible cases as follows, based on the absolute values.

First case: $\frac{4x+3}{x+6}\eq>1$

Second case: $\frac{4x+3}{x+6} \eq< -1$

Let's find the inequality of the first case.

$\frac{4x+3}{x+6} \eq>1$

Multiply both sides by x + 6.

$4x+3 \eq>x+6$

Subtract x from both sides, then subtract 3 from both sides.

$3x \eq>3$

Divide both sides by 3.

$x\eq>1$

Let's find the inequality of the second case.

$\frac{4x+3}{x+6} \eq< -1$

Multiply both sides by x + 6.

$4x+3 \eq< -1(x+6)$

Simplify.

4x+3< -x-6

Add x to both sides, then subtract 3 from both sides.

$5x\eq< -9$

Divide both sides by 5.

$x \eq<-\frac{9}{5}$

So the range of x-values is $x<-\frac{9}{5}$ and $x \eq>1$ .

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Example Question #1 : Solving Absolute Value Equations

Solve for :

$\left |2x+3 \right |>7$

Possible Answers:

x>2, x<-5

x<2, x>-5

x>2, x<-10

x>3.5, x<-5

x>2

Correct answer:

x>2, x<-5

Explanation:

Solve for positive values by ignoring the absolute value. Solve for negative values by switching the inequality and adding a negative sign to 7.

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Example Question #1 : Absolute Value Inequalities

Give the solution set for the following equation:

| 4x - 7 | + 5 = 86

Possible Answers:

$\begin{Bmatrix} -24.5, 22 \end{Bmatrix}$

$\begin{Bmatrix} 22 \end{Bmatrix}$

$\begin{Bmatrix} 22, 24.5 \end{Bmatrix}$

$\begin{Bmatrix} -18.5, 22 \end{Bmatrix}$

$\begin{Bmatrix} 18.5, 22 \end{Bmatrix}$

Correct answer:

$\begin{Bmatrix} -18.5, 22 \end{Bmatrix}$

Explanation:

First, subtract 5 from both sides to get the absolute value expression alone.

| 4x - 7 | + 5 = 86

| 4x - 7 | = 81

Split this into two linear equations:

4x - 7 = 81

4x = 88

x = 22

4x - 7 = -81

4x = -74

$x = -74 \div 4 = -18.5$

The solution set is $\begin{Bmatrix} -18.5, 22 \end{Bmatrix}$

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Example Question #3 : Absolute Value Inequalities

Solve for in the inequality below.

$|x-2| \leq 4$

Possible Answers:

$-2 \leq x\leq 6$

$x\leq 4$

No solutions

$x\leq -2\; \; or\; \; x\geq 6$

All real numbers

Correct answer:

$-2 \leq x\leq 6$

Explanation:

The absolute value gives two problems to solve. Remember to switch the "less than" to "greater than" when comparing the negative term.

$x-2\leq 4$ or $x-2\geq -4$

Solve each inequality separately by adding to all sides.

$x\leq 6$ or $x\geq -2$

This can be simplified to the format $-2 \leq x\leq 6$ .

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Example Question #10 : Solving Absolute Value Equations

$\textup{If }\left | 2x+4\right |\leq10\textup{ , what are all possible values of }x\textup{?}$

Possible Answers:

$-7\leq x\leq3$

x=-7,x=3

$-3\leq x\leq7$

$-3\leq x\leq3$

$x\leq-7\textup{ and }x\geq3$

Correct answer:

$-7\leq x\leq3$

Explanation:

$\textup{Set up two inequalities. Reverse the inequality sign for the negative one.}$

$2x+4\leq10\;\;\;\;\;\;\;\;\;\;\;\;\;2x+4\geq-10$

$2x\leq6\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \; 2x\geq-14$

$x\leq3\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x\geq-7$

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Example Question #3 : Absolute Value Inequalities

Solve the inequality.

$\small \left | x+2 \right |>5$

Possible Answers:

$\small x<-7\ or\ x>3$

$\small -7<x<3$

$\small -3<x<7$

$\small x<-3\ or\ x>7$

Correct answer:

$\small x<-7\ or\ x>3$

Explanation:

$\small \left | x+2 \right |>5$

Remove the absolute value by setting the term equal to either $\small 5$ or $\small -5$ . Remember to flip the inequality for the negative term!

$\small x+2>5\ or\ x+2<-5$

Solve each scenario independently by subtracting $\small 2$ from both sides.

$\small x+2-2>5-2\ or\ x+2-2<-5-2$

$\small x>3\ or\ x<-7$

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Example Question #1 : Absolute Value Inequalities

Solve for :

| 4y + 7 | > -5

Possible Answers:

$(-\infty , -3) \cup \left ( \frac{1}{2}, \infty \right)$

$\left(-\infty , -\frac{1}{2}\right) \cup \left ( \frac{1}{2}, \infty \right)$

$(-\infty , \infty )$

$\left(-\infty , -\frac{1}{2}\right) \cup \left (3, \infty \right)$

$(-\infty , -3) \cup (3, \infty )$

Correct answer:

$(-\infty , \infty )$

Explanation:

The absolute value of any number is nonnegative, so | 4y + 7 | must always be greater than . Therefore, any value of makes this a true statement.

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Example Question #81 : Linear Inequalities

Solve the following inequality.

$\left | x-6\right |+8>12$

Possible Answers:

x>2

$x>-10\:or\:x<2$

$x>10\:or\:x<-14$

$x>10\:or\:x<2$

2<x<10

Correct answer:

$x>10\:or\:x<2$

Explanation:

Inequalities involving $> \: or\: \geq$ generate two separate inequalities and can't be combined into a single inequality.

First isolate the absolute value expression on one side of the inequality

$\left | x-6\right |+8>12$

Subtract eight from each side.

$\left |x-6 \right |>4$

From here separate the expression into two expressions for which we will need to solve.

$x-6>4\: \: \: \:or\: \: \: \:x-6<-4$

Add six to get side.

$x>10\: \: \: \:or\: \: \: \:x<2$

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Example Question #1411 : Algebra 1

Solve the inequality for .

$\left | 3x+5\right |\geq 1$

Possible Answers:

$x\geq -2$

$x\leq -2$ and $x\geq -\frac{4}{3}$

$x\geq -\frac{4}{3}$

$-2\leq x\leq -\frac{4}{3}$

$x\geq -\frac{4}{3}$ and $x\leq \frac{4}{3}$

Correct answer:

$x\leq -2$ and $x\geq -\frac{4}{3}$

Explanation:

To solve an inequality with absolute value you have to consider the two equations it creates.

$\left | 3x+5\right |\geq 1$ becomes $3x+5\geq 1$ and $3x+5\leq -1$

Solve for both inequalities by following the balancing rules. Be careful of division or multiplication of a negative number; if that happens, flip the inequality sign.

$3x+5\geq 1$

$3x\geq -4$

$x\geq -\frac{4}{3}$

$3x+5\leq -1$

$3x\leq -6$

$x\leq -2$

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Example Question #83 : Linear Inequalities

$\left | 3x\right |<27$

Possible Answers:

-30<x<30

-9<x<9

x<-8,x>8

x<-9,x>9

x>9

Correct answer:

-9<x<9

Explanation:

We first need to eliminate the absolute value sign by making two inequalities:

3x<27 and 3x>-27 . Remember that when the number becomes negative we must flip the inequality symbol.

From there, it is a simple one-step inequality. We divide both sides by to get:

x<9

x>-9 .

Because x lies between and , we can combine these into one inequality:

-9<x<9

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Algebra 1 : Linear Inequalities

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #1 : Solving Inequalities

Example Question #1 : Solving Absolute Value Equations

Example Question #1 : Absolute Value Inequalities

Example Question #3 : Absolute Value Inequalities

Example Question #10 : Solving Absolute Value Equations

Example Question #3 : Absolute Value Inequalities

Example Question #1 : Absolute Value Inequalities

Example Question #81 : Linear Inequalities

Example Question #1411 : Algebra 1

Example Question #83 : Linear Inequalities

All Algebra 1 Resources

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