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Algebra II : Solving Inequalities

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Example Question #91 : Solving Inequalities

Solve the inequality: $2(3-2x)\leq 9(2-3x)$

Possible Answers:

$x\leq -\frac{12}{23}$

$x\geq \frac{12}{23}$

$x\leq \frac{12}{23}$

$x\geq \frac{12}{31}$

$x\geq -\frac{12}{23}$

Correct answer:

$x\leq \frac{12}{23}$

Explanation:

Use the distributive property to simplify the binomials.

$2(3)-2(2x)\leq 9(2)-9(3x)$

Simplify both sides.

$6-4x\leq 18-27x$

Add 27x on both sides.

$6-4x+27x\leq 18-27x+27x$

$6+23x\leq 18$

Subtract six from both sides.

$6+23x-6\leq 18-6$

$23x\leq 12$

Divide by 23 on both sides.

$\frac{23x}{23}\leq \frac{12}{23}$

The answer is: $x\leq \frac{12}{23}$

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Example Question #92 : Solving Inequalities

Solve: 3x>-5(2x-3)

Possible Answers:

$x>-\frac{15}{7}$

$x>\frac{15}{13}$

$x>-\frac{15}{13}$

$x>\frac{15}{7}$

$x>-\frac{13}{15}$

Correct answer:

$x>\frac{15}{13}$

Explanation:

Use distribution to simplify the right side.

3x>-5(2x)-5(-3)

Simplify the right side.

3x>-10x+15

Add 10x on both sides of the equation.

3x+10x>-10x+15+10x

13x>15

Divide by 13 on both sides.

$\frac{13x}{13}>\frac{15}{13}$

The answer is: $x>\frac{15}{13}$

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Example Question #93 : Solving Inequalities

Solve: 2-4x>6x-8

Possible Answers:

x<-5

x>1

x>5

x<1

x<5

Correct answer:

x<1

Explanation:

Add on both sides.

2-4x+4x>6x-8+4x

2>10x-8

Add 8 on both sides.

2+8>10x-8+8

10>10x

Divide by 10 on both sides.

$\frac{10}{10}>\frac{10x}{10}$

1>x

The answer is: x<1

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Example Question #94 : Solving Inequalities

Solve the inequality: -9x+7>8+9x

Possible Answers:

$x>-\frac{1}{18}$

$x>-\frac{15}{18}$

$x<-\frac{1}{18}$

$x<-\frac{15}{18}$

$\textup{No solution.}$

Correct answer:

$x<-\frac{1}{18}$

Explanation:

Add on both sides to avoid dividing by a negative later in the calculation. Dividing by a negative value will require switching the sign.

-9x+7+(9x)>8+9x+(9x)

The inequality becomes:

7> 18x+8

Subtract 8 from both sides.

7-8> 18x+8-8

-1>18x

Divide by 18 on both sides.

$\frac{-1}{18}>\frac{18x}{18}$

The answer is: $x<-\frac{1}{18}$

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Example Question #95 : Solving Inequalities

Solve the inequality: $\sqrt{5x-7}<\sqrt{2x}$

Possible Answers:

$(-\infty,\frac{7}{3})$

$(0,\frac{7}{3})$

$[\frac{7}{5},\frac{7}{3})$

$(0,\frac{7}{5}]$

$(\frac{7}{5},\frac{7}{3})$

Correct answer:

$[\frac{7}{5},\frac{7}{3})$

Explanation:

There is a domain restriction with the square root functions.

Both the functions must be zero or greater.

Square both sides.

$(\sqrt{5x-7}) ^2<(\sqrt{2x})^2$

5x-7<2x

Subtract from both sides, and add 7 on both sides.

5x-7+(7)-[2x]<2x-[2x]+(7)

3x<7

Divide by three on both sides to isolate x.

$x<\frac{7}{3}$

Notice that this value will also include all the values that are negative, which cannot satisfy the original equation. We will need to identify the largest possible x-value of both radicals to determine a viable domain.

Set the two radicals greater or equal to zero and solve for .

$\sqrt{2x}\geq0 \rightarrow 2x\geq 0 \rightarrow x\geq0$

$\sqrt{5x-7}\geq0 \rightarrow5x-7\geq0 \rightarrow 5x\geq7\rightarrow x\geq \frac{7}{5}$

The value of must also be greater than $\frac{7}{5}$ .

Rewrite $\frac{7}{5}\leq x<\frac{7}{3}$ in interval notation.

The answer is: $[\frac{7}{5},\frac{7}{3})$

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Example Question #96 : Solving Inequalities

Solve: 2x^2>18

Possible Answers:

[-3,3]

(-3,3)

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

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

$(-\infty, -3] \bigcup[3,\infty)$

Correct answer:

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

Explanation:

Divide by two on both sides.

$\frac{2x^2}{2}>\frac{18}{2}$

The inequality becomes:

x^2>9

Square root both sides.

$\sqrt{x^2}>\sqrt{9}$

$\pm x >3$

Split the inequalities.

x>3 is one solution.

Simplify -x>3 by dividing by negative one on both sides. This will change the sign.

$\frac{-x}{-1}>\frac{3}{-1}$

x<-3 is the second solution.

Write the terms in interval notation.

The answer is: $(-\infty, -3) \bigcup(3,\infty)$

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Example Question #97 : Solving Inequalities

Solve: $-3x+7\leq 8x-4$

Possible Answers:

$\textup{The answer is not given.}$

$x\leq-1$

$x\geq1$

$x\leq1$

$x\geq \frac{3}{11}$

Correct answer:

$x\geq1$

Explanation:

Add on both sides.

$-3x+7+(3x)\leq 8x-4+(3x)$

$7\leq 11x-4$

Add four on both sides.

$7+4\leq 11x-4+4$

$11\leq 11x$

Divide by eleven on both sides.

$\frac{11}{11}\leq \frac{11x}{11}$

$1\leq x$

The answer is: $x\geq1$

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Example Question #98 : Solving Inequalities

Solve: -3x-8<-4x-9

Possible Answers:

x>-17

x<-17

$x\leq-1$

$x<-\frac{7}{17}$

x<-1

Correct answer:

x<-1

Explanation:

Add on both sides.

-3x-8+4x<-4x-9+4x

x-8<-9

Add eight on both sides.

x-8+8<-9+8

Simplify both sides.

The answer is: x<-1

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Example Question #99 : Solving Inequalities

Solve: -8(2x-3)<17

Possible Answers:

$x<\frac{41}{16}$

$x>\frac{7}{16}$

$x>-\frac{41}{16}$

$x>\frac{41}{16}$

$x<\frac{7}{16}$

Correct answer:

$x<\frac{7}{16}$

Explanation:

Distribute the negative eight into the binomial.

-16x+24<17

Subtract 24 on both sides.

-16x+24-24<17-24

-16x<-7

Divide by negative 16 on both sides. The sign will change direction since we are dividing by a negative value.

$\frac{-16x}{-16}<\frac{-7}{-16}$

The answer is: $x<\frac{7}{16}$

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Example Question #100 : Solving Inequalities

Solve: -3(3x-8)>9x+7

Possible Answers:

$x<\frac{17}{18}$

$x>\frac{31}{18}$

$x>\frac{17}{18}$

$x<-\frac{31}{18}$

$x<\frac{31}{18}$

Correct answer:

$x<\frac{17}{18}$

Explanation:

Distribute the negative three through both terms of the binomial.

-9x+24>9x+7

Add on both sides.

-9x+24+9x>9x+7+9x

24>18x+7

Subtract seven on both sides.

24-7>18x+7-7

17>18x

Divide by 18 on both sides.

$\frac{17}{18}>\frac{18x}{18}$

The answer is: $x<\frac{17}{18}$

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Algebra II : Solving Inequalities

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #91 : Solving Inequalities

Example Question #92 : Solving Inequalities

Example Question #93 : Solving Inequalities

Example Question #94 : Solving Inequalities

Example Question #95 : Solving Inequalities

Example Question #96 : Solving Inequalities

Example Question #97 : Solving Inequalities

Example Question #98 : Solving Inequalities

Example Question #99 : Solving Inequalities

Example Question #100 : Solving Inequalities

All Algebra II Resources

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