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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #85 : Solving Inequalities

Solve the inequality: -3(9-x)>7x-4

Possible Answers:

$x<-\frac{23}{4}$

$x<-\frac{31}{10}$

$x>-\frac{23}{4}$

$x<-\frac{31}{4}$

$x>-\frac{31}{4}$

Correct answer:

$x<-\frac{23}{4}$

Explanation:

Simplify the left side by distribution.

-27+3x>7x-4

Subtract from both sides.

-27+3x-3x>7x-4-3x

-27>4x-4

Add 4 on both sides.

-27+4>4x-4+4

-23>4x

Divide by four on both sides.

$\frac{-23}{4}>\frac{4x}{4}$

The answer is: $x<-\frac{23}{4}$

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Example Question #86 : Solving Inequalities

Solve the inequalities: $\frac{1}{10}x-\frac{1}{100}>\frac{3}{1000}$

Possible Answers:

$x<\frac{7}{100}$

$x>-\frac{17}{100}$

$x>\frac{13}{100}$

$x>\frac{103}{100}$

$x>-\frac{7}{100}$

Correct answer:

$x>\frac{13}{100}$

Explanation:

Multiply both sides by the least common denominator in order to eliminate the fractions.

$1000(\frac{1}{10}x-\frac{1}{100})>\frac{3}{1000}\cdot 1000$

The inequality becomes:

100x-10>3

Add on both sides of the inequality.

100x-10+10>3+10

The inequality becomes:

100x>13

Divide by 100 on both sides.

$\frac{100x}{100}>\frac{13}{100}$

The answer is: $x>\frac{13}{100}$

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Example Question #87 : Solving Inequalities

Solve the inequality: 9(-3x+8)>9x+4

Possible Answers:

$x>\frac{9}{17}$

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

$x>\frac{19}{9}$

$x>-\frac{19}{9}$

$x<\frac{17}{9}$

Correct answer:

$x<\frac{17}{9}$

Explanation:

Expand the terms on the left by distribution.

9(-3x+8) = -27x+72

Rewrite the inequality.

-27x+72>9x+4

Add 27x on both sides.

-27x+72+27x>9x+4+27x

72>36x+4

Subtract 4 from both sides.

72-4>36x+4-4

68>36x

Divide by 36 on both sides.

$\frac{68}{36}>\frac{36x}{36}$

Reduce the fractions.

The answer is: $x<\frac{17}{9}$

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Example Question #88 : Solving Inequalities

Solve the inequality: 9>5x-17

Possible Answers:

$x<-\frac{8}{5}$

$x>-\frac{8}{5}$

$x>\frac{26}{5}$

$x<\frac{26}{5}$

$x>\frac{8}{5}$

Correct answer:

$x<\frac{26}{5}$

Explanation:

Add 17 on both sides.

9+17>5x-17+17

Simplify the inequality.

26>5x

Divide by five on boths ides

$\frac{26}{5}>\frac{5x}{5}$

The answer is: $x<\frac{26}{5}$

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Example Question #89 : Solving Inequalities

Solve the inequality: -9x+5>8(3+x)

Possible Answers:

$x< -\frac{29}{17}$

$x< -\frac{19}{17}$

$x> -\frac{19}{17}$

$x> -\frac{29}{17}$

x>19

Correct answer:

$x< -\frac{19}{17}$

Explanation:

Use the distributive property to expand the right side.

8(3+x) = 8(3)+8x = 8x+24

The inequality becomes: -9x+5>8x+24

To avoid having to divide by a negative value later in the calculation, we should add nine on both sides to move to the right.

-9x+5+9x>8x+24+9x

5>17x+24

Subtract 24 from both sides.

5-24>17x+24-24

-19>17x

Divide by 17 on both sides.

$\frac{-19}{17}>\frac{17x}{17}$

The answer is: $x< -\frac{19}{17}$

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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 #91 : Solving Inequalities

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

Possible Answers:

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

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

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

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

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

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

Report an Error

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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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #85 : Solving Inequalities

Example Question #86 : Solving Inequalities

Example Question #87 : Solving Inequalities

Example Question #88 : Solving Inequalities

Example Question #89 : Solving Inequalities

Example Question #91 : Solving Inequalities

Example Question #91 : Solving Inequalities

Example Question #93 : Solving Inequalities

Example Question #94 : Solving Inequalities

Example Question #95 : Solving Inequalities

All Algebra II Resources

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