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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #91 : Solving Inequalities

Solve: 2x^2>18

Possible Answers:

[-3,3]

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

(-3,3)

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

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

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 #91 : 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: $x<\frac{17}{18}$

Report an Error

Example Question #331 : Basic Single Variable Algebra

Solve: $\frac{2}{3}x-\frac{8}{9}<\frac{1}{12}$

Possible Answers:

$x<\frac{35}{24}$

$x>\frac{29}{24}$

$x>\frac{35}{24}$

$x<-\frac{29}{24}$

$x>-\frac{29}{24}$

Correct answer:

$x<\frac{35}{24}$

Explanation:

Multiply both sides by the least common denominator. This will save extra work converting all the fractions.

The LCD is 36 since this value is divisible by all three denominators in the inequality.

$36(\frac{2}{3}x-\frac{8}{9})<\frac{1}{12} \cdot 36$

24x-32<3

Add 32 on both sides.

24x-32+32<3+32

24x<35

Divide by 24 on both sides.

$\frac{24x}{24}<\frac{35}{24}$

The answer is: $x<\frac{35}{24}$

Report an Error

Example Question #2172 : Algebra Ii

Solve the inequality: $3x> \sqrt x$

Possible Answers:

$x\geq0$

x>0

$\textup{There is no solution.}$

$x\leq\frac{1}{9}$

$x>\frac{1}{9}$

Correct answer:

$x>\frac{1}{9}$

Explanation:

Square both sides of the inequality to eliminate the radical.

$(3x)^2> (\sqrt x)^2$

9x^2>x

Subtract from both sides.

9x^2-x>x-x

9x^2-x>0

Factor out an on the left side.

x(9x-1)>0

The factors possible are:

x>0

$9x-1>0 \rightarrow 9x>1 \rightarrow x>\frac{1}{9}$

The values from zero to $\frac{1}{9}$ will not satisfy since the curve will lie beneath the $\sqrt x$ curve.

This means that only $x>\frac{1}{9}$ is valid.

The answer is: $x>\frac{1}{9}$

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Example Question #332 : Basic Single Variable Algebra

Solve the inequalities: $\frac{1}{8}x-\frac{3}{64}\leq \frac{5}{16}$

Possible Answers:

$x\leq\frac{ 17}{8}$

$x\geq-\frac{ 23}{8}$

$x\leq\frac{ 23}{8}$

$x\leq-\frac{ 23}{8}$

$x\geq-\frac{ 17}{8}$

Correct answer:

$x\leq\frac{ 23}{8}$

Explanation:

Multiply both sides by the least common denominator.

The LCD is 64 because this value is divisible by all of the numbers in the denominator.

$64(\frac{1}{8}x-\frac{3}{64})\leq \frac{5}{16} \cdot 64$

$8x-3\leq 20$

Add 3 on both sides.

$8x-3+3\leq 20+3$

$8x\leq 23$

Divide by 8 on both sides.

$\frac{8x}{8}\leq\frac{ 23}{8}$

The answer is: $x\leq\frac{ 23}{8}$

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Example Question #333 : Basic Single Variable Algebra

Solve the inequality: $-\frac{x}{9}> 20$

Possible Answers:

$x>-\frac{20}{9}$

$x<-\frac{20}{9}$

x<-180

x>-180

$x<-\frac{9}{20}$

Correct answer:

x<-180

Explanation:

Multiply by negative nine on both sides of the inequality.

Since we still deal with negative sign, we will need to switch the direction of the sign.

$-\frac{x}{-9}\cdot 9> 20\cdot-9$

The answer is: x<-180

Report an Error

Example Question #334 : Basic Single Variable Algebra

Solve the inequality: $-3(-2x+3)\leq 14$

Possible Answers:

$x\leq\frac{10}{3}$

$x\leq \frac{23}{6}$

$x<\frac{23}{6}$

$x\geq \frac{5}{6}$

$x\leq \frac{5}{6}$

Correct answer:

$x\leq \frac{23}{6}$

Explanation:

Use distribution to simplify the left side.

$-3(-2x)+-3(3)\leq 14$

$6x-9\leq 14$

Add 9 on both sides.

$6x-9+9\leq 14+9$

$6x\leq 23$

Divide by 6 on both sides.

$\frac{6x}{6}\leq \frac{23}{6}$

The answer is: $x\leq \frac{23}{6}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #91 : Solving Inequalities

Example Question #97 : Solving Inequalities

Example Question #98 : Solving Inequalities

Example Question #99 : Solving Inequalities

Example Question #91 : Solving Inequalities

Example Question #331 : Basic Single Variable Algebra

Example Question #2172 : Algebra Ii

Example Question #332 : Basic Single Variable Algebra

Example Question #333 : Basic Single Variable Algebra

Example Question #334 : Basic Single Variable Algebra

All Algebra II Resources

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