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Algebra II : Mathematical Relationships and Basic Graphs

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Example Questions

Example Question #43 : Absolute Value

Solve the absolute value: $\left |8x+1 \right |<49$

Possible Answers:

$-\frac{25}{4}<x<6$

$x<-\frac{25}{4}\:\:\textup{OR} \:\:x>6$

$\textup{No solution.}$

x<6

$\frac{25}{4}<x<6$

Correct answer:

$-\frac{25}{4}<x<6$

Explanation:

Break up this absolute value into its positive and negative components.

8x+1<49

-(8x+1)<49

Evaluate the first equation. Subtract one from both sides and simplify.

8x+1-1<49-1

8x<48

Divide by eight on both sides.

$\frac{8x}{8}<\frac{48}{8}$

Simplify both sides. There is no need to switch the sign because we are not dividing both sides by a negative number.

The first solution is: x<6

Evaluate the second equation by dividing both sides by a negative one.

$\frac{-(8x+1)}{-1}<\frac{49}{-1}$

Simplify and switch the direction of the sign.

8x+1>-49

Subtract one from both sides.

8x+1-1>-49-1

8x>-50

Divide by eight on both sides.

$\frac{8x}{8}>-\frac{50}{8}$

$x>-\frac{25}{4}$

The second solution is: $x>-\frac{25}{4}$

The answer is: $-\frac{25}{4}<x<6$

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Example Question #44 : Solving Absolute Value Equations

Solve the inequality: $\left | 8x+1\right |<-25$

Possible Answers:

$-\frac{13}{4}<x<\frac{13}{4}$

-3<x<3

$-\frac{13}{4}<x<3$

$-3<x<\frac{13}{4}$

$\textup{No solution.}$

Correct answer:

$\textup{No solution.}$

Explanation:

According to absolute value rules, an absolute value will be converted into a positive number.

In the given problem $\left | 8x+1\right |<-25$ ,

we are asked to determine a certain value inside an absolute value that must be less than negative twenty five. However, there are no values of an absolute value that will give a negative number.

The answer is: $\textup{No solution.}$

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Example Question #43 : Solving Absolute Value Equations

Solve the absolute value equation: $\left | 8x-10\right |=5$

Possible Answers:

$-\frac{5}{8},\frac{5}{8}$

$\frac{5}{8},\frac{15}{8}$

$-\frac{5}{8},\frac{15}{8}$

$-\frac{5}{8},-\frac{15}{8}$

$\textup{There is no solution.}$

Correct answer:

$\frac{5}{8},\frac{15}{8}$

Explanation:

Break up the absolute value and rewrite the equations in their positive and negative components.

The equations are:

8x-10=5

-(8x-10)=5

Evaluate the first equation. Add 10 on both sides.

8x-10+10=5+10

8x=15

Divide by 8 on both sides.

The first solution is: $x=\frac{15}{8}$

Evaluate the second equation. Divide by negative one on both sides.

$\frac{-(8x-10)}{-1}=\frac{5}{-1}$

8x-10=-5

Add 10 on both sides.

8x-10+10=-5+10

8x=5

Divide by 8 on both sides.

The second solution is: $x=\frac{5}{8}$

The answers are: $\frac{5}{8},\frac{15}{8}$

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Example Question #46 : Solving Absolute Value Equations

Solve the absolute value equation: $\left | 9x-9\right | = -1$

Possible Answers:

$\frac{8}{9},\frac{10}{9}$

$\frac{8}{9},-\frac{10}{9}$

$-\frac{8}{9},-\frac{10}{9}$

$\textup{No solution.}$

$-\frac{8}{9},\frac{10}{9}$

Correct answer:

$\textup{No solution.}$

Explanation:

Recall that the absolute value sign will convert any value to a positive sign. There will be no occurrences of that will evaluate into a negative one as a final solution.

There are no solutions for this equation.

The answer is: $\textup{No solution.}$

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Example Question #45 : Absolute Value

Solve the equation: $\left | 9x+3\right |= 30$

Possible Answers:

$-\frac{11}{3},3$

$-\frac{1}{3},\frac{19}{3}$

$\textup{No solution.}$

$\frac{11}{3},3$

$-3,-\frac{11}{3}$

Correct answer:

$-\frac{11}{3},3$

Explanation:

Break up the absolute value into its positive and negative components.

The equations become:

9x+3=30

-(9x+3)=30

Solve the first equation. Subtract three on both sides.

9x+3-3=30-3

9x=27

Divide by nine on both sides.

$\frac{9x}{9}=\frac{27}{9}$

The first solution is x=3 .

Solve the second equation. Divide by negative one on both sides.

$\frac{-(9x+3)}{-1}=\frac{30}{-1}$

9x+3=-30

Subtract three on both sides.

9x+3-3=-30-3

9x=-33

Divide by nine on both sides.

$\frac{9x}{9}=\frac{-33}{9}$

Reduce the fractions.

The second solution is: $x=-\frac{11}{3}$

The answer is: $-\frac{11}{3},3$

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Example Question #46 : Absolute Value

Solve the following equation in real numbers:

|x^2-12|=13

Possible Answers:

x=-1

x=5

x=5,-1

x=-5

$x=\pm5$

Correct answer:

$x=\pm5$

Explanation:

Recall the definition of the absolute value operation:

|x|=x whenever $x\geq0$ , and |x|=-x whenever x<0 .

Hence, this equation can be split into two separate equations:

x^2-12=13 and x^2-12=-13 .

Solving the first equation yields:

x^2=25 ,

$\Rightarrow x=\pm5$

Solving the second equation yields:

x^2=-1 ,

$\Rightarrowx=\pm i$

Substituting each of these solutions into the original equation yields a true statement. However, we are only looking for real values of , and so must disregard the complex solutions $x=\pm i$ . Hence, the desired solutions to the equation are

$x=\pm5$ .

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Example Question #51 : Absolute Value

Solve the equation: $\left | 2x+3\right | = 78$

Possible Answers:

$\frac{75}{2},\frac{81}{2}$

$-\frac{81}{2},\frac{75}{2}$

$-\frac{39}{2},\frac{75}{2}$

$-\frac{81}{2},-\frac{75}{2},\frac{75}{2},\frac{81}{2}$

$\textup{No solution.}$

Correct answer:

$-\frac{81}{2},\frac{75}{2}$

Explanation:

Break up the inequality and write out its positive and negative components.

2x+3=78

-(2x+3)=78

Solve the first equation. Subtract 3 from both sides.

2x+3-3=78-3

2x=75

Divide both sides by 2.

$\frac{2x}{2}=\frac{75}{2}$

The first solution is: $\frac{75}{2}$

Solve the second equation. Divide the equation by negative one, which will move the negative sign from left to right.

$\frac{-(2x+3)}{-1}=\frac{78}{-1}$

2x+3=-78

Subtract three from both sides.

2x+3-3=-78-3

2x=-81

Divide by 2 on both sides.

$\frac{2x}{2}=\frac{-81}{2}$

The second answer is: $x=-\frac{81}{2}$

The answers are: $-\frac{81}{2},\frac{75}{2}$

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Example Question #52 : Absolute Value

Solve the equation: $\left | 8x\right | -40= 35$

Possible Answers:

$\textup{There is no solution.}$

$\pm \frac{25}{4}$

$\pm \frac{75}{8}$

$\pm \frac{15}{8}$

$\pm \frac{5}{8}$

Correct answer:

$\pm \frac{75}{8}$

Explanation:

Add 40 on both sides of the equation.

$\left | 8x\right | -40+40= 35+40$

$\left | 8x\right | =75$

Split this absolute value into its positive and negative components.

8x=75

-(8x)=75

For the first equation, divide by eight to isolate the x-variable.

$\frac{8x}{8}=\frac{75}{8}$

The first answer is: $x=\frac{75}{8}$

For the second equation, divide by negative eight on both sides.

$\frac{-(8x)}{-8}=\frac{75}{-8}$

The second answer is: $x=-\frac{75}{8}$

The answer is: $\pm \frac{75}{8}$

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Example Question #53 : Absolute Value

Solve the equation: $\left |-9x \right | -18= 18$

Possible Answers:

$\pm\frac{1}{4}$

$\pm4$

$\textup{All real numbers.}$

$\textup{No solution.}$

Correct answer:

$\pm4$

Explanation:

Add 18 on both sides.

$\left |-9x \right | -18+(18)= 18+(18)$

Simplify both sides.

$\left |-9x \right | =36$

Separate the equation into its positive and negative components and eliminate the absolute value.

-9x=36

-(-9x)=36

Simplify the first equation by dividing both sides by negative nine.

$\frac{-9x}{-9}=\frac{36}{-9}$

x=-4

Simplify the second equation. Double negatives negate, turning the true sign to positive.

9x=36

Divide by nine on both sides.

$\frac{9x}{9}=\frac{36}{9}$

x=4

The answer is: $\pm4$

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Example Question #4841 : Algebra Ii

Solve for x: |2x-1| = 7

Possible Answers:

x = 4, -3

x = 4, -4

x=-4, -3

x=3, -4

Correct answer:

x = 4, -3

Explanation:

Whatever is inside the absolute value brackets could either be positive or negative.

Positive: 2x - 1 = 7

2x = 8

x = 4

Negative: 2x-1=-7

2x = -6

x=-3

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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #43 : Absolute Value

Example Question #44 : Solving Absolute Value Equations

Example Question #43 : Solving Absolute Value Equations

Example Question #46 : Solving Absolute Value Equations

Example Question #45 : Absolute Value

Example Question #46 : Absolute Value

Example Question #51 : Absolute Value

Example Question #52 : Absolute Value

Example Question #53 : Absolute Value

Example Question #4841 : Algebra Ii

All Algebra II Resources

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