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College Algebra : Solving Equations and Inequallities

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Example Questions

Example Question #81 : Solving Equations And Inequallities

Solve the following

$\left | 3x-5\right |=20$

Possible Answers:

There are no solutions

x=-5

$x=\frac{25}{3}$

x=-5

$x=\frac{25}{3}$

Correct answer:

x=-5

$x=\frac{25}{3}$

Explanation:

We have to set up two equations, which are.

3x-5=20

3x-5=-20

Now lets solve for x in each equation.

3x-5=20

$3x=25\rightarrow x=\frac{25}{3}$

3x-5=-20

$3x=-15\rightarrow x=-5$

So the solutions are x=-5 and $x=\frac{25}{3}$

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Example Question #82 : Solving Equations And Inequallities

Solve:

|x-10|=20

Possible Answers:

x=30

x=10

x=-30

x=10

x=-30

x=-10

x=30

x=-10

Correct answer:

x=30

x=-10

Explanation:

We need to set up two equations since we are dealing with absolute value

$\\x-10=20\\x-10=-20$

We solve for , in each equation to get the solutions.

x=20+10=30

x=-20+10=-10

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Example Question #83 : Solving Equations And Inequallities

Solve:

|x-45|=2

Possible Answers:

x=43

x=47

x=-43

x=47

x=43

x=-47

x=-43

x=-47

Correct answer:

x=43

x=47

Explanation:

We need to set up two equations since we are dealing with absolute value

$\\x-45=2\\x-45=-2$

We solve for , in each equation to get the solutions.

x=2+45=47

x=-2+45=43

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Example Question #84 : Solving Equations And Inequallities

Solve the following:

$\left | x-2 \right |>1$

Possible Answers:

$(-\infty,-3)\cup (-1, \infty)$

(1,3)

$(-\infty,-1)\cup (3, \infty)$

$(-\infty,1)\cup (3, \infty)$

Correct answer:

$(-\infty,1)\cup (3, \infty)$

Explanation:

To solve, you must split the absolute value into the two following equations.

x-2>1 and x-2<-1

Now, solve for x and right it in interval form.

$x-2>1\Rightarrow x>3$

$x-2<-1\Rightarrow x<1$

$(-\infty,1)\cup (3,\infty)$

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Example Question #85 : Solving Equations And Inequallities

Solve for x:

$\left | x-7 \right | = 3$

Possible Answers:

$x=-10\;; \; x=4$

$x = 10, \; \; x=4$

$x=4, \;\; x=-4$

$x=10\; ; \; x=-10$

Correct answer:

$x = 10, \; \; x=4$

Explanation:

To solve $\left | x-7\right | = 3$ we need to find a way to reverse the operation of taking the absolute value. What we need to do is think about what the absolute value operation does to an expression. Since it makes everything positive, $\left | x-7\right | = \left | (-1) (x-7)\right |$ . So actually solving the original equation comes down to solving the following two equations:

x-7=3

-(x-7)=3

So we get the two solutions as:

x=7+3=10

x=7-3=4

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Example Question #4 : Absolute Value Equations

Solve the following equation for .

$\left |x^2-8 \right |=1$

Possible Answers:

$x=\pm4,\pm2$

$x=\pm3,\pm\sqrt{7}$

$x=\pm7,\pm\sqrt{3}$

$x=3,\sqrt{7}$

$x=\sqrt3,{-7}$

Correct answer:

$x=\pm3,\pm\sqrt{7}$

Explanation:

We first need to get rid of the absolute value symbol to solve the equation. TO break this absolute value, we assign two values to the right hand side, as shown below.

x^2-8=1

x^2-8=-1

We now proceed to solve each equation independently.

Starting with the first equation, we get

x^2=9

$x=\pm3$

Now for the second equation,

x^2=7

$x=\pm\sqrt{7}$

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Example Question #86 : Solving Equations And Inequallities

Simplify the radical:

$\frac{2\sqrt{20}}{\sqrt{4}}$

Possible Answers:

None of the other answers.

$2\sqrt{5}$

$\sqrt{5}$

$-5\sqrt{2}$

Correct answer:

$2\sqrt{5}$

Explanation:

To simplify the radical, break the radical in the numerator down into its factors. When doing so, the radical in the bottom will call with one from the numerator.

$\frac{2\sqrt{20}}{\sqrt{4}}=\frac{2\sqrt{4}\sqrt{5}}{\sqrt{4}}=2{\sqrt{5}}$

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Example Question #8 : Absolute Value Equations

Solve the following equation:

$-3-\left | 8x-6\right |=3$

Possible Answers:

$x=\frac{3}{4}$

$x=-2\:or\:x=0$

$x=0 \:or\:x=\frac{3}{2}$

$x=-3\:or\:x=1$

$x=\frac{3}{2}\:or\:x=10$

Correct answer:

$x=0 \:or\:x=\frac{3}{2}$

Explanation:

To solve for first isolate the absolute value portion on one side of the equation and all other constants on the other side.

$-3-\left | 8x-6\right |=3$

$-\left | 8x-6\right |=6$

$\left | 8x-6\right |=-6$

Recall that the absolute value can come from either a negative or positive value therefore two possible equations are set up.

$8x-6=6\: \:or\: \:8x-6=-6$

$8x=12\: \:or\: \:8x=0$

$x=\frac{12}{8}=\frac{3}{2}\: \:or\: \:x=0$

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Example Question #7 : Absolute Value Equations

Solve the equation:

$-5\left | 4+2x\right |=-100$

Possible Answers:

$x=-\frac{1}{2}\:or\:x=4$

$x=0\:or\:x=-8$

$x=-8\:or\:x=9$

$x=-8\:or\:x=12$

$x=-12\:or\:x=8$

Correct answer:

$x=-12\:or\:x=8$

Explanation:

To solve for first isolate the absolute value portion on one side of the equation and all other constants on the other side.

$-5\left | 4+2x\right |=-100$

$\left | 4+2x\right |=20$

Since absolute values can come from either negative or positive values, two equations need to be set up and solved for.

$4+2x=20\: \:or\: \:4+2x=-20$

$2x=16\: \:or\: \:2x=-24$

$x=-12\:or\:x=8$

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Example Question #1 : Absolute Value Inequalities

Solve: -2x-8<-4x+20

Possible Answers:

$x>\frac{14}{3}$

$x<-\frac{14}{3}$

$x<\frac{14}{3}$

x<14

x>14

Correct answer:

x<14

Explanation:

Add on both sides.

-2x-8+4x<-4x+20+4x

2x-8< 20

Add 8 on both sides.

2x-8+8< 20+8

2x<28

Divide by two on both sides.

$\frac{2x}{2}<\frac{28}{2}$

The answer is: x<14

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College Algebra : Solving Equations and Inequallities

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #81 : Solving Equations And Inequallities

Example Question #82 : Solving Equations And Inequallities

Example Question #83 : Solving Equations And Inequallities

Example Question #84 : Solving Equations And Inequallities

Example Question #85 : Solving Equations And Inequallities

Example Question #4 : Absolute Value Equations

Example Question #86 : Solving Equations And Inequallities

Example Question #8 : Absolute Value Equations

Example Question #7 : Absolute Value Equations

Example Question #1 : Absolute Value Inequalities

All College Algebra Resources

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