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Algebra 1 : Algebra 1

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Example Questions

Example Question #77 : How To Solve Absolute Value Equations

Which of the following is equivalent to 88 - |45 - x | if is less than 45?

Possible Answers:

133-x

133+x

None of the other responses is correct.

43 + x

45-x

Correct answer:

43 + x

Explanation:

If x < 45 , then 45 - x > 0 , and

|45 - x | = 45 - x

Therefore,

88 - |45 - x | = 88 - (45-x) = 88 - 45 + x = 43 + x

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Example Question #78 : How To Solve Absolute Value Equations

Which of the following is equivalent to |x+65| -3 7 if is less than 65?

Possible Answers:

-x -102

None of the other responses is correct.

-x-28

-x +102

-x+28

Correct answer:

-x -102

Explanation:

If x < 65 , it follows that

x - 65 < 65 - 65 = 0 .

Since the expression between the abosolute value bars is negative, it follows that

|x+65| = -(x+65) = -x-65

and

|x+65| -3 7 = -x-65 - 37 = -x -102

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Example Question #71 : How To Solve Absolute Value Equations

Solve for x.

|2x-7|=11

Possible Answers:

x=9 and x=-2

x=4

x=3 and x=-1

x=0 and x=5

Correct answer:

x=9 and x=-2

Explanation:

To solve for we set up the two equations.

2x-7=11 and 2x-7=-11

For the first equation

2x-7+7=11+7

2x=18

$\frac{1}{2}*2x=\frac{1}{2}*18$

x=9

And for the second equation

2x-7+7=-11+7

2x=-4

$\frac{1}{2}*2x=\frac{1}{2}*(-4)$

x=-2

And so the solutions are

x=9 and x=-2

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Example Question #71 : How To Solve Absolute Value Equations

Solve: $\left | 3x-4\right |=9$

Possible Answers:

$\frac{5}{3}, -\frac{13}{3}$

$-\frac{5}{3}, -\frac{13}{3}$

$\frac{5}{3}, \frac{13}{3}$

$\frac{13}{3}$

$-\frac{5}{3}, \frac{13}{3}$

Correct answer:

$-\frac{5}{3}, \frac{13}{3}$

Explanation:

Break up the absolute value and rewrite the equation in its positive and negative components.

3x-4=9

-(3x-4) = 9

Solve the first equation. Start by adding four on both sides.

3x-4+4=9+4

Simplify.

3x=13

Divide by three on both sides.

$x=\frac{13}{3}$

This is one of the solutions.

Solve the second equation. Start by dividing a negative one on both sides.

$\frac{-(3x-4) }{-1}= \frac{9}{-1}$

Simplify both sides.

3x-4=-9

Add four on both sides.

3x-4+4=-9+4

Simplify both sides.

3x=-5

Divide by three on both sides.

$\frac{3x}{3}=\frac{-5}{3}$

The answer for this equation is: $-\frac{5}{3}$

The answers to this absolute value equation are: $-\frac{5}{3}, \frac{13}{3}$

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Example Question #911 : Algebra 1

Solve for x:

|5x-2|+1=9

Possible Answers:

x=1.2 $ and $ x=2

x=-1.6 $ and $ x=2

x=-1.6 $ and $ x=2.4

x=1.6 $ and $ x = -2.4

x=-1.2 $ and $ x=2

Correct answer:

x=-1.2 $ and $ x=2

Explanation:

|5x-2|+1=9

Subtract 1 from both sides:

$(|5x-2|+1)-1=(9)-1\n\n$

|5x-2| = 8

Remove absolute Value Signs.

(5x-2)=8$ and $ (5x-2)=-8

Solve for both solutions of x:

(5x-2)+2=8+2

5x=10

x=10/5=2

AND

(5x-2)+2=-8+2

5x=-6

x=-6/5=-1.2

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Example Question #82 : How To Solve Absolute Value Equations

|6x-1| = |3x+8|

Possible Answers:

x=3 $ and $ x=7/9

x=3

x=-3 $ and $ x=7/9

x=-3 $ and $x=-7/9

x=3 $ and $x=-7/9

Correct answer:

x=3 $ and $x=-7/9

Explanation:

Removing the absolute value signs we get the four equations:

(1) $ $ (6x-1) = (3x+8)

(2)$ $(6x-1) = -(3x+8)

(3)$ $-(6x-1) = (3x+8)

(4)$ $-(6x-1) = -(3x+8)

Since 1&4 are identical and 2&3 are identical, we only need to solve for (1) and (2) to get the complete answer:

Equation 1 solution

(6x-1) = (3x+8)

(6x-1) +1= (3x+8)+1

(6x)-3x= (3x+9)-3x

3x=9

x=9/3=3

Equation 2 Solution

(6x-1) = -(3x+8)

(6x-1)+1 = (-3x-8)+1

(6x) +3x= (-3x-7)+3x

9x= -7

x= -7/9

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Example Question #1 : How To Write Expressions And Equations

Write in simplest form:

2 (4x + 5) - 3 (2x - 7)

Possible Answers:

2x + 3

2x + 31

2x + 12

2x - 2

2x - 11

Correct answer:

2x + 31

Explanation:

Rewrite, then distribute:

$\small \small 2 (4x + 5) - 3 (2x - 7)$

$\small = 2 (4x + 5) + (- 3) (2x - 7)$

$\small = 2 \cdot 4x + 2 \cdot 5 + (- 3) \cdot 2x - (-3) \cdot7$

$\small = \small 8x + 10 + (- 6) x - (-21)$

$\small \small = \small 8x + (- 6) x + 10 - (-21)$

$\small = 8x - 6 x + 10 + 21 = 2x + 31$

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Example Question #2 : How To Write Expressions And Equations

A car travels at a speed of 60 miles per hour. It is driven for 2.5 hours. How many miles does it travel?

Possible Answers:

120

100

150

180

Correct answer:

150

Explanation:

To solve this problem, you need to construct an algebraic equation. If is the distance traveled, then must equal to the speed multiplied by the time travelled. In this case, x=2.5*60 , which gives you a result of 150 miles.

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Example Question #1 : How To Write Expressions And Equations

Rewrite the expression in simplest terms, where is the imaginary number $\sqrt-1$ .

$i* \sqrt{-12}$

Possible Answers:

$-2\sqrt3$

$-\sqrt{12}$

$2\sqrt3$

12i

Correct answer:

$-2\sqrt3$

Explanation:

Writing this expression in simplest terms can be achieved by first factoring the radical into its smallest factors.

$i*\sqrt{-12}=\sqrt{-1} * (\sqrt{-1}\sqrt{4}\sqrt{3})$

Multiplying the two $\sqrt{-1}$ together results in . Multiplying this by $\sqrt4\sqrt3$ (which is simplified to $2\sqrt3$ ) results in the answer $-2\sqrt3$ .

$\sqrt{-1}*(\sqrt{-1}\sqrt{4}\sqrt{3})=(\sqrt{-1})^2*2\sqrt{3}=-2\sqrt{3}$

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Example Question #2 : How To Write Expressions And Equations

Rewrite the equation for in terms of .

$\frac{5x^5y}{12y^3} = \frac{4x}{7x^3y}$

Possible Answers:

$y = \frac{35x^7}{48}$

$x = \sqrt[7]{\frac{35y}{48}}$

$y = \frac{48}{35x^7}$

$x = \sqrt[7]{\frac{48y}{35}}$

Correct answer:

$y = \frac{35x^7}{48}$

Explanation:

The goal in expressing in terms of is to isolate on one side of the equation. One way to do this is to factor out of the fraction on the right side of the equation, then divide the entire equation by the fraction that remains after factoring. Remember that dividing by a fraction is the same as multiplying by the reciprocal of that fraction.

$\frac{5x^5y}{12y^3} = \frac{4x}{7x^3y} \rightarrow y (\frac{5x^5}{12y^3}) = \frac{4x}{7x^3y}$

$y (\frac{5x^5}{12y^3}) * (\frac{12y^3}{5x^5}) = \frac{4x}{7x^3y} * (\frac{12y^3}{5x^5})$

The left side of this equation will simply resolve into , although there are still variables on the right, so this is not yet in terms of . The right side resolves based on the rules for multiplying and dividing variables with exponents (add the exponents of like variables being multiplied, subtract the smaller exponent from the larger in the case of division, and change the variable to a if the resulting exponent is ).

$y = \frac{4x}{7x^3y} * (\frac{12y^3}{5x^5})$

$y = \frac{48xy^3}{35x^8y}$

$y = \frac{48y^2}{35x^7}$

Since there is still a y^2 in the numerator on the right side of the equation, we will need to divide both sides of the equation by y^2 .

$y = \frac{48y^2}{35x^7} \rightarrow (y) * \frac{1}{y^2}= (\frac{48y^2}{35x^7}) * \frac{1}{y^2}$

$\frac{y}{y^2}= \frac{1}{y} = \frac{48}{35x^7}$

We have no solved for the reciprocal of in terms of . We simply flip both sides of the equation to get our answer.

$y = \frac{35x^7}{48}$

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Algebra 1 : Algebra 1

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #77 : How To Solve Absolute Value Equations

Example Question #78 : How To Solve Absolute Value Equations

Example Question #71 : How To Solve Absolute Value Equations

Example Question #71 : How To Solve Absolute Value Equations

Example Question #911 : Algebra 1

Example Question #82 : How To Solve Absolute Value Equations

Example Question #1 : How To Write Expressions And Equations

Example Question #2 : How To Write Expressions And Equations

Example Question #1 : How To Write Expressions And Equations

Example Question #2 : How To Write Expressions And Equations

All Algebra 1 Resources

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