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Algebra II : Basic Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #271 : Solving Equations

Solve the equation: 9(2-3x)=-(9-3x)

Possible Answers:

$\frac{3}{10}$

$-\frac{9}{8}$

$\frac{9}{10}$

$-\frac{3}{10}$

$\frac{3}{8}$

Correct answer:

$\frac{9}{10}$

Explanation:

Distribute the nine on the left side through both terms of the binomial.

9(2-3x) = 18-27x

Evaluate the right side of the equation.

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

Set the two simplified terms equal.

18-27x=3x-9

Add 27x on both sides.

18-27x+27x=3x-9+27x

18=30x-9

Add 9 on both sides.

18+9=30x-9+9

27=30x

Divide by thirty on both sides.

$\frac{27}{30}=\frac{30x}{30}$

Reduce both fractions.

The answer is: $\frac{9}{10}$

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Example Question #272 : Solving Equations

Set up the equation: -8x+18=46

Possible Answers:

$-\frac{7}{9}$

$-\frac{1}{8}$

$-\frac{29}{4}$

$-\frac{7}{2}$

Correct answer:

$-\frac{7}{2}$

Explanation:

Subtract 18 from both sides.

-8x+18-18=46-18

Simplify the equation.

-8x = 28

Divide by negative eight on both sides.

$\frac{-8x }{-8}= \frac{28}{-8}$

Reduce both fractions.

The answer is: $-\frac{7}{2}$

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Example Question #273 : Solving Equations

Solve the equation: 6-9x=-2(4-7x)

Possible Answers:

$-\frac{2}{23}$

$\frac{12}{41}$

$\frac{14}{23}$

$\frac{7}{46}$

$\frac{14}{9}$

Correct answer:

$\frac{14}{23}$

Explanation:

Distribute the negative two on the right side.

The equation becomes:

6-9x= -8+14x

Add on both sides.

6-9x+9x= -8+14x+9x

6=23x-8

Add 8 on both sides.

6+8=23x-8+8

14= 23x

Divide by 23 on both sides.

$\frac{14}{23}=\frac{ 23x}{23}$

The answer is: $\frac{14}{23}$

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Example Question #274 : Solving Equations

Solve the equation: -8(3x+5)=-14+2x

Possible Answers:

$-\frac{27}{13}$

$\frac{27}{13}$

$\frac{1}{7}$

$\frac{13}{11}$

Correct answer:

Explanation:

Distribute the negative eight through the binomial.

-8(3x)+(-8)(5)=-14+2x

Simplify the left side.

-24x-40 = -14+2x

Add 24x on both sides.

-24x-40+24x = -14+2x+24x

Simplify both sides.

-40 =-14+26x

Add 14 on both sides.

-40 +14=-14+26x+14

Simplify both sides.

-26=26x

Divide by 26 on both sides

$\frac{-26}{26}=\frac{26x}{26}$

The answer is:

Report an Error

Example Question #751 : Basic Single Variable Algebra

Solve the equation: $-8(2x-3) = \frac{1}{5}$

Possible Answers:

$\frac{29}{16}$

$\frac{119}{80}$

$-\frac{121}{80}$

$\frac{121}{80}$

$-\frac{29}{16}$

Correct answer:

$\frac{119}{80}$

Explanation:

Multiply the negative eight through both terms of the binomial.

$-8(2x)-(-8)(3) = \frac{1}{5}$

Simplify the equation.

$-16x+24 = \frac{1}{5}$

Multiply by five on both sides to eliminate the fraction on the right side.

$5(-16x+24 )= \frac{1}{5} \cdot 5$

-80x+120 = 1

Subtract 120 on both sides.

-80x=-119

Divide by negative 80 on both sides. The double negatives will negate.

The answer is: $\frac{119}{80}$

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Example Question #276 : Solving Equations

Solve the equation: 8x-34 = -13

Possible Answers:

$-\frac{47}{8}$

$-\frac{21}{8}$

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

$\frac{47}{8}$

$\frac{21}{8}$

Correct answer:

$\frac{21}{8}$

Explanation:

Add 34 on both sides of the equation.

8x-34 +34= -13+34

The equation becomes:

8x=21

Divide by eight on both sides.

$\frac{8x}{8}=\frac{21}{8}$

The answer is: $\frac{21}{8}$

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Example Question #752 : Basic Single Variable Algebra

Solve the equation: 5x-7 = 35-4x

Possible Answers:

$\frac{14}{3}$

$\frac{28}{3}$

$\frac{52}{9}$

Correct answer:

$\frac{14}{3}$

Explanation:

Add on both sides.

5x-7+4x = 35-4x+4x

Combine like terms. The equation will become:

9x-7 = 35

Add seven on both sides.

9x-7+7 = 35+7

9x=42

Divide by nine on both sides.

$\frac{9x}{9}=\frac{42}{9}$

Reduce the fraction.

The answer is: $\frac{14}{3}$

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Example Question #751 : Basic Single Variable Algebra

Solve the equation: $\frac{1}{20}x+\frac{2}{15}= \frac{1}{30}$

Possible Answers:

$\frac{5}{2}$

Correct answer:

Explanation:

In order to eliminate the complications of converting fractions, we can multiply both sides by the least common denominator.

Write out the multiples of the denominators.

15-[15,30,45,60]

20-[20,40,60]

30-[30,60]

Multiply both sides of the equation by 60.

$60(\frac{1}{20}x+\frac{2}{15})= 60(\frac{1}{30})$

Simplify both sides.

3x+8 = 2

Subtract eight from both sides.

3x+8 -8= 2-8

3x=-6

Divide by three on both sides.

$\frac{3x}{3}=\frac{-6}{3}$

Simplify both sides.

The answer is:

Report an Error

Example Question #279 : Solving Equations

Solve the equation: -8x = -8(-x+3)

Possible Answers:

$\textup{No solution.}$

$-\frac{3}{2}$

$-\frac{3}{16}$

$\frac{3}{2}$

Correct answer:

$\frac{3}{2}$

Explanation:

Use distribution to simplify the right side.

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

Simplify the parentheses.

-8x =8x-24

Subtract on both sides.

-8x -8x=8x-24-8x

-16x = -24

Divide by negative 16 on both sides.

$\frac{-16x }{-16}= \frac{-24}{-16}$

Reduce both fractions.

The answer is: $\frac{3}{2}$

Report an Error

Example Question #280 : Solving Equations

Solve the equation: $\frac{x}{5}+\frac{5}{3} = 5$

Possible Answers:

$\frac{25}{3}$

$\frac{75}{4}$

$\frac{50}{3}$

$\frac{1}{5}$

$\frac{35}{3}$

Correct answer:

$\frac{50}{3}$

Explanation:

In order to eliminate the fractions, we can multiply both sides by the least common denominator.

The least common denominator can be determined by multiplying the two denominators together.

$5\times 3 = 15$

$(15)(\frac{x}{5}+\frac{5}{3}) = 5(15)$

The equation becomes:

3x+25 = 75

Subtract both sides by 25.

3x+25 -25= 75-25

3x=50

Divide by three on both sides.

$\frac{3x}{3}=\frac{50}{3}$

The answer is: $\frac{50}{3}$

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Algebra II : Basic Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #271 : Solving Equations

Example Question #272 : Solving Equations

Example Question #273 : Solving Equations

Example Question #274 : Solving Equations

Example Question #751 : Basic Single Variable Algebra

Example Question #276 : Solving Equations

Example Question #752 : Basic Single Variable Algebra

Example Question #751 : Basic Single Variable Algebra

Example Question #279 : Solving Equations

Example Question #280 : Solving Equations

All Algebra II Resources

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