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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 #2622 : Algebra Ii

Solve the equation: 9x-17 = -41+3x

Possible Answers:

$\frac{1}{2}$

$\frac{5}{4}$

Correct answer:

Explanation:

Subtract on both sides.

9x-17 -3x= -41+3x-3x

6x-17 = -41

Add 17 on both sides.

6x-17+17 = -41+17

6x=-24

Divide by six on both sides.

$\frac{6x}{6}=\frac{-24}{6}$

The answer is:

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Example Question #432 : Equations

Solve: $9x-6= -\frac{1}{20}$

Possible Answers:

$\frac{119}{180}$

$-\frac{119}{180}$

$\frac{3}{40}$

$-\frac{3}{40}$

$-\frac{121}{180}$

Correct answer:

$\frac{119}{180}$

Explanation:

Multiply both sides by the least common denominator, 20.

$20(9x-6)= -\frac{1}{20}\cdot20$

Simplify this equation.

180x-120 = -1

Add 120 on both sides.

180x-120 +120= -1+120

180x = 119

Divide by 180 on both sides.

$\frac{180x }{180}= \frac{119}{180}$

The answer is: $\frac{119}{180}$

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Example Question #431 : Equations

Solve the equation: $\frac{3x}{2}-4=5$

Possible Answers:

$\frac{11}{3}$

$\frac{2}{3}$

$\frac{1}{18}$

$\frac{1}{5}$

Correct answer:

Explanation:

Add four on both sides.

$\frac{3x}{2}-4+4=5+4$

$\frac{3x}{2} = 9$

To isolate the x-variable, we will need to multiply by two thirds on both sides.

$\frac{3x}{2} \cdot \frac{2}{3} = 9 \cdot \frac{2}{3}$

x=6

The answer is:

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Example Question #301 : Solving Equations

Solve the equation: 5-7x=6+8x

Possible Answers:

$\frac{1}{15}$

$-\frac{1}{15}$

Correct answer:

$-\frac{1}{15}$

Explanation:

Add on both sides.

5-7x+7x=6+8x+7x

5= 6+15x

Subtract six from both sides.

5-6= 6+15x-6

-1=15x

Divide by 15 on both sides.

$\frac{-1}{15}=\frac{15x}{15}$

The answer is: $-\frac{1}{15}$

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Example Question #2626 : Algebra Ii

Solve the equation: -3x+8 = -4x+34

Possible Answers:

$\frac{6}{7}$

$\frac{7}{6}$

Correct answer:

Explanation:

Add on both sides.

-3x+8 +4x= -4x+34+4x

x+8 = 34

Subtract eight on both sides.

x+8 -8= 34-8

The answer is:

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Example Question #301 : Solving Equations

Solve: $\frac{3}{7}+\frac{1}{2x}-\frac{1}{14}=\frac{1}{14x}$

Possible Answers:

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

$-\frac{6}{5}$

$\frac{2}{3}$

$\frac{7}{3}$

$\frac{5}{6}$

Correct answer:

$-\frac{6}{5}$

Explanation:

To reduce calculation while solving this equation, we can multiply both sides by the least common denominator.

Every denominator must have an x-term, and each term is divisible by 14. This means that the LCD is 14x .

Multiply 14x on both sides.

$14x(\frac{3}{7}+\frac{1}{2x}-\frac{1}{14})=(\frac{1}{14x})\cdot 14x$

6x+7-x = 1

Combine like-terms.

5x+7 = 1

Subtract seven from both sides.

5x+7 -7= 1-7

5x=-6

Divide both sides by five.

$\frac{5x}{5}=\frac{-6}{5}$

The answer is: $-\frac{6}{5}$

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Example Question #2628 : Algebra Ii

Solve the equation: 6x= -7(3x+7)

Possible Answers:

$-\frac{49}{27}$

$-\frac{7}{3}$

$\frac{1}{7}$

$-\frac{1}{14}$

$\frac{49}{27}$

Correct answer:

$-\frac{49}{27}$

Explanation:

Distribute the negative seven through the binomial on the right side.

6x= -21x-49

Add 21x on both sides.

6x+21x= -21x-49+21x

Combine like-terms.

27x =- 49

Divide by 27 on both sides.

$\frac{27x}{27} = \frac{-49}{27}$

The answer is: $-\frac{49}{27}$

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Example Question #302 : Solving Equations

Solve: $\frac{7x}{x-7}=x$

Possible Answers:

1,14

0,14

$\textup{There is no solution.}$

$\pm\frac{1}{7}$

Correct answer:

0,14

Explanation:

Multiply both sides by (x-7) to cancel the denominator.

$\frac{7x}{x-7}\cdot (x-7)=x\cdot (x-7)$

Simplify both sides.

7x = x^2-7x

Subtract from both sides.

7x-7x = x^2-7x-7x

0 =x^2-14x

Factor an in the equation.

0 =x(x-14)

The roots are: 0,14

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Example Question #781 : Basic Single Variable Algebra

Solve the equation: $\frac{3x}{2}(\frac{5}{x}+3)=10$

Possible Answers:

$-\frac{7}{3}$

$-\frac{10}{9}$

$\frac{5}{9}$

$-\frac{9}{2}$

$-\frac{3}{2}$

Correct answer:

$\frac{5}{9}$

Explanation:

Use distribution to simplify the left side.

$\frac{3x}{2}(\frac{5}{x})+(\frac{3x}{2})(3)=10$

$\frac{15}{2}+\frac{9x}{2} =10$

Multiply by two on both sides.

$2(\frac{15}{2}+\frac{9x}{2}) =10\cdot 2$

15+9x=20

Subtract 15 on both sides.

15+9x-15=20-15

9x=5

Divide by nine on both sides.

$\frac{9x}{9}=\frac{5}{9}$

The answer is: $\frac{5}{9}$

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Example Question #782 : Basic Single Variable Algebra

Solve: 8(-8x+6) = -x+2

Possible Answers:

$\frac{50}{63}$

$-\frac{16}{9}$

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

$-\frac{46}{63}$

$\frac{46}{63}$

Correct answer:

$\frac{46}{63}$

Explanation:

Solve the left side by distribution.

8(-8x)+8(6) = -x+2

The equation becomes:

-64x+48 =-x+2

Add 64x on both sides.

-64x+48 +64x=-x+2+64x

48= 63x+2

Subtract 2 from both sides.

48-2= 63x+2-2

46= 63x

Divide by 63 on both sides.

$\frac{46}{63}= \frac{63x}{63}$

The answer is: $\frac{46}{63}$

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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 #2622 : Algebra Ii

Example Question #432 : Equations

Example Question #431 : Equations

Example Question #301 : Solving Equations

Example Question #2626 : Algebra Ii

Example Question #301 : Solving Equations

Example Question #2628 : Algebra Ii

Example Question #302 : Solving Equations

Example Question #781 : Basic Single Variable Algebra

Example Question #782 : Basic Single Variable Algebra

All Algebra II Resources

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