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Algebra II : Solving Equations

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #2561 : Algebra Ii

Solve the equation: -5x-80 = 15

Possible Answers:

Correct answer:

Explanation:

Add 80 on both sides.

-5x-80+80 = 15+80

Simplify both sides.

-5x=95

Divide by negative five on both sides.

$\frac{-5x}{5}=\frac{95}{5}$

The answer is:

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Example Question #2562 : Algebra Ii

Solve the equation: 6x-19 = -36

Possible Answers:

$\frac{17}{3}$

$-\frac{2}{3}$

$-\frac{17}{3}$

$-\frac{17}{6}$

Correct answer:

$-\frac{17}{6}$

Explanation:

Add 19 on both sides.

6x-19 +19= -36+19

6x=-17

Divide by six on both sides.

$\frac{6x}{6}=\frac{-17}{6}$

The answer is: $-\frac{17}{6}$

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Example Question #2563 : Algebra Ii

Solve: 9x-23=41

Possible Answers:

$\frac{64}{9}$

$\frac{7}{3}$

$-\frac{7}{3}$

Correct answer:

$\frac{64}{9}$

Explanation:

Add 23 on both sides.

9x-23+23=41+23

Simplify both sides.

9x=64

Divide by nine on both sides.

$\frac{9x}{9}=\frac{64}{9}$

The answer is: $\frac{64}{9}$

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Example Question #2564 : Algebra Ii

Solve the equation: 8x-18= 4x-7

Possible Answers:

$\frac{9}{4}$

$\frac{11}{4}$

$\frac{11}{12}$

$\frac{25}{12}$

$-\frac{25}{12}$

Correct answer:

$\frac{11}{4}$

Explanation:

Subtract on both sides.

8x-18-4x= 4x-7-4x

Simplify both sides of the equation.

4x-18=-7

Add 18 on both sides.

4x-18+18=-7+18

4x=11

Divide by four on both sides.

$\frac{4x}{4}=\frac{11}{4}$

The answer is: $\frac{11}{4}$

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Example Question #2565 : Algebra Ii

Solve the equation: $\frac{3}{x^2}-3 =6$

Possible Answers:

$\pm \frac{\sqrt3}{3}$

$\pm \frac{\sqrt3}{6}$

$\pm \sqrt3$

$\textup{There is no solution.}$

$\pm1$

Correct answer:

$\pm \frac{\sqrt3}{3}$

Explanation:

Add three on both sides.

$\frac{3}{x^2}-3+3 =6+3$

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

Multiply by x^2 on both sides.

3=9x^2

Divide by 9 on both sides.

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

$x^2 =\frac{1}{3}$

Square root both sides.

$\sqrt{x^2 }=\pm\sqrt{\frac{1}{3}}$

Rationalize the denominator.

$x=\pm \frac{1}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}} = \pm \frac{\sqrt3}{3}$

The answer is: $\pm \frac{\sqrt3}{3}$

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Example Question #2566 : Algebra Ii

Solve the equation: 12x-2 = 48x+7

Possible Answers:

$-\frac{1}{4}$

$-\frac{7}{18}$

$\frac{1}{4}$

$\frac{5}{36}$

$-\frac{3}{25}$

Correct answer:

$-\frac{1}{4}$

Explanation:

Subtract 12x on both sides.

12x-2 -12x= 48x+7-12x

-2 = 36x+7

Subtract seven on both sides.

-2-7 = 36x+7-7

-9=36x

Divide by 36 on both sides.

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

The answer is: $-\frac{1}{4}$

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Example Question #2567 : Algebra Ii

Solve: 3+5x=7x+4

Possible Answers:

$-\frac{1}{2}$

$-\frac{1}{3}$

$\frac{1}{12}$

$-\frac{1}{4}$

$-\frac{1}{12}$

Correct answer:

$-\frac{1}{2}$

Explanation:

Isolate the x-variables and integers on separate sides of the equation.

Subtract on both sides.

3+5x-5x=7x+4-5x

3= 2x+4

Subtract on both sides.

3-4= 2x+4-4

-1=2x

Divide by two on both sides.

$\frac{-1}{2}=\frac{2x}{2}$

The answer is: $-\frac{1}{2}$

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Example Question #2568 : Algebra Ii

Solve the equation: 4(3-8x) = 2x

Possible Answers:

$\frac{6}{5}$

$\frac{3}{32}$

$\frac{6}{17}$

$\frac{3}{17}$

$\frac{3}{5}$

Correct answer:

$\frac{6}{17}$

Explanation:

Distribute the four through the binomial.

The equation becomes: 12-32x=2x

Add 32x on both sides.

12-32x+32x=2x+32x

12=34x

Divide by 34 on both sides.

$\frac{12}{34}=\frac{34x}{34}$

Reduce both fractions.

$\frac{12}{34}=\frac{6\times 2}{17\times 2}$

The answer is: $\frac{6}{17}$

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Example Question #2569 : Algebra Ii

Solve the equation: 10x-31 = 40-5x

Possible Answers:

$\frac{9}{5}$

$\frac{3}{5}$

$-\frac{3}{5}$

$\frac{71}{15}$

$-\frac{9}{5}$

Correct answer:

$\frac{71}{15}$

Explanation:

Add on both sides of the equation.

10x-31+(5x) = 40-5x+(5x)

The equation becomes:

15x-31= 40

Add 31 on both sides.

15x-31+31= 40+31

15x=71

Divide by 15 on both sides.

$\frac{15x}{15}=\frac{71}{15}$

The answer is: $\frac{71}{15}$

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Example Question #241 : Solving Equations

Solve for : $\begin{Bmatrix} 2x-3y=8\\ 6x-y=4 \end{Bmatrix}$

Possible Answers:

$\frac{1}{4}$

$-\frac{5}{2}$

$\frac{1}{3}$

$-\frac{2}{5}$

Correct answer:

$\frac{1}{4}$

Explanation:

Use the elimination method to eliminate the y-variable.

Multiply the second equation by three.

$\begin{Bmatrix} 2x-3y=8\\ 3[6x-y=4] \end{Bmatrix}\rightarrow\begin{Bmatrix} 2x-3y=8\\ 18x-3y =12 \end{Bmatrix}$

By subtracting the first and second equations, we can eliminate the y-variable.

The equation becomes:

2x-18x = 8-12

-16x = -4

Divide by negative sixteen on both sides.

$\frac{-16x }{-16}=\frac{ -4}{-16}$

The answer is: $\frac{1}{4}$

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Algebra II : Solving Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #2561 : Algebra Ii

Example Question #2562 : Algebra Ii

Example Question #2563 : Algebra Ii

Example Question #2564 : Algebra Ii

Example Question #2565 : Algebra Ii

Example Question #2566 : Algebra Ii

Example Question #2567 : Algebra Ii

Example Question #2568 : Algebra Ii

Example Question #2569 : Algebra Ii

Example Question #241 : Solving Equations

All Algebra II Resources

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