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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #250 : Solving Equations

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

Possible Answers:

$-\frac{2}{5}$

$-\frac{5}{2}$

$\frac{1}{4}$

$\frac{1}{3}$

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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Example Question #251 : Solving Equations

Solve the equation: $-8+\frac{x}{3}= \frac{x}{6}$

Possible Answers:

$\frac{1}{48}$

$\textup{Cannot be determined.}$

$\frac{1}{6}$

Correct answer:

Explanation:

To eliminate both the denominators, we should multiply six on both sides. This will give integers as coefficients instead of fractions, which will require more steps to simplify.

$6(-8+\frac{x}{3})= 6(\frac{x}{6})$

Use the distributive property to simplify the equation.

-48+2x=x

Subtract on both sides.

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

-48=-x

Divide by negative on both sides.

$\frac{-48}{-1}=\frac{-x}{-1}$

The answer is:

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Example Question #252 : Solving Equations

Solve the equation: $\frac{1}{x}+9 = -\frac{8}{3x}$

Possible Answers:

$\frac{5}{27}$

$-\frac{18}{31}$

$-\frac{11}{27}$

$\frac{1}{3}$

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

Correct answer:

$-\frac{11}{27}$

Explanation:

Without having to determine the least common denominator, one way to eliminate the fractions is by multiplying both sides by .

$3x(\frac{1}{x}+9 )= -(\frac{8}{3x})(3x)$

Simplify both sides.

3+27x = -8

Subtract three on both sides.

3+27x -3= -8-3

27x=-11

Divide by on both sides.

$\frac{27x}{27}=\frac{-11}{27}$

The answer is: $-\frac{11}{27}$

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Example Question #381 : Equations

Solve the equation: 8x-30 = -8x - 16

Possible Answers:

$\frac{17}{8}$

$-\frac{23}{4}$

$\frac{7}{8}$

$\textup{No solution.}$

$\frac{17}{4}$

Correct answer:

$\frac{7}{8}$

Explanation:

Add on both sides.

8x-30+8x = -8x - 16+8x

16x-30 = -16

Add 30 on both sides.

16x-30+30 = -16+30

Simplify both sides.

16x = 14

Divide by 16 on both sides.

$\frac{16x}{16} = \frac{14}{16} = \frac{7}{8}$

The answer is: $\frac{7}{8}$

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Example Question #2571 : Algebra Ii

Solve: 2x-4x-8=16

Possible Answers:

None of the other answers.

x=-8

x=-4

x=-12

Correct answer:

x=-12

Explanation:

2x-4x-8=16

Combine x terms:

-2x-8=16

Isolate the x term:

-2x=24

Divide both sides by -2:

$\boldsymbol{x=-12}$

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Example Question #732 : Basic Single Variable Algebra

$\frac{1}{5}x-5=20$

Possible Answers:

x=7.5

x=125

x=5

x=12.5

x=75

Correct answer:

x=125

Explanation:

$\frac{1}{5}x-5=20$

Isolate the x term:

$\frac{1}{5}x=25$

Multiply by the reciprocal of 1/5:

$\frac{5}{1}*\frac{1}{5}x=25*\frac{5}{1}\rightarrow \boldsymbol{x=125}$

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Example Question #255 : Solving Equations

Solve: $\frac{6}{6}x+5x+12=24$

Possible Answers:

None of the other answers.

x=4

x=0

x=6

x=2

Correct answer:

x=2

Explanation:

$\frac{6}{6}x+5x+12=24$

Combine the x terms:

$x+5x+12=24\rightarrow 6x+12=24$

Isolate the x term:

6x=12

Solve for x:

$\boldsymbol{x=2}$

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Example Question #256 : Solving Equations

$\frac{1}{3}x+\frac{2}{9}x-12=16$

Possible Answers:

No solution

x=45.0

x=50.4

None of these

x=54.0

Correct answer:

x=50.4

Explanation:

$\frac{1}{3}x+\frac{2}{9}x-12=16$

Find a common denominator between the fractions:

$\frac{3}{9}x+\frac{2}{9}x-12=16$

Combine x terms:

$\frac{5}{9}x-12=16$

Isolate the x term:

$\frac{5}{9}x=28$

Solve for x by multiplying by the reciprocal:

$x=\frac{252}{5}$ or $\boldsymbol{x=50.4}$

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Example Question #2572 : Algebra Ii

5x-4.9x=10

Possible Answers:

x=10

x=100

x=1

x=.01

x=.1

Correct answer:

x=100

Explanation:

5x-4.9x=10

Combine x terms:

.1x=10

Multiply by the reciprocal to isolate x:

$\frac{1}{10}x=10\rightarrow 10*\frac{1}{10}x=10*10\rightarrow \boldsymbol{x=100}$

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Example Question #258 : Solving Equations

Solve for : $\begin{Bmatrix} y=3x-7\\2y = 4x-8 \end{Bmatrix}$

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

In order to solve for y, we will want to eliminate the x-variable.

Multiply the top equation by four and the the bottom equation by three. This will allow us to subtract both equations and eliminate the x-terms.

$\begin{Bmatrix} 4[y=3x-7]\\3[2y = 4x-8] \end{Bmatrix}\rightarrow\begin{Bmatrix} 4y=12x-28\\6y = 12x-24 \end{Bmatrix}$

Subtract the first and second equation. The terms will become one equation:

-2y = -4

Divide by negative two on both sides.

$\frac{-2y}{-2} = \frac{-4}{-2}$

The answer is:

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #250 : Solving Equations

Example Question #251 : Solving Equations

Example Question #252 : Solving Equations

Example Question #381 : Equations

Example Question #2571 : Algebra Ii

Example Question #732 : Basic Single Variable Algebra

Example Question #255 : Solving Equations

Example Question #256 : Solving Equations

Example Question #2572 : Algebra Ii

Example Question #258 : Solving Equations

All Algebra II Resources

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