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Algebra II : Basic Single-Variable Algebra

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Example Questions

Example Question #422 : Equations

Solve the equation: $-2x+\frac{1}{9} = \frac{1}{6}$

Possible Answers:

$-\frac{1}{18}$

$\frac{5}{36}$

$-\frac{5}{72}$

$-\frac{1}{36}$

$-\frac{5}{36}$

Correct answer:

$-\frac{1}{36}$

Explanation:

Multiply both sides by the least common denominator in order to eliminate the fractions. The least common denominator is 18.

$18(-2x+\frac{1}{9}) = (\frac{1}{6})\cdot 18$

Simplify both sides.

-36x+2 =3

Subtract two from both sides.

-36x+2-2 =3-2

-36x = 1

Divide by negative 36 on both sides.

$\frac{-36x}{-36} = \frac{1}{-36}$

The answer is: $-\frac{1}{36}$

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Example Question #421 : Equations

Solve the equation: $\frac{1}{3x}+8= \frac{1}{6}$

Possible Answers:

$-\frac{2}{47}$

$-\frac{1}{7}$

$\frac{2}{49}$

$\frac{2}{47}$

$-\frac{2}{49}$

Correct answer:

$-\frac{2}{47}$

Explanation:

In order to solve this equation, we can start by multiplying both sides by the least common denominator in order to avoid having to convert fractions.

$6x(\frac{1}{3x}+8)= \frac{1}{6}\cdot 6x$

Simplify both sides.

2+48x = x

Subtract 48x on both sides to isolate the x-variable.

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

2=-47x

Divide by negative 47 on both sides.

$\frac{2}{-47}=\frac{-47x}{-47}$

The answer is: $-\frac{2}{47}$

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Example Question #422 : Equations

Solve the equation: 2x-16 = -8x+32

Possible Answers:

$\frac{8}{3}$

$\frac{12}{5}$

$\frac{24}{5}$

$-\frac{8}{3}$

Correct answer:

$\frac{24}{5}$

Explanation:

Add on both sides.

2x-16 +8x= -8x+32+8x

10x-16 = 32

Add 16 on both sides.

10x-16 +16= 32+16

10x=48

Divide by 10 on both sides.

$\frac{10x}{10}=\frac{48}{10}$

Simplify both fractions.

The answer is: $\frac{24}{5}$

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Example Question #772 : Basic Single Variable Algebra

Solve the equation: -5x+7 = 12-14x

Possible Answers:

$-\frac{19}{5}$

$-\frac{5}{19}$

$-\frac{5}{9}$

$\frac{5}{9}$

Correct answer:

$\frac{5}{9}$

Explanation:

Add 14x on both sides.

-5x+7 +14x= 12-14x+14x

Simplify both sides.

9x+7 = 12

Subtract seven on both sides.

9x+7 -7= 12-7

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 #421 : Equations

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

Possible Answers:

$-\frac{1}{16}$

$-\frac{1}{4}$

$-\frac{1}{32}$

$-\frac{1}{8}$

Correct answer:

$-\frac{1}{16}$

Explanation:

In order to solve this equation, we will need to first multiply by on both sides.

$-\frac{3}{8x} \cdot 8x= 6\cdot 8x$

-3 = 48x

Divide by 48 on both sides.

$\frac{-3}{48} = \frac{48x}{48}$

Reduce the fractions.

The answer is: $-\frac{1}{16}$

Report an Error

Example Question #2612 : Algebra Ii

Solve the equation: 9-x =6x-7

Possible Answers:

$\frac{2}{7}$

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

$\frac{16}{5}$

$\frac{16}{7}$

$-\frac{16}{5}$

Correct answer:

$\frac{16}{7}$

Explanation:

Add on both sides of the equation.

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

9=7x-7

Add 7 on both sides.

9+(7)=7x-7+(7)

Simplify both sides.

16=7x

Divide by seven on both sides.

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

The answer is: $\frac{16}{7}$

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Example Question #428 : Equations

Solve the equation: $\frac{3}{8x}-\frac{1}{24} = \frac{3}{4}$

Possible Answers:

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

$\frac{19}{9}$

$\frac{17}{9}$

$\frac{9}{17}$

$\frac{9}{19}$

Correct answer:

$\frac{9}{19}$

Explanation:

To eliminate all the fractional terms, we can simplify the equation by multiplying both sides by the least common denominator.

The least common denominator is 24x since this is the minimal term divisible by all the denominators provided.

$24x(\frac{3}{8x}-\frac{1}{24}) = 24x(\frac{3}{4})$

Simplify the equation.

9-x =18x

Add x on both sides.

9-x +x=18x+x

9=19x

Divide by 19 on both sides to isolate x.

$\frac{9}{19}=\frac{19x}{19}$

The answer is: $\frac{9}{19}$

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Example Question #429 : Equations

Solve the equation: $\frac{1}{3x}+\frac{2}{x} = \frac{1}{2}$

Possible Answers:

$\textup{There is no solution.}$

$\frac{7}{6}$

$\frac{14}{3}$

$\frac{17}{6}$

$\frac{8}{3}$

Correct answer:

$\frac{14}{3}$

Explanation:

In order to eliminate the fractions, multiply both sides by the least common denominator. The least common denominator is .

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

Simplify the equation.

$1+6 = \frac{3x}{2}$

Simplify the left side and multiply by two thirds on both sides to isolate x.

$7= \frac{3x}{2}$

$7\cdot \frac{2}{3}= \frac{3x}{2}\cdot \frac{2}{3}$

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

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Example Question #291 : Solving Equations

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

Possible Answers:

$-\frac{5}{3}$

$-\frac{9}{2}$

$\frac{3}{5}$

$\frac{15}{8}$

$\frac{15}{2}$

Correct answer:

$\frac{15}{2}$

Explanation:

Add three on both sides. To add and combine the terms on the right side, this is the same as adding $\frac{12}{4}$ on the right.

$\frac{x}{2}-3 +(3)=\frac{3}{4}+(\frac{12}{4})$

Simplify both sides.

$\frac{x}{2} = \frac{15}{4}$

Multiply by two on both sides to isolate the x-variable.

$\frac{x}{2} \cdot 2= \frac{15}{4}\cdot2$

The answer is: $\frac{15}{2}$

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Example Question #431 : Equations

Solve the equation: $\frac{2}{7x}+8 = \frac{1}{7}$

Possible Answers:

$-\frac{2}{11}$

$-\frac{1}{11}$

$\frac{2}{57}$

$-\frac{2}{55}$

$-\frac{13}{57}$

Correct answer:

$-\frac{2}{55}$

Explanation:

Multiply both sides by the least common denominator to eliminate the fractions.

The least common denominator is:

$7x(\frac{2}{7x}+8 )=( \frac{1}{7})\cdot 7x$

2+56x =x

Subtract 56x on both sides.

2+56x-56x =x-56x

2=-55x

Divide by negative 55 on both sides.

$\frac{2}{-55}=\frac{-55x}{-55}$

The answer is: $-\frac{2}{55}$

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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 #422 : Equations

Example Question #421 : Equations

Example Question #422 : Equations

Example Question #772 : Basic Single Variable Algebra

Example Question #421 : Equations

Example Question #2612 : Algebra Ii

Example Question #428 : Equations

Example Question #429 : Equations

Example Question #291 : Solving Equations

Example Question #431 : Equations

All Algebra II Resources

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