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Algebra II : Algebra II

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 #162 : Solving Equations

Solve for .

x+178.7=205

Possible Answers:

198.4

383.7

43.1

26.3

269.4

Correct answer:

26.3

Explanation:

To solve for the variable isolate it on one side of the equation by moving all other constants to the other side. To do this, perform opposite operations to manipulate the equation.

x+178.7=205

Subtract 178.7 on both sides.

x=26.3

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Example Question #171 : Solving Equations

Solve the equation: 3x-6= 14x-7

Possible Answers:

$-\frac{13}{17}$

$\frac{13}{11}$

$-\frac{13}{11}$

$-\frac{1}{11}$

$\frac{1}{11}$

Correct answer:

$\frac{1}{11}$

Explanation:

Group the integers and the x-values on one side of the equation. Subtract on both sides.

3x-6-(3x)= 14x-7-(3x)

Simplify both sides.

-6=11x-7

Add seven on both sides.

-6+(7)=11x-7+(7)

Simplify both sides.

1=11x

Divide by 11 on both sides to isolate the x-variable.

The answer is: $x=\frac{1}{11}$

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Example Question #172 : Solving Equations

Set up the equation and solve: Three times the quantity of six more than a number is ten.

Possible Answers:

$\frac{8}{3}$

$-\frac{8}{3}$

$\frac{4}{3}$

Correct answer:

$-\frac{8}{3}$

Explanation:

Break up the statement into parts.

Six more than a number: x+6

The quantity of six more than a number: (x+6)

Three times the quantity of six more than a number: 3(x+6)

Is ten: =10

The equation is: 3(x+6)=10

Simplify by distribution.

3x+18 = 10

Subtract 18 from both sides.

3x+18 -18= 10-18

Simplify both sides.

3x=-8

Divide by three on both sides.

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

Simplify both fractions.

The answer is: $x=-\frac{8}{3}$

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Example Question #172 : Solving Equations

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

Possible Answers:

None of these answers.

$x=\frac{3}{4}x$

x=4

x=6

x=1

Correct answer:

x=4

Explanation:

$2x+\frac{3}{2}x=14$

Combine the x terms by finding a common denominator and adding:

$\frac{7}{2}x=14$

Multiply both sides by 2:

7x=28

Divide both sides by 7:

x=4

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Example Question #173 : Solving Equations

Solve: $6x+\frac{5}{2}=\frac{29}{2}$

Possible Answers:

x=4

x=2

x=4.5

x=3

x=2.5

Correct answer:

x=2

Explanation:

$6x+\frac{5}{2}=\frac{29}{2}$

Isolate the x term by subtracting:

$6x=\frac{29}{2}-\frac{5}{2}\rightarrow 6x=\frac{24}{2}$

Reduce the fraction and divide both sides by 6:

6x=12

$\mathbf{x=2}$

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Example Question #174 : Solving Equations

Solve: $\sqrt{2x-2}=4$

Possible Answers:

None of the other answers

x=9

x=6

x=8

x=6

Correct answer:

x=9

Explanation:

$\sqrt{2x-2}=4$

Square both sides to eliminate the square root:

$(\sqrt{2x-2})^2=(4)^2$

2x-2=16

Isolate the x term by subtracting:

2x=18

Divide both sides by 2:

$\boldsymbol{x=9}$

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Example Question #2491 : Algebra Ii

Solve: $\frac{1}{2}x-\frac{1}{4}=\frac{1}{8}$

Possible Answers:

$x=\frac{3}{4}$

$x=\frac{1}{4}$

None of these.

$x=\frac{1}{6}$

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

Correct answer:

$x=\frac{3}{4}$

Explanation:

$\frac{1}{2}x-\frac{1}{4}=\frac{1}{8}$

Isolate the x term:

$\frac{1}{2}x=\frac{1}{8}+\frac{1}{4}$

Find a common denominator to add the fractions:

$\frac{1}{2}x=\frac{1}{8}+\frac{2}{8}$

$\frac{1}{2}x=\frac{3}{8}$

Multiply both sides by 2 to completely isolate x:

$\boldsymbol{x=\frac{3}{4}}$

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Example Question #176 : Solving Equations

Solve: $(\frac{1}{3})^2x+\frac{1}{3}=3$

Possible Answers:

x=24

x=20

x=12

None of these.

x=16

Correct answer:

x=24

Explanation:

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

Simplify squared terms:

$\frac{1}{9}x+\frac{1}{3}=3$

Isolate the x term:

$\frac{1}{9}x=3-\frac{1}{3}$

Simplify right side:

$\frac{1}{9}x=\frac{8}{3}$

Multiply both sides by 9:

$\boldsymbol{x=24}$

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Example Question #2492 : Algebra Ii

Solve: $\sqrt{x-9}=\sqrt{9}$

Possible Answers:

x=18

x=0

x=-18

x=9

x=-9

Correct answer:

x=18

Explanation:

$\sqrt{x-9}=\sqrt{9}$

Square both sides:

x-9=9

Add 9 to both sides:

$\mathbf{x=18}$

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Example Question #2499 : Algebra Ii

Solve for : xyz+2y+x=y

Possible Answers:

$y=\frac{x}{xz+1}$

$y=-\frac{x}{xz+1}$

$y=-\frac{x}{x+z+1}$

$y=-\frac{x}{z+1}$

$y=-\frac{1}{xz+1}$

Correct answer:

$y=-\frac{x}{xz+1}$

Explanation:

In order to solve for the variable, we will need to isolate all terms on one side of the equation so that we can pull out a common factor.

Subtract and on both sides to move the from the right side of the equation, and the from the left to the right side.

xyz+2y+x-[x]-(y)=y-(y)-[x]

Simplify both sides.

xyz+2y-y=-x

xyz+y=-x

Pull out a common factor on the left side.

y(xz+1)=-x

Divide both sides by the quantity (xz+1) on both sides of the equation.

$\frac{y(xz+1)}{(xz+1)}=\frac{-x}{(xz+1)}$

The answer is: $y=-\frac{x}{xz+1}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #162 : Solving Equations

Example Question #171 : Solving Equations

Example Question #172 : Solving Equations

Example Question #172 : Solving Equations

Example Question #173 : Solving Equations

Example Question #174 : Solving Equations

Example Question #2491 : Algebra Ii

Example Question #176 : Solving Equations

Example Question #2492 : Algebra Ii

Example Question #2499 : Algebra Ii

All Algebra II Resources

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