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Algebra II : Intermediate 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 #1829 : Algebra Ii

Solve for .

$\frac{x+1}{3}=\frac{7}{2}$

Possible Answers:

9.5

10.5

Correct answer:

9.5

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.

$\frac{x+1}{3}=\frac{7}{2}$

Cross multiply.

2(x+1)=7*3

Remember you are multiplying with the expression. Now distribute.

2x+2=21

Subtract on both sides.

2x=19

Divide on both sides.

x=9.5

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Example Question #129 : Solving Rational Expressions

Solve for .

$\frac{x+3}{4}=\frac{7}{16}$

Possible Answers:

2.25

-1.25

1.5625

-1.5625

-1.75

Correct answer:

-1.25

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.

$\frac{x+3}{4}=\frac{7}{16}$

Cross multiply.

16(x+3)=7*4

Remember we are multiplying with the expression. Now distribute.

16x+48=28

Subtract on both sides.

16x=-20

Divide on both sides.

$x=-\frac{5}{4}=-1.25$

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Example Question #1831 : Algebra Ii

Solve for .

$\frac{x-6}{5}=\frac{21}{20}$

Possible Answers:

10.75

8.55

5.55

11.25

Correct answer:

11.25

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.

$\frac{x-6}{5}=\frac{21}{20}$

Cross multiply.

20(x-6)=21*5

Remember we are multiplying with the expression. Now distribute.

20x-120=105

Add 120 on both sides.

20x=225

Divide on both sides.

$x=\frac{45}{4}=10.25$

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Example Question #32 : Solving Rational Expressions

Solve for .

$\frac{5}{x-5}=\frac{15}{72}$

Possible Answers:

Correct answer:

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.

$\frac{5}{x-5}=\frac{15}{72}$

Cross multiply.

15(x-5)=72*5

Remember we are multiplying to the expression. Now distribute.

15x-75=360

Add on both sides.

15x=435

Divide on both sides.

x=29

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Example Question #33 : Solving Rational Expressions

Solve for .

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

Possible Answers:

Correct answer:

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.

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

Cross multiply.

3(x+3)=4(x+4)

Remember to multiply 3, 4 to each of the expressions respectively. Then distribute.

3x+9=4x+16

Subtract and on both sides.

x=-7

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Example Question #34 : Solving Rational Expressions

Solve for .

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

Possible Answers:

Correct answer:

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.

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

Cross multiply.

3(x+1)=2(x+2)

Remember we are multiplying 2, 3 to the expressions respectively. Then distribute.

3x+3=2x+4

Subtract and on both sides.

x=1

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Example Question #35 : Solving Rational Expressions

Solve for .

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

Possible Answers:

-3.75

3.25

-1.75

-1.25

2.25

Correct answer:

-3.75

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.

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

Cross multiply.

10x=3(2x-5)

Remember we are multiplying to the expression. Then we distribute.

10x=6x-15

Subtract on both sides.

4x=-15

Divide on both sides.

$x=-\frac{15}{4}=-3.75$

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Example Question #36 : Solving Rational Expressions

Solve for .

$\frac{x+1}{x+2}=\frac{x+4}{x+3}$

Possible Answers:

2.5

1.333

-2.5

2.4

-2.25

Correct answer:

-2.5

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.

$\frac{x+1}{x+2}=\frac{x+4}{x+3}$

Distribute. Remember to apply FOIL.

(x+1)(x+3)=(x+2)(x+4)

x^2+3x+x+3=x^2+4x+2x+8

x^2+4x+3=x^2+6x+8

Subtract x^2 , , and on both sides.

-2x=5

Divide on both sides.

x=-2.5

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Example Question #136 : Solving Rational Expressions

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

Possible Answers:

$\frac{7x-12}{5x^2+20}$

$-\frac{2x-3}{5x^2+20}$

$\frac{12x-12}{5x^2+20}$

$-\frac{1}{6x+4}$

$\frac{13x-12}{5x^2+20}$

Correct answer:

$\frac{7x-12}{5x^2+20}$

Explanation:

To simplify the expressions, we will need a least common denominator.

Multiply the two denominators together to obtain the least common denominator.

5x(4+x) = 20x+5x^2

Convert the fractions.

$\frac{2}{4+x}-\frac{3}{5x} = \frac{2(5x)}{20x+5x^2}-\frac{3(4+x)}{20x+5x^2}$

Combine the fractions as one fraction.

$\frac{2(5x)-3(4+x)}{20x+5x^2}$

Simplify the numerator and combine like-terms.

$\frac{10x-12-3x}{20x+5x^2}= \frac{7x-12}{5x^2+20}$

The answer is: $\frac{7x-12}{5x^2+20}$

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Example Question #137 : Solving Rational Expressions

$Given\ the\ function\ f(x)=\frac{x+3}{x^2-16};$

$which\ values\ of\ x\ could\ not\ be\ part\ of\ any\ solution\ space?$

Possible Answers:

x=3

x=-3

$x=4\ or\ x=-4$

x=4

Correct answer:

$x=4\ or\ x=-4$

Explanation:

When considering the solution space for a rational function, we must look at the denominator.

Any value of x in the denominator that results in a zero cannot be part of the solution space because it is a mathematical impossibility to divide by 0.

$Solving\ the\ denominator\ for\ zero\ gives\ us:$

x^2-16=0 (add 16 to both sides)

x^2=16 (take the square root of both sides)

$x=\pm4$

If we were to plug in a positive or negative 4 into the function, both of these would result in a zero in the denominator, which is a mathematical impossibility.

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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1829 : Algebra Ii

Example Question #129 : Solving Rational Expressions

Example Question #1831 : Algebra Ii

Example Question #32 : Solving Rational Expressions

Example Question #33 : Solving Rational Expressions

Example Question #34 : Solving Rational Expressions

Example Question #35 : Solving Rational Expressions

Example Question #36 : Solving Rational Expressions

Example Question #136 : Solving Rational Expressions

Example Question #137 : Solving Rational Expressions

All Algebra II Resources

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