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Algebra II : Solving Rational Expressions

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Example Questions

Example Question #113 : Solving Rational Expressions

If $f(x)=\frac{(x+3)}{(x-2)}$ and $g(x)=\frac{(x-2)}{(x+4)}$ , what is f(x)+g(x) ?

Possible Answers:

$\frac{2x^2 -3x+16}{x^2 +2x-8}$

$\frac{x^{2}+7x+12}{x^2+4x-8}$

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

$\frac{x^2+x-6}{x^2+2x-8}$

2x^2 +4x -5

Correct answer:

$\frac{2x^2 -3x+16}{x^2 +2x-8}$

Explanation:

To begin, let's write the whole problem out plainly:

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

From here, in order to add the two fractions, we need to get the denominators to be the same. To do this, we multiply the numerator and denominator of each fraction by the denominator in the OTHER fraction:

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

Now we can add the fractions together:

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

Now we can expand using FOIL:

$\frac{(x^2+7x+12)+(x^2 -4x+4)}{(x^2+2x-8)}$

Finally, we collect like terms in the numerator for a final answer of:

$f(x)+g(x)=\frac{2x^2-3x+16}{x^2+2x-8}$

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Example Question #681 : Intermediate Single Variable Algebra

If $f(x)=\frac{x^2 +3x +8}{x+2}$ , what is f(4) ?

Possible Answers:

$\frac{14}{9}$

$\frac{32}{5}$

Undefined

Correct answer:

Explanation:

First we're going to give the function a quick glance to see if we can simplify by factoring. We can't, so in order to solve, we're going to replace all the 's with the number we want to solve for, :

$f(4)=\frac{(4)^2+3(4)+8}{(4)+2}$

$f(4)=\frac{16+12+8}{4+2}$

$f(4)=\frac{36}{6}$

f(4)=6

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Example Question #1823 : Algebra Ii

Solve for x: $\sqrt{x+20}=x$

Possible Answers:

x=-4,-5

x=-4,5

x=4,-5

x=4,5

Correct answer:

x=-4,5

Explanation:

Step 1: We need to square both sides of the equation. Squaring a square root will just leave me with whatever is inside the root.

$(\sqrt{x+20})^2=x^2$

$\rightarrow x+20=x^2$

Step 2: Move all terms on the left side to the right side.

0=x^2-x-20

Step 3: Factor the right side. We want two numbers that multiply to and add to .

$4\cdot-5=-20, 4-5=-1$

Rewrite the trinomial as two binomals:

(x+4)(x-5)=0

Step 4: Set each parentheses equal to and solve for :

(x+4)=0, x=-4
(x-5)=0, x=5

The solutions to this equation are -4 and 5.

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Example Question #1824 : Algebra Ii

Solve for .

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

Possible Answers:

Correct answer:

Explanation:

To solve for the variable , isolate the variable on one side of the equation with all other constants on the other side. To accomplish this perform the opposite operation to manipulate the equation.

First cross multiply.

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

Next, divide by four on both sides.

4x=4*3=12

$\frac{4x}{4}=\frac{12}{4}=\frac{4\cdot 3}{4}=3$

x=3

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Example Question #1825 : Algebra Ii

Solve: $\frac{x+1}{x}-\frac{4-x}{x^2}$

Possible Answers:

2x+3

$-\frac{x^2-6}{x}$

$\frac{x^2-4}{x^2}$

$\frac{x^2-2x-4}{x^2}$

$\frac{x^2+2x-4}{x^2}$

Correct answer:

$\frac{x^2+2x-4}{x^2}$

Explanation:

The terms of the numerator cannot be added or subtracted without a common denominator.

Convert the first term with a denominator of x^2 .

$\frac{x+1}{x}-\frac{4-x}{x^2} = \frac{(x)(x+1)}{x(x)}-\frac{4-x}{x^2}$

Expand the first term and combine the fractions as one fraction. Be sure to enclose the numerator of the second term in parentheses.

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

Simplify this fraction.

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

The answer is: $\frac{x^2+2x-4}{x^2}$

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Example Question #1826 : Algebra Ii

Solve for .

$\frac{x}{5}=\frac{7}{10}$

Possible Answers:

2.5

3.5

6.5

Correct answer:

3.5

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.

$\frac{x}{5}=\frac{7}{10}$

Cross multiply.

10x=7*5=35

Divide on both sides.

x=3.5

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Example Question #1821 : Algebra Ii

Solve for .

$\frac{2x}{5}=\frac{18}{5}$

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.

$\frac{2x}{5}=\frac{18}{5}$

Cross multiply.

10x=5*18=90

Divide on both sides.

x=9

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Example Question #1828 : Algebra Ii

Solve for .

$\frac{3}{4x}=\frac{9}{144}$

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.

$\frac{3}{4x}=\frac{9}{144}$

Cross multiply.

36x=3*144=432

Divide on both sides.

x=12

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Example Question #691 : Intermediate Single Variable Algebra

Solve for .

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

Possible Answers:

10.5

9.5

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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Algebra II : Solving Rational Expressions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #113 : Solving Rational Expressions

Example Question #681 : Intermediate Single Variable Algebra

Example Question #1823 : Algebra Ii

Example Question #1824 : Algebra Ii

Example Question #1825 : Algebra Ii

Example Question #1826 : Algebra Ii

Example Question #1821 : Algebra Ii

Example Question #1828 : Algebra Ii

Example Question #691 : Intermediate Single Variable Algebra

Example Question #129 : Solving Rational Expressions

All Algebra II Resources

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