Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #121 : Solving Rational Expressions

Solve for \displaystyle x.

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

Possible Answers:

\displaystyle 5.55

\displaystyle 8.55

\displaystyle 10.75

\displaystyle 11.25

\displaystyle 12

Correct answer:

\displaystyle 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.

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

Cross multiply.

\displaystyle 20(x-6)=21*5 

Remember we are multiplying \displaystyle 20 with the expression. Now distribute.

\displaystyle 20x-120=105 

Add \displaystyle 120 on both sides.

\displaystyle 20x=225 

Divide \displaystyle 20 on both sides.

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

Example Question #32 : Solving Rational Expressions

Solve for \displaystyle x.

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

Possible Answers:

\displaystyle 31

\displaystyle 26

\displaystyle 29

\displaystyle 33

\displaystyle 24

Correct answer:

\displaystyle 29

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.

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

Cross multiply.

\displaystyle 15(x-5)=72*5 

Remember we are multiplying \displaystyle 15 to the expression. Now distribute.

\displaystyle 15x-75=360 

Add \displaystyle 75 on both sides.

\displaystyle 15x=435 

Divide \displaystyle 15 on both sides.

\displaystyle x=29

Example Question #131 : Solving Rational Expressions

Solve for \displaystyle x.

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

Possible Answers:

\displaystyle 6

\displaystyle -7

\displaystyle 1

\displaystyle -1

\displaystyle -5

Correct answer:

\displaystyle -7

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.

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

Cross multiply.

\displaystyle 3(x+3)=4(x+4) 

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

\displaystyle 3x+9=4x+16 

Subtract \displaystyle 3x and \displaystyle 16 on both sides.

\displaystyle x=-7

Example Question #171 : Rational Expressions

Solve for \displaystyle x.

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

Possible Answers:

\displaystyle -2

\displaystyle -1

\displaystyle 1

\displaystyle 0

\displaystyle 2

Correct answer:

\displaystyle 1

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.

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

Cross multiply.

\displaystyle 3(x+1)=2(x+2) 

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

\displaystyle 3x+3=2x+4 

Subtract \displaystyle 2x and \displaystyle 3 on both sides.

\displaystyle x=1

Example Question #172 : Rational Expressions

Solve for \displaystyle x.

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

Possible Answers:

\displaystyle -1.75

\displaystyle 3.25

\displaystyle 2.25

\displaystyle -3.75

\displaystyle -1.25

Correct answer:

\displaystyle -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.

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

Cross multiply.

\displaystyle 10x=3(2x-5) 

Remember we are multiplying \displaystyle 3 to the expression. Then we distribute.

\displaystyle 10x=6x-15 

Subtract \displaystyle 6x on both sides.

\displaystyle 4x=-15 

Divide \displaystyle 4 on both sides.

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

Example Question #173 : Rational Expressions

Solve for \displaystyle x.

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

Possible Answers:

\displaystyle 2.5

\displaystyle -2.25

\displaystyle 2.4

\displaystyle -2.5

\displaystyle 1.333

Correct answer:

\displaystyle -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.

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

Distribute. Remember to apply FOIL.

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

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

\displaystyle x^2+4x+3=x^2+6x+8 

Subtract \displaystyle x^2 , \displaystyle 6x, and \displaystyle 3 on both sides. 

\displaystyle -2x=5 

Divide \displaystyle -2 on both sides.

\displaystyle x=-2.5

Example Question #174 : Rational Expressions

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

Possible Answers:

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

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

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

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

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

Correct answer:

\displaystyle \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.

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

Convert the fractions.

\displaystyle \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.

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

Simplify the numerator and combine like-terms.

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

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

Example Question #691 : Intermediate Single Variable Algebra

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

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

Possible Answers:

\displaystyle x=4

\displaystyle x=4\ or\ x=-4

\displaystyle x=3

\displaystyle x=-3

Correct answer:

\displaystyle 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. 

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

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

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

\displaystyle 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. 

Example Question #173 : Rational Expressions

Solve:  \displaystyle \frac{2}{5x}+\frac{1}{6x}+\frac{x}{30}

Possible Answers:

\displaystyle \frac{x+17}{30}

\displaystyle \frac{x^3+17x}{30}

\displaystyle \frac{x^2+17}{30x}

\displaystyle \frac{3}{5x}

\displaystyle \frac{x^3+17}{30x}

Correct answer:

\displaystyle \frac{x^2+17}{30x}

Explanation:

Convert the fractions to a common denominator.

\displaystyle \frac{2}{5x}+\frac{1}{6x}+\frac{x}{30} = \frac{2(6)}{5x(6)}+\frac{1(5)}{6x(5)}+\frac{x(x)}{30(x)}

Simplify the top and bottom and combine like terms on the numerator.

\displaystyle \frac{12}{30x}+\frac{5}{30x}+\frac{x^2}{30x} = \frac{x^2+17}{30x}

The answer is:  \displaystyle \frac{x^2+17}{30x}

Example Question #1831 : Algebra Ii

Solve:  \displaystyle \frac{1}{2-x}-\frac{5}{6x}

Possible Answers:

\displaystyle \frac{11x+3}{3x-6}

\displaystyle -\frac{11x-10}{6x^2-12x}

\displaystyle \frac{11x+10}{6x-12}

\displaystyle \frac{11x+10}{6x^2-12x}

\displaystyle -\frac{11x-10}{6x^2-2x}

Correct answer:

\displaystyle -\frac{11x-10}{6x^2-12x}

Explanation:

Find the least common denominator by multiplying both denominators together.

\displaystyle 6x(2-x) =12x-6x^2

Convert the fractions.

\displaystyle \frac{1}{2-x}-\frac{5}{6x} = \frac{1(6x)}{6x(2-x)}-\frac{5(2-x)}{6x(2-x)}

Simplify the numerator and denominator.

\displaystyle \frac{6x}{12x-6x^2}-\frac{10-5x}{12x-6x^2}

Combine both fractions together.  Remember to brace the second numerator in parentheses.

\displaystyle \frac{6x}{12x-6x^2}-\frac{10-5x}{12x-6x^2}=\frac{6x-(10-5x)}{12x-6x^2}

Simplify the fraction.

\displaystyle \frac{11x-10}{12x-6x^2}

Factor out a negative one in the denominator.

\displaystyle \frac{11x-10}{12x-6x^2} =\frac{11x-10}{-1(-12x+6x^2)} = -\frac{11x-10}{6x^2-12x}

The answer is:  \displaystyle -\frac{11x-10}{6x^2-12x}

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