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Algebra II : Algebra II

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Example Questions

Example Question #153 : Basic Single Variable Algebra

Simplify: $\frac{\frac{y}{x(x+1)}}{1+\frac{y}{x(x+1)}}$

Possible Answers:

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

$\frac{y}{x+y}$

$\frac{x}{x+y}$

$\frac{y}{x^2+x+y}$

$\frac{x}{x^2+x+y}$

Correct answer:

$\frac{y}{x^2+x+y}$

Explanation:

Simplify the complex fraction by parts.

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

Replace the terms.

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

Simplify the denominator. Convert the integer using the least common denominator, which is x^2+x .

$1+\frac{y}{x^2+x}=\frac{x^2+x}{x^2+x}+\frac{y}{x^2+x} = \frac{x^2+x+y}{x^2+x}$

Replace this term in the denominator.

$\frac{\frac{y}{x^2+x}}{1+\frac{y}{x^2+x}}=\frac{\frac{y}{x^2+x}}{ \frac{x^2+x+y}{x^2+x}}$

Rewrite this using a division sign.

$\frac{\frac{y}{x^2+x}}{ \frac{x^2+x+y}{x^2+x}} = \frac{y}{x^2+x} \div\frac{x^2+x+y}{x^2+x}$

Change the sign to multiplication and take the reciprocal of the second term.

$\frac{y}{x^2+x} \cdot \frac{x^2+x}{x^2+x+y}$

Cancel the common terms.

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

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Example Question #2001 : Algebra Ii

Evaluate a^2+b^2 if 2a+2 =4 and b=5 .

Possible Answers:

Correct answer:

Explanation:

The value of will need to be solved for. Subtract two from both sides given the following equation.

2a+2 =4

2a+2 -2=4-2

2a=2

Divide by two on both sides.

$\frac{2a}{2}=\frac{2}{2}$

a=1

Now that we know the value of a and b, we substitute both values in the expression a^2+b^2 .

a^2+b^2 = 1^2+5^2 = 1+25= 26

The answer is:

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Example Question #2001 : Algebra Ii

Simplify: $\frac{b}{a^2}(a^2b-2a)$

Possible Answers:

$b-\frac{2b}{a}$

$b^2-\frac{2b}{a}$

$b-\frac{2}{a}$

$2b-\frac{2b}{a}$

a^2b^2-1

Correct answer:

$b^2-\frac{2b}{a}$

Explanation:

Distribute the fraction through both terms in the parentheses.

$\frac{b}{a^2}(a^2b-2a)=\frac{b}{a^2}(a^2b)-\frac{b}{a^2}(2a)$

Simplify the terms.

The answer is: $b^2-\frac{2b}{a}$

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Example Question #1 : Variables

Factor this expression: x^2 - 4x - 12

Possible Answers:

Cannot be factored

(x + 2) (x - 6)

(x + 6) (x - 2)

(x + 2) (x - 3)

(x + 2) (x - 4)

Correct answer:

(x + 2) (x - 6)

Explanation:

When we factor, we have to remember to check the signs in the trinomial. In this case, we have minus 4 and minus 12. That automatically tells us the signs in the factors must be opposite, one plus and one minus.

Next, we ask ourselves what are the factors of 12? We get 2 & 6, 1 & 12, and 4 & 3.

Then, we ask ourselves, which of these, when subtracted in a given order, will give us -4? The answer is 6 and 2! So we place these in the parentheses with the 's that we know go there so it looks like

$\left ( x\ \ \ 2 \right )\left ( x\ \ \ 6 \right )$

Finally, we ask ourselves, what signs do we need to put in to get negative 4 and negative 12? We need a positive but a negative so we put the addition sign in with the 2 and the negative sign in with the 6!

$\left ( x+2 \right )(x-6)$

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Example Question #163 : Basic Single Variable Algebra

Given that $x\neq 0$ or , simplify:

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

Possible Answers:

(x+1)^2

$\frac{1}{x}$

x+1

Correct answer:

$\frac{1}{x}$

Explanation:

Step 1: Find a common denominator for terms in numerator

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

Step 2: Divide

$\frac{\frac{x+1}{x}}{x+1}=\frac{x+1}{x}*\frac{1}{x+1}=\frac{1}{x}$

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Example Question #1 : Solving Expressions

Solve for x.

3x(8)-5=-2(3x+5)

Possible Answers:

-1/6

-5/18

1/6

Correct answer:

-1/6

Explanation:

a. Simplify each side of the equation using the distributive property.

24x-5=-6x-10

b. Add 6x to both sides of the equation to move all terms with "x" to the left side of the equation.

30x-5=-10

c. Add 5 to both sides of the equation to move all constants to the right side of the equation.

30x=-5

d. Divide both sides of the equation by 30 to isolate the variable. Simplify the resulting fraction

x=-5/30=-1/6

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Example Question #121 : Expressions

If x=4 , simplify $x^{2}+3x-4$ .

Possible Answers:

Correct answer:

Explanation:

First, you substitute for : $(4^{2})+3(4)-4$

Next, use PEMDAS (Parentheses, Exponents, Multiplication, Dividion, Addition, and Subtraction) to preform the algebraic operations in the correct order. When we apply this rule to simplify we get the following:

16+12-4

28-4

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Example Question #2002 : Algebra Ii

Solve for if z=2 . $3z+y^{2}=22$ .

Possible Answers:

$\pm 4$

$\pm 8$

Correct answer:

$\pm 4$

Explanation:

First, substitute 2 for z: $3(2)+y^{2}=22$ .

Then, simplify: $6+y^{2}=22$ .

Next, you must isolate y by moving all other numbers and variables to the other side of the equation: $6+y^{2}-6=22-6$ , which gives you $y^{2}=22-6$ .

And simplify: $y^{2}=16$ .

Here, we then take the square root of both sides: $\sqrt{y^{2}}=\sqrt{16}$ .

Simplfy: $y=\pm 4$ , because both $-4\cdot-4=16$ and $4\cdot4=16$ .

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Example Question #122 : Expressions

Simplify $z^{3}+x^{4}y^{2}-y^{3}+13$ given z=1, x=2, and y=3 .

Possible Answers:

131

149

Correct answer:

131

Explanation:

First, substitute 1 for z, 2 for x and 3 for y: $1^{3}+2^{4}3^{2}-3^{3}+13$ and simplify: $1+16\cdot9-27+13$ .

Using PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction), we simplify the multiplication: 1+144-27+13 .

Then add and subtract from left to right: 145-27+13=118+13=131 .

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Example Question #2003 : Algebra Ii

Emily buys a rose plant when it is inches tall. The tag indicates that it will grow inches every year. She also buys a tulip plant when it is inches tall. The tag indicates that it will grow inches a year. After how many years are the two plants the same height?

Possible Answers:

$23 \text{ years}$

They will never be the same height at the same time.

$6 \text{ years}$

$16 \text{ years}$

$7 \text{ years}$

Correct answer:

$6 \text{ years}$

Explanation:

We can express each plant's growth as a function of years in the following equations:

Rose height after x years = 3x+5

Tulip height after x years = 2x+11

Since we are looking for the year when the two plants are of equal height, we set these expressions for height equal to each other, and solve for x:

3x+5=2x+11

Combining like terms by subtracting 2x from both sides and subtracting 5 from both sides gives us:

x=6

The plants will reach the same height after 6 years of growth.

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #153 : Basic Single Variable Algebra

Example Question #2001 : Algebra Ii

Example Question #2001 : Algebra Ii

Example Question #1 : Variables

Example Question #163 : Basic Single Variable Algebra

Example Question #1 : Solving Expressions

Example Question #121 : Expressions

Example Question #2002 : Algebra Ii

Example Question #122 : Expressions

Example Question #2003 : Algebra Ii

All Algebra II Resources

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