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High School Math : Algebra II

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Example Questions

Example Question #23 : Solving Equations

is what percent of ?

Possible Answers:

Correct answer:

Explanation:

Verbal cues include "IS" means equals and "OF" means multiplication.

So the equation to solve becomes

$15=P\cdot 75$ and dividing both sides by gives $P=0.20 \; or \; 20\%$

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Example Question #1 : Solving Equations

Solve the system of equations.

4x+y=16

2x+3y=18

Possible Answers:

(-3,4)

(-4,3)

(4,3)

(3,4)

None of the other answers are correct.

Correct answer:

(3,4)

Explanation:

Isolate in the first equation.

y=16-4x

Plug into the second equation to solve for .

2x+3(16-4x)=18

2x+48-12x=18

-10x+48=30

-10x=-30

x=3

Plug into the first equation to solve for .

4(3)+y=16

12+y=16

y=4

Now we have both the and values and can express them as a point: (3,4) .

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Example Question #2 : Solving Equations

Solve for and .

3x+2y=2

2x+2y=4

Possible Answers:

$x=-2\ and\ y=4$

$x=-2\ and\ y=8$

$x=2\ and\ y=-4$

$x=2\ and\ y=-8$

Cannot be determined.

Correct answer:

$x=-2\ and\ y=4$

Explanation:

1st equation: 3x+2y=2

2nd equation: 2x+2y=4

Subtract the 2nd equation from the 1st equation to eliminate the "2y" from both equations and get an answer for x:

Plug the value of into either equation and solve for :

2(-2)+2y=4

-4+2y=4

2y=4+4

2y=8

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Example Question #31 : Equations

Solve for :

$\left | 5x + 39 \right | = 44$

Possible Answers:

$x = 16\frac{3}{5}$

$x = -1 \textrm{ or } x = 1$

$x = -16\frac{3}{5} \textrm{ or } x = 1$

$x = 1 \textrm{ or }x = 16\frac{3}{5}$

$x = -16\frac{3}{5} \textrm{ or }x = 16\frac{3}{5}$

Correct answer:

$x = -16\frac{3}{5} \textrm{ or } x = 1$

Explanation:

Rewrite $\left | 5x + 39 \right | = 44$ as a compound statement and solve each part separately:

$\left | 5x + 39 \right | = 44$

$5x + 39 = - 44 \textrm{ or }5x + 39 = 44$

5x + 39 = - 44

5x + 39 - 39 = - 44- 39

5x = - 83

$5x \div 5 = - 83 \div 5$

$x = -16\frac{3}{5}$

5x + 39 = 44

5x + 39 - 39 = 44- 39

5x = 5

$5x \div 5 =5 \div 5$

x = 1

Therefore the solution set is $\left \{-16\frac{3}{5}, 1 \right \}$ .

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Example Question #32 : Equations

Solve the following equation for :

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

Possible Answers:

x = -9

x = -18

x = 18

x = -6

x = 2

Correct answer:

x = 18

Explanation:

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

The first step in solving this equation is to distribute the 3 and the 4 through the parentheses:

$3\cdot x+3\cdot2 = 4\cdot x-4\cdot 3$

Simplify:

3x+6 = 4x-12

Now, we want to get like terms on the same sides of the equation. That is, all of the terms with an should be on one side, and those without an should be on the other. To do this, we first subtract from both sides:

3x+6 -4x= 4x-12-4x

Simplify:

-x+6 = -12

Now, we subtract 6 from both sides:

-x+6-6 = -12-6

-x = -18

x = 18

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Example Question #31 : Equations

Goldenrod paint is made by mixing one part red with three parts yellow. How many gallons of yellow paint should be mixed with two quarts of red paint?

Possible Answers:

$1.25\; gallons$

$1.00\; gallon$

$1.75\; gallons$

$1.50\; gallons$

$0.75\; gallons$

Correct answer:

$1.50\; gallons$

Explanation:

This problem is solved using proportions and the following conversion factor:

$4\; quarts=1\; gallon$

Let yellow quarts of paint.

$\frac{red}{yellow}=\frac{1}{3}=\frac{2\; quarts}{x\; quarts}$

Then cross multiply to get $x=6\; quarts = 1.50 \; gallons$ .

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Example Question #32 : Equations

Solve for :

$\frac{1}{2}(x+3)=\frac{1}{3}(x+5)$

Possible Answers:

Correct answer:

Explanation:

Multiply both sides by to eliminate the fractions to get 3(x+3)=2(x+5) .

Then use the distributive property to get 3x+9=2x+10 .

Now subtract from both sides to get x+9=10 .

Now subtract from both sides to get x=1 .

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Example Question #33 : Basic Single Variable Algebra

To mail a package, there is an initial charge of $55\cent$ to cover the first ounce, with another $25\cent$ for each additional ounce. How much does it cost to mail a half pound package?

Possible Answers:

$\$ 3.95$

$\$ 4.40$

$\$ 2.30$

$\$ 3.50$

$\$ 2.00$

Correct answer:

$\$ 2.30$

Explanation:

$1\; pound = 16 \; ounces$ , so we are trying to mail ounces. The total postage becomes $0.55+7(0.25)=0.55+1.75=\$ 2.30$ .

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Example Question #1 : Expressions

In April, the price of a t-shirt is $\$X$ . In May, the store increases the price by 50%, so that the new price is $\$Y$ . Then in June, the store decreases the price by 50%, so that the t-shirt price is now $\$Z$ . What is the ratio of to ?

Possible Answers:

$\frac{4}{3}$

$\frac{1}{2}$

$\frac{3}{4}$

Correct answer:

$\frac{3}{4}$

Explanation:

If the original price of the T-shirt is , increasing the price by 50% means that the new price is 150% of , or 1.5X .

If the price is then decreased by 50%, the new price is 50% of

$.5\cdot (1.5X)=.75X$

The ratio of to is then:

$\frac{.5*1.5*X}{X}$

The 's in the numerator and denominator cancel, leaving $.5\cdot 1.5$ , or

$\frac{3}{4}$ .

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Example Question #1 : Simplifying Expressions

Simplify $\frac{x^4-y^4}{(x-y)^2}\cdot\frac{x-y}{4x^2-4y^2}$ .

Possible Answers:

$\frac{x-y}{4(x+y)}$

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

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

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

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

Correct answer:

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

Explanation:

When multiplying rational expressions, we simply have to multiply the numerators together and the denominators together. (Warning: you only need to find a lowest common denominator when adding or subtracting, but not when multiplying or dividing rational expression.)

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

In order to simplify this, we will need to factor x^4-y^4 and 4x^2-4y^2 . Because looks a little simpler, let's start with it first.

We can easily factor out a four from both terms.

4x^2-4y^2=4(x^2-y^2) .

Next, notice that x^2-y^2 fits the form of our difference of squares factoring formula. In general, we can factor a^2-b^2 as (a-b)(a+b) . In the polynomial x^2-y^2 we will let a=x and b=y . Thus, x^2-y^2=(x-y)(x+y) .

Now, we can see that 4x^2-4y^2=4(x^2-y^2)=4(x-y)(x+y) .

We then factor x^4-y^4 . This also fits our difference of squares formula; however, this time a=x^2 and b=y^2 . In other words, x^4-y^4=(x^2)^2-(y^2)^2 . Applying the formula, we see that

x^4-y^4=(x^2)^2-(y^2)^2=(x^2-y^2)(x^2+y^2) . Now, let's take our factorization one step further and factor x^2-y^2 , which we already did above.

(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2) .

Be careful here. A common mistake that students make is to try to factor x^2+y^2 . There is no sum of squares factoring formula. In other words, in general, if we have a^2+b^2 , we can't factor it any further. (It is considered prime.)

We will then put all of these pieces of information in order to simplify our rational expression.

$\frac{(x-y)(x^4-y^4)}{(x-y)^2\cdot(4x^2-4y^2)}=\frac{(x-y)(x-y)(x+y)(x^2+y^2)}{(x-y)^2\cdot(4(x-y)(x+y))}\\ \\ \\ =\frac{(x-y)^2(x+y) (x^2+y^2)}{4(x-y)^3(x+y)}$

Lastly, we cancel the factors that appear in both the numerator and the denominator. We can cancel an x+y and a (x-y)^2 term.

$\frac{(x-y)^2(x+y) (x^2+y^2)}{4(x-y)^3(x+y)}=\frac{(x^2+y^2)}{4(x-y)}$ .

The answer is $\frac{(x^2+y^2)}{4(x-y)}$ .

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High School Math : Algebra II

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #23 : Solving Equations

Example Question #1 : Solving Equations

Example Question #2 : Solving Equations

Example Question #31 : Equations

Example Question #32 : Equations

Example Question #31 : Equations

Example Question #32 : Equations

Example Question #33 : Basic Single Variable Algebra

Example Question #1 : Expressions

Example Question #1 : Simplifying Expressions

All High School Math Resources

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