### All High School Math Resources

## Example Questions

### Example Question #6 : Basic Single Variable Algebra

Cindy's Cotton Candy sells cotton candy by the bag. Her monthly fixed costs are . It costs to make each bag and she sells them for .

What is the monthly break-even point?

**Possible Answers:**

**Correct answer:**

The break-even point occurs when the .

The equation to solve becomes

so the break-even point is .

### Example Question #7 : Basic Single Variable Algebra

Cindy's Cotton Candy sells cotton candy by the bag. Her monthly fixed costs are . It costs to make each bag and she sells them for .

To make a profit of , how many bags of cotton candy must be sold?

**Possible Answers:**

**Correct answer:**

So the equation to solve becomes , or must be sold to make a profit of .

### Example Question #8 : Basic Single Variable Algebra

Solve for and to satisfy both equations in the system:

**Possible Answers:**

,

,

,

,

,

**Correct answer:**

,

The two equations in this system can be combined by addition or subtraction to solve for and . Isolate the variable to solve for it by multiplying the top equation by so that when the equations are combined the term disappears.

Divide both sides by to find as the value for .

Substituting for in both of the two equations in the system and solving for gives a value of for .

### Example Question #9 : Basic Single Variable Algebra

Solve for :

**Possible Answers:**

**Correct answer:**

Rewrite as a compound statement and solve each part separately:

The solution set is

### Example Question #10 : Basic Single Variable Algebra

and . What is the value of ?

**Possible Answers:**

25

112

130

33

65

**Correct answer:**

65

First, notice that we can factor into the form (a-b)(a+b). We are told that a-b=3, so we can substitute that into the first equation.

If we divide both sides by 3, we can obtain the value of a+b.

We now have a system of equations: a-b = 3, a+b = 11. We will solve this system by elimination. If we add the two equations together, we obtain the following:

.

Divide both sides by 2.

Going back to the equation a-b = 7, we can solve for b.

Add b to both sides.

Subtract 3 from both sides.

Ultimately, the question asks us to determine the value of .

= .

The answer is 65.

### Example Question #1 : Solving Equations

Solve the pair of equations for x and y:

**Possible Answers:**

**Correct answer:**

Equation 1:

Equation 2:

Solve equation 2 for X:

Substitute into Equation 1:

Solve for y: ,

Take the answer for y and plug it back into either original equation to find x:

### Example Question #2 : Solving Equations

Solve the pair of equations for x and y:

**Possible Answers:**

**Correct answer:**

Equation 1:

Equation 2:

Solve Equation 2 for y:

Substitute into Equation 1:

Plug x back into either original equations and solve for y:

### Example Question #3 : Solving Equations

For which of the following functions is the result of a positive integer?

**Possible Answers:**

**Correct answer:**

Simply plugging in -2 into each answer choice will determine the correct answer:

The key to solving this problem is to remember the order of operations and that negative numbers squared are positive, while negative numbers raised to the third power are negative.

### Example Question #4 : Solving Equations

Given the equation , what is the value of ?

**Possible Answers:**

15

12

18

8

4

**Correct answer:**

15

When solving the equation , observe that . Taking the cube root of 216 gives 6. Thus,

### Example Question #5 : Solving Equations

A cardboard packing box contains footballs and baseballs. The ratio by weight of baseballs to footballs is 7 to 9. How many kilograms of footballs will there be in the box if the total weight of the box is 48 kilograms?

**Possible Answers:**

27

30

26

20

15

**Correct answer:**

27

In the box, there are kilograms of baseballs and kilograms of footballs. In total, there are kilograms of balls. The total weight of the box is 48 kilograms, so Since there are kilograms of footballs, the total weight of the footballs in the box is equal to: