### All High School Math Resources

## Example Questions

### Example Question #1 : Adding And Subtracting Fractions

Simplify

**Possible Answers:**

**Correct answer:**

Find the least common denominator (LCD) and convert each fraction to the LCD and then add. Simplify as necessary.

The result is an improper fraction because the numerator is larger than the denominator and can be simplified and converted to a mix numeral.

### Example Question #2 : Algebra Ii

Simplify .

**Possible Answers:**

**Correct answer:**

Chenge the mixed numbers into improper fractions by multiplying the whole number by the denominator and adding the numerator to get

.

### Example Question #1 : Algebra Ii

All of the following matrix products are defined EXCEPT:

**Possible Answers:**

**Correct answer:**

Every matrix has a dimension, which is represented as the number of rows and columns. For example, a matrix with three rows and two columns is said to have dimension 3 x 2.

The matrix

has two rows and three columns, so its dimension is 2 x 3. (Remember that rows go from left to right, while columns run up and down.)

Matrix multiplication is defined only if the number of columns in the first matrix is equal to the number of rows on the second matrix. The easiest way to determine this is to write the dimension of each matrix. For example, let's say that one matrix has dimension a x b, and the second matrix has dimension c x d. We can only multiply the first matrix by the second matrix if the values of b and c are equal. It doesn't matter what the values of a and d are, as long as b (the number of columns in the first matrix) matches c (the number of rows in the second matrix).

Let's go back to the problem and analyze the choice .

The dimension of the first matrix is 2 x 3, because it has two rows and three columns. The second matrix has dimension 2 x 2, because it has two rows and two columns.

We can't multiply these matrices because the number of columns in the first matrix (3) is not equal to the number of rows in the second matrix (2). Thus, this product is not defined.

The answer is .

### Example Question #1 : Understanding Multiplication And Division

Simplify.

**Possible Answers:**

**Correct answer:**

Convert the mixed numbers into improper fractions by multiplying the whole number by the denominator and adding the numerator to get

Dividing by a fraction is the same as multiplying by its reciprocal so the problem becomes

### Example Question #1 : Understanding Absolute Value

Expression 1:

Expression 2:

Find the set of values for where Expression 1 is greater than Expression 2.

**Possible Answers:**

All values where

All values where

All values where

All real numbers

All values where

**Correct answer:**

All values where

In finding the values for where , break the comparison of these two absolute value expressions into the four possible ways this could potentially be satisfied.

The first possibility is described by the inequality:

If you think of a number line, it is evident that there is no solution to this inequality since there will never be a case where subtracting from will lead to a greater number than adding to .

The second possibility, wherein is negative and converted to its opposite to being an absolute value expression but is positive and requires no conversion, can be represented by the inequality (where the sign is inverted due to multiplication by a negative):

We can simplify this inequality to find that satisfies the conditions where .

The third possibility can be represented by the following inequality (where the sign is inverted due to multiplication by a negative):

This is again simplified to and is redundant with the above inequality.

The final possibility is represented by the inequality

This inequality simplifies to . Rewriting this as makes it evident that this inequality is true of all real numbers. This does not provide any additional conditions on how to satisfy the original inequality.

The only possible condition that satisfies the inequality is that which arises in two of the tested cases, when .

### Example Question #2 : Understanding Absolute Value

What is the absolute value of -3?

**Possible Answers:**

10

-3

1

3

9

**Correct answer:**

3

The absolute value is the distance from a given number to 0. In our example, we are given -3. This number is 3 units away from 0, and thus the absolute value of -3 is 3.

If a number is negative, its absolute value will be the positive number with the same magnitude. If a number is positive, it will be its own absolute value.

### Example Question #1 : Solving Absolute Value Equations

**Possible Answers:**

**Correct answer:**

Notice that the equation has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) *must *be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

Simplifying and solving this equation for gives the answer:

### Example Question #2 : Solving Absolute Value Equations

What are the possible values for ?

**Possible Answers:**

**Correct answer:**

The absolute value measures the distance from zero to the given point.

In this case, since , or , as both values are twelve units away from zero.

### Example Question #3 : Solving Absolute Value Equations

**Possible Answers:**

**Correct answer:**

### Example Question #4 : Solving Absolute Value Equations

Solve:

**Possible Answers:**

All real numbers

No solution

**Correct answer:**

The absolute value can never be negative, so the equation is ONLY valid at zero.

The equation to solve becomes .

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