Algebraic Fractions - PSAT Math

The absolute value will always be positive or 0, therefore all values of z will create a true statement as long as . Thus all values except for 2 will work.

The domain of is restricted by two different denominators, neither of which can be equal to 0, so the excluded values are:

The correct response is therefore .

cannot equal a value which would make the denominator equal to 0. In order to figure out what that value is, we must first simplify this fraction, then set each factor of the denominator equal to 0. As follows:

First, simplify by finding the original binomial multiples:

Now, set each factor of the denominator equal to 0

If OR , the denominator will equal 0. is the only choice provided by the answers.

Note: Even though you can cancel out from the numerator and denominator, still cannot be equal to two. The graph would have a hole at and an aymptote at .

The domain of is restricted by two things.

First, the expression within the radical in the numerator must be nonnegative. We therefore solve for in the inequality

,

or, in interval notation,

Second, the expression in the denominator must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded value(s):

We exclude 4 from , so the correct response is

The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:

Therefore, the domain of is the set of all real numbers except - that is,

There are two things restricting the domain of .

One is the radical symbol in the numerator. The expression inside the radical must be nonnegative, so solve the inequality:

,

or, in interval notation,

The other is the denominator, which must be equal to 0, so set, and solve for in, the equation:

is already excluded from the domain; we exclude 4, so the domain is

.

The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the expression in the denominator, which must be nonzero. Furthermore, the radicand must be nonnegative. Combined, these facts mean that the radicand must be positive, and that the following inequality be solved:

or, equivalently, .

In interval form, this is

The numerator, being a polynomial, does not restrict our domain. The denominator, however, does restrict it to the values for which it is not equal to 0. We set the denominator equal to 0 to find the excluded values:

The domain, in interval notation, is therefore

.

The domain of the product of two functions is the intersection of the domains of the individual functions.

The domain of is restricted to all values of that yield a nonzero denominator. Since this means that

,

then, subsequently,

,

so the domain is the set of all real numbers except 4.

Similarly, the domain of is restricted to all values of that yield a nonzero denominator. This set is found to be the set of all real numbers except 7.

The intersection of these sets is the set of all real numbers except 4 and 7, or

.

The definition of has two denominators of two fractions (one within the other), so we must exclude the values of that make either denominator equal to zero.

One denominator is . Since it cannot be zero, we have

,

and, subsequently,

.

The value 2 is excluded from the domain.

The other denominator is . If it is equal to zero, then

Therefore, this value is also excluded from the domain.

The correct domain is the set .

The numerator , being a polynomial, does not restrict the domain.

appears as a radicand of a square root in the definition of , so must be restricted to nonnegative numbers. Also, is a denominator, which means 0 must also be excluded. Therefore, we are restricted so far to positive numbers.

There is one more denominator, which is ; it must be nonzero. We set this equal to zero to determine any additional value(s) that must be excluded:

Therefore, the domain is the set of all positive numbers except 9 - or

.

Given a fractional algebraic equation with variables in the numerator and denominator of one side and the other side equal to zero, we rely on a simple concept. Zero divided by anything equals zero. That means we can focus in on what values make the numerator (the top part of the fraction) zero, or in other words,

The expression is a difference of squares that can be factored as

Solving this for gives either or . That means either of these values will make our numerator equal zero. We might be tempted to conclude that both are valid answers. However, our statement earlier that zero divided by anything is zero has one caveat. We can never divide by zero itself. That means that any values that make our denominator zero must be rejected. Therefore we must also look at the denominator.

The left side factors as follows

This means that if is or , we end up dividing by zero. That means that cannot be a valid solution, leaving as the only valid answer. Therefore only #3 is correct.

To find the excluded values of a algebraic fraction you need to find when the denominator is zero. To find when the denominator is zero you need to factor it. This denominator factors into

so this is zero when x=4,7 so our answer is

Simplifying this expression is similar to 1/2 – 1/5. The denominators are relatively prime (have no common factors) so the least common denominator (LCD) is 2 * 5 = 10. So the problem becomes 1/2 – 1/5 = 5/10 – 2/10 = 3/10.

$\frac{0}{6}$=0, which is an integer (a number with no fraction or decimal part). All the other choices reduce to non-integers.

$$\frac{4x^{5}$$$y^{3}$$z}{12x^{3}$$y^{6}$$z^{2}$$}=\frac{x^{2}$$$}{3y^{3}$z}

First, let's simplify $\frac{4}{12}$. The greatest common factor of 4 and 12 is 4. 4 divided by 4 is 1 and 12 divided by 4 is 3. Therefore $\frac{4}{12}$=\frac{1}{3}$.

To simply fractions with exponents, subtract the exponent in the numerator from the exponent in the denominator. That leaves us with $$\frac{1}{3}$x^{2}$$y^{-3}$$z^{-1}$ or $$\frac{x^{2}$$$}{3y^{3}$z}

Let us simplify $\frac{6}{45}$:

$\frac{6}{45}$=\frac{3times 2}{3times 15}$=\frac{2}{15}$

We can get alternate forms of the same fraction by multiplying the denominator and the numerator by the same number:

$\frac{2times 2}{15times 2}$=\frac{4}{30}$

$\frac{2times 1.5}{15times 1.5}$=\frac{3}{22.5}$

Now let's look at $\frac{12}{89}$:

, but .

Therefore, $\frac{12}{89}$ is the correct answer, as it is not equivalent to $\frac{6}{45}$.

24/32 = 1.33

16/12 =1.33

224/168 =1.33

4/3 = 1.33

96/72 = 1.33

160/96 = 1.67

The root occurs where . So we substitute 0 for .

This means that the root is at .