Algebraic Fractions

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PSAT Math › Algebraic Fractions

Questions 1 - 10

Define the function as follows:

$f (x) = \frac{x^{2}- 1}{3x- 2}$

Give the domain of .

$\left ( - \infty, \frac{2}{3}\right ) \cup \left ( \frac{2}{3}, \infty \right )$

$\left ( - \infty, -1 }\right ) \cup \left (1, \infty \right )$

$\left ( - \infty,-1\right ) \cup \left ( \frac{2}{3}, 1 \right ) \cup \left ( 1, \infty \right )$

$\left ( - \infty, \frac{2}{3}\right ) \cup \left (1, \infty \right )$

$\left ( - \infty, 1 }\right ) \cup \left (1, \infty \right )$

Explanation

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:

$x= \frac{2}{3}$

Therefore, the domain of is the set of all real numbers except $\frac{2}{3}$ - that is,

$\left ( - \infty, \frac{2}{3}\right ) \cup \left ( \frac{2}{3}, \infty \right )$

Define the function as follows:

$f (x) = \frac{x^{2}- 1}{3x- 2}$

Give the domain of .

$\left ( - \infty, \frac{2}{3}\right ) \cup \left ( \frac{2}{3}, \infty \right )$

$\left ( - \infty, -1 }\right ) \cup \left (1, \infty \right )$

$\left ( - \infty,-1\right ) \cup \left ( \frac{2}{3}, 1 \right ) \cup \left ( 1, \infty \right )$

$\left ( - \infty, \frac{2}{3}\right ) \cup \left (1, \infty \right )$

$\left ( - \infty, 1 }\right ) \cup \left (1, \infty \right )$

Explanation

$x= \frac{2}{3}$

Therefore, the domain of is the set of all real numbers except $\frac{2}{3}$ - that is,

$\left ( - \infty, \frac{2}{3}\right ) \cup \left ( \frac{2}{3}, \infty \right )$

The maximum number of sweaters that Lauren can sew every day is equal to s, and the amount, in cents, that she charges for each sweater is equal to c. Which of the following expressions is equivalent to the maximum amount of money that Lauren can make, in dollars, after three weeks?

3_c_/(100_s_)

2100_sc_

21_sc_/100

2100_s_/c

300_c_/s

Explanation

The amount of money that Lauren can make depends on the number of sweaters that she can make. If she makes at most s sweaters a day, then we can multiply the number of days that she works by s to determine the total number of sweaters she makes.

total number of sweaters = (s)(number of days)

We are told to consider a time interval of three weeks. Because there are seven days in one week, the number of days over this period of time would equal 3(7), or 21 days. In other words, there are 21 days in three weeks Thus, the number of sweaters is equal to the product of s and 21.

total number of sweaters = (s)(21)

Now that we have the number of sweaters Lauren can make, we can multiply this by the cost of each sweater, which is equal to c cents, in order to obtain the amount of money she eared.

amount of money earned = (number of sweaters)(cost of each sweater)

amount of money earned = s(21)(c)

However, because the price of each sweater is given in terms of cents, the amount of money s(21)(c) will be equal to the number of cents she makes. The question, though, asks us to find the amount of money in dollars. We must use a conversion factor to change the number of cents to dollars. Remember that there are 100 cents per dollar.

Sweaters

If $\dpi{100} \small z>0$ , what is 40 percent of $\dpi{100} \small 30z$ ?

$\dpi{100} \small 12z$

$\dpi{100} \small 5z$

$\dpi{100} \small 20z$

$\dpi{100} \small 30z$

$\dpi{100} \small 120z$

Explanation

To find 40 percent of $\dpi{100} \small 30z$ multiply $\dpi{100} \small 30z\times 0.4$

The result is $\dpi{100} \small 12z$

Solve for .

$\frac{1}{3}(3x-6)+\frac{1}{2}(2x+4)=\frac{1}{5}(15x -5)$

$-2$

$-3$

$0$

$2$

Explanation

First distribute the fractions:

Combine like terms:

Evaluate the following equation when and round your answer to the nearest hundredth.

$\frac{x^{3}-2x^{2}-3x+5}{5x+3}$

Explanation

$\frac{x^{3}-2x^{2}-3x+5}{5x+3}$

1. Plug in wherever there is an in the above equation.

$\frac{5^{3}-(2)(5^{2})-(3)(5)+5}{(5)(5)+3}$

2. Perform the above operations.

$\frac{5^{3}-(2)(25)-(3)(5)+5}{(5)(5)+3}$

$\frac{125-50-15+5}{25+3}$

$\frac{65}{28}=2.32$

Simplify:

$\frac{8x^3y^2z^2}{4xy^4z^3}$

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

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

Explanation

$\frac{8x^3y^2z^2}{4xy^4z^3}$

To simply a fraction with variables, subtract exponents of like bases:

This leaves us with the expression:

$2x^2y^{-2}z^{-1}$

Next, we know that to change negative exponents in the numerator into positive exponents, we place them in the denominator. Thus, our expression simplifies to:

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

Simplify:

$\frac{8x^3y^2z^2}{4xy^4z^3}$

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

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

Explanation

$\frac{8x^3y^2z^2}{4xy^4z^3}$

To simply a fraction with variables, subtract exponents of like bases:

This leaves us with the expression:

$2x^2y^{-2}z^{-1}$

Next, we know that to change negative exponents in the numerator into positive exponents, we place them in the denominator. Thus, our expression simplifies to:

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

Simplify the expression, $\left(\frac{x^2}{y^{-3}}\right)\left(\frac{x^{-1}}{y^3}\right)$ .

Can't be simplified

$\frac{x}{y^6}$

Explanation

The and variables with negative exponents can be rewritten with positive exponents by moving them from the denominator to the numerator, and vice versa. Therefore, the expression can be rewritten as

$\left(x^2y^3 \right) \left(\frac{1}{xy^3}\right) = \frac{x^2y^3}{xy^3}$ .

The exponents on the denominator can then be subtracted from the exponent in the numerator to give

$x^{2-1}y^{3-3} = x^1y^0 = x \cdot1 = x$

Simplify the expression

$\left(\frac{A^{\frac{3}{2}}\sqrt{B}}{C^{-\frac{1}{2}}}\right)\left(\frac{B^{-\frac{1}{2}}}{2\sqrt{AC}}\right)$ .

$\frac{A}{2}$

$\frac{A^2}{2}$

Explanation

The expression can be rewritten as

$\left(\frac{A^{\frac{3}{2}}B^\frac{1}{2}}{C^{-\frac{1}{2}}}\right)\left(\frac{B^{-\frac{1}{2}}}{2A^\frac{1}{2}C^\frac{1}{2}}\right)$

We can now move the variables to the numerator and combine alike variables

$(A^\frac{3}{2}B^\frac{1}{2}C^\frac{1}{2})\left(\frac{A^{-\frac{1}{2}}B^{-\frac{1}{2}}C^{-\frac{1}{2}}}{2}\right) = \frac{A^{(\frac{3}{2}-\frac{1}{2})}B^{(\frac{1}{2}-\frac{1}{2})}C^{(\frac{1}{2}-\frac{1}{2})}}{2}$

This becomes

$\frac{A^{\frac{2}{2}}B^0C^0}{2} = \frac{A^1}{2} = \frac{A}{2}$

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