Precalculus : Rational Equations and Partial Fractions

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #1 : Rational Equations And Partial Fractions

What is the LCD (least common denominator) of the rational expressions in the following equation?

Possible Answers:

24(a+1)(a-1)

12(a+1)2

12(a+1)(a-1)

12(a+1)

(a+1)(a-1)

Correct answer:

12(a+1)(a-1)

Explanation:

To find the least common denominator (LCD), we must consider each individual denominator: 3(a+1), 2(a-1), and 4(a+1). The smallest number that 3, 2, and 4 all go into is 12. Of the items in parentheses, we must represent both (a+1) and (a-1). Therefore, the LCD of these three terms is 12(a+1)(a-1).

Example Question #1727 : Pre Calculus

The Southern Springs swimming pool can be filled by a pump in 12 hours. It can be drained in 20 hours. If the drain pipe was accidentally left open while the pool was being filled, how long would it take to fill the pool?

Possible Answers:

32 hours

8 hours

30 hours

240 hours

Correct answer:

32 hours

Explanation:

Let h represent the number of hours that it will take to fill the pool. Because it takes 12 hours to fill the pool, in 1 hour, 1/12 of the pool will be filled. Because it takes 20 hours to drain the pool, in one hour, 1/20 of the pool will be drained. Additionally, a full pool can be represented by 1. Therefore, the equation that represents this situation is:

In order to solve this equation, we want to get the least common denominator, or the smallest number that both 12 and 20 go into. This is 60. Multiply to get:

Therefore, it will take 30 hours to fill the pool if you were filling it while the drain pipe was open. 

Example Question #1728 : Pre Calculus

Decompose  into partial fractions.

Possible Answers:

Correct answer:

Explanation:

Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.

 

Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y. 

Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD. 

Looking back to our original fraction, note that the values b = 3 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.

First, let b=3.

Second, let b=-1

Now, substitute these values in for X and Y. 

Therefore, .

Example Question #2 : Rational Equations And Partial Fractions

Solve 

Possible Answers:

-4 < x < -1 or x > 2

x < -4 or -1 < x < 2

-1 < x < 2

x < -4 or x > 2

Correct answer:

x < -4 or -1 < x < 2

Explanation:

Draw a number line, and label each value that causes  to be equal to zero. Then, draw a vertical line through each to separate it into "zones."

Screen shot 2020 05 28 at 10.00.29 pm

There are five "zones." We want to select a sample value within each zone and plug it into our expression to see if it yields a positive or negative result. You don't need to fully find each answer, only if it will be positive or negative. For example:

Plug in f(0): 

Therefore, in the zone that includes 0, all values of x are negative, or less than zero, and therefore satisfy this equation.

Continue plugging in values in the other four zones to determine other possible solutions of the equation.

Plug in f(-5): 

Plug in f(-2): 

Plug in f(2.5): 

Plug in f(4): 

Therefore, the zones that include -5 and 0 are the ones that satisfy .

The solution to this equation is x < -4 or -1 < x < 2.

Example Question #1730 : Pre Calculus

Solve the inequality: 

Possible Answers:

 or 

Correct answer:

 or 

Explanation:

To start, get all of the fractions on one side. Then find the zeroes of the function by multiplying all fractions by the LCD and solving for a.

If this were an equality, we'd have our answer, but because this is an inequality, the point  is a point of discontinuity. Draw a number line, and plot both 0 and :

Screen shot 2020 05 28 at 10.30.11 pm

Next, we'll test a number in each range and look to see if it is less than .

f(-1): 

Since , any values in this zone will satisfy this equation.

f(1): 

Since , any values in this zone will not satisfy this equation.

f(3): 

Since , any values in this zone will satisfy this equation.

Therefore the solution to this equation is  or .

Example Question #1 : Rational Equations And Partial Fractions

This graph could be which of the following?

Screen shot 2020 05 28 at 10.54.18 pm

Possible Answers:

 and 

 or 

Correct answer:

Explanation:

Looking at the graph, we see a line at x=3, as well as a curved function. Looking ahead to the answer choices, it is reasonable to assume that this other function is .

Notice that the region that is below, or less than, x=3 is shaded. Notice that the region above, or greater than,  is shaded. 

Therefore we can conclude that  and . However, we don't have an exact answer match to this. (Don't be fooled by the answer that looks similar but says "or" instead of "and. These aren't the same!)

Examine the answer choices  and  as well as the graph. In the shaded area, x=3 is always above the square root function. Therefore  is the correct answer. 

Example Question #1732 : Pre Calculus

Solve this equation and check your answer:

Possible Answers:

No solution

Correct answer:

Explanation:

To solve this, first, find the common denominator. It is (n+1)(n-2). Multiply the entire equation by this: 

Simplify to get: 

Expand to get: 

Move all terms to one side and combine to get: 

Use the quadratic formula to get:

 

Example Question #2 : Rational Equations And Partial Fractions

Solve the inequality: 

Possible Answers:

No solution

1 < z < 31

z < 1 or z > 31

z < 1

Correct answer:

z < 1 or z > 31

Explanation:

First, find the LCD, which is 6(z-1). Then, multiply the entire equation by this: 

This simplifies to:

Now, make a number line with points at 1 and 31, and vertical lines in between zones: 

Screen shot 2020 05 28 at 11.32.35 pm

Let's test three points, one in each zone:

f(0): 

f(10):

f(35): 

 

Therefore, the inequality is true for values of z such that z < 1 or z > 31. 

Example Question #1734 : Pre Calculus

Decompose  into partial fractions.

Possible Answers:

Correct answer:

Explanation:

Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.

Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y. 

Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD. 

Looking back to our original fraction, note that the values p = 1 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.

First, let p=1.

Second, let p=-1

Now, substitute these values in for X and Y. 

Therefore, .

Example Question #1735 : Pre Calculus

Decompose  into partial fractions.

Possible Answers:

Correct answer:

Explanation:

Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.

Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y. 

Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD. 

Looking back to our original fraction, note that the values b = 0 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.

First, let b=0.

Second, let b=-1

Now, substitute these values in for X and Y. 

Therefore, .

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