Rational Equations and Partial Fractions - Pre-Calculus

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.

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, .

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."

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.

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 :

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 .

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.

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:

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:

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.

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, .

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, .

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).

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.

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, .

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."

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.

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 :

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 .

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.

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:

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:

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.

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, .

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, .