Mastering Rational Equations and Inequalities: Comprehensive Guide

Rational Equation	Method
$\frac{x}{2 x + 7} = \frac{2}{x - 1}$	Cross Products Property
$\frac{x}{x - 5} + \frac{x}{x + 5} = \frac{2}{x ^{2} - 2 5}$	LCD

Rational Equation

Method

\frac{x}{2 x + 7} = \frac{2}{x - 1}

Cross Products Property

\frac{x}{x - 5} + \frac{x}{x + 5} = \frac{2}{x ^{2} - 2 5}

LCD

	Original	Added	New
Amount of Acid	$0.15 (15)$	$0.6 (x)$	$0.15 (15) + 0.6 (x)$
Total Solution	$15$	$x$	$15 + x$

Original

Added

New

Amount of Acid

0.15 (15)

0.6 (x)

0.15 (15) + 0.6 (x)

Total Solution

15

x

15 + x

Rational Expression	LCD
$\frac{5}{x + 2}$	$(x + 2) (x + 5)$
$\frac{4}{x + 5}$
$\frac{1 6}{( x + 2 ) ( x + 5 )}$

Rational Expression

LCD

\frac{5}{x + 2}

(x + 2) (x + 5)

\frac{4}{x + 5}

\frac{1 6}{( x + 2 ) ( x + 5 )}

	Going	Returning
Distance (km)	$1$	$1$
Speed (km/h)	$x + 3$	$x - 3$
Time (h)	$\frac{1}{x + 3}$	$\frac{1}{x - 3}$

Going

Returning

Distance (km)

1

1

Speed (km/h)

x + 3

x - 3

Time (h)

\frac{1}{x + 3}

\frac{1}{x - 3}

$x = \frac{2 \pm 8 5}{3}$
$x_{1} = \frac{2 + 8 5}{3}$	$x_{2} = \frac{2 - 8 5}{3}$
$x_{1} = \frac{2}{3} + \frac{8 5}{3}$	$x_{2} = \frac{2}{3} - \frac{8 5}{3}$
$x_{1} \approx 3.74$	$x_{2} \approx - 2.41$

x = \frac{2 \pm 8 5}{3}

x_{1} = \frac{2 + 8 5}{3}

x_{2} = \frac{2 - 8 5}{3}

x_{1} = \frac{2}{3} + \frac{8 5}{3}

x_{2} = \frac{2}{3} - \frac{8 5}{3}

x_{1} \approx 3.74

x_{2} \approx - 2.41

If I clear the denominators I find that the only solution is $x = 1,$ but when I substitute $x = 1$ into the equation, it does not make any sense.

$t = \frac{1 4 \pm 1 0}{2}$
$t = \frac{1 4 + 1 0}{2}$	$t = \frac{1 4 - 1 0}{2}$
$t = 12$	$t = 2$

t = \frac{1 4 \pm 1 0}{2}

t = \frac{1 4 + 1 0}{2}

t = \frac{1 4 - 1 0}{2}

t = 12

t = 2

Denominator	Factored Form	Excluded Values
$x + 1$	$-$	$x = - 1$
$x^{2} - 4 x - 5$	$(x + 1) (x - 5)$	$x = - 1$ and $x = 5$
$x - 5$	$-$	$x = 5$

Denominator

Factored Form

Excluded Values

x + 1

-

x = - 1

x^{2} - 4 x - 5

(x + 1) (x - 5)

x = - 1

and

x = 5

x - 5

-

x = 5

	$x < - 1$	$- 1 < x < 2$	$2 < x < 5$	$x > 5$
Test Value	$- 2$	$0$	$3$	$6$
Statement	$- \frac{2 9}{7} ≱ \frac{3}{7}$	$\frac{1 9}{5} \geq \frac{3}{5}$	$\frac{1}{2} ≱ \frac{3}{2}$	$\frac{1 1}{7} \geq 3$

x < - 1

- 1 < x < 2

2 < x < 5

x > 5

Test Value

- 2

0

3

6

Statement

- \frac{2 9}{7} ≱ \frac{3}{7}

\frac{1 9}{5} \geq \frac{3}{5}

\frac{1}{2} ≱ \frac{3}{2}

\frac{1 1}{7} \geq 3

	$x < 0$	$0 < x < 80$	$x > 80$
Choose a Test Value	$- 1$	$1$	$100$
Statement	$- 1995 < 30$	$2005 < 30$	$25 < 30$

x < 0

0 < x < 80

x > 80

Choose a Test Value

- 1

1

100

Statement

- 1995 < 30

2005 < 30

25 < 30

Dominika
Win	Loss
$24$	$16$

Dominika

Win

Loss

24

16

Ramsha can clean a room in $5$ hours. Ramsha, working together with her sister Maya, can clean the room in just $3$ hours.

Find the work done by Ramsha when she is working with her sister.

Find how long Maya would take to clean the room when working alone.

To find the work done by Ramsha when working together with her sister Maya, we can use the formula below. ( Work done)=( Work rate)*( Time) We know that Ramsha can clean a room in 5 hours when working alone. Therefore, her work rate is 15 room/hours. We also know that Ramsha and Maya can clean the room in 3 hours when working together. We can find the work done by Ramsha when working together with Maya by substituting these values into the formula.

The work done by Ramsha is 35 room when working together with Maya.

Ramsha and Maya do 100 % of the work when they work together. Therefore, the sum of the work done by Ramsha and Maya is 100 % = 1. We found in Part A that the work done by Ramsha is 35 room when working together with Maya. Therefore, we can find the work done d by Maya by solving the following equation. 3/5+d= 1 Now, we will solve the equation for d.

The work done by Maya when working together with Ramsha is 25. Next, using the formula in Part A, we can find the work rate of Maya.

The work rate can be expressed as 1t room/hours. Work rate=1/t ⇒ 1/t=2/15 Finally, by solving this equation for t, we can find how long Maya would take to clean the room when working alone.

It takes 7.5 hours for Maya to clean the room when working alone.

Consider the following rational equation.

\frac{3}{x - 4} + \frac{1}{x} = \frac{x - 1}{x - 4}

Solve the equation for

x .

Consider now the graphs of the functions related to both sides of the equation on the same coordinate plane for

0 < x < 7 .

y_{1} = \frac{3}{x - 4} + \frac{1}{x} and y_{2} = \frac{x - 1}{x - 4}

Which statements are correct?

I . II . III . IV . The graphs of the functions intersect at x = 1 . The graphs of the functions do not intersect at x = 4 . The line x = 3 is a vertical asymptote for both graphs . The line x = 4 is a vertical asymptote for both graphs .

We want to solve the given rational equation. 3/x-4 + 1/x = x-1/x-4 To do so, we will first determine the least common denominator (LCD) so that we can clear the denominators. Note that the denominators are already factored. Therefore, in this case the LCD is the product of (x-4) and x.

We have a quadratic equation in standard form. Let's identify the values of a, b, and c. x^2 -5x+4=0 ⇔ 1x^2 +( -5x)+ 4=0 We can see that a= 1, b= -5, and c= 4. Let's substitute these values into the Quadratic Formula to find the possible values of x.

Let's calculate both solutions by using the positive and negative signs.

x=5 ± 3/2
x_1=5+ 3/2	x_2=5- 3/2
x_1=8/2	x_2=2/2
x_1=4	x_2=1

Finally, we will check whether either x=4 or x=1 are extraneous solutions. To do so, we will substitute 4 and 1 for x in the original equation. Let's start with the first one.

Since the denominator cannot be 0, x=3 is an extraneous solution. Now, we will check our second solution, x=1.

The only solution to the given equation is x=1.

We need to graph the given rational functions to determine which statements are correct. y_1 = 3/x-4+1/x and y_2 = x-1/x-4 Let's take a look at their graphs drawn on the same coordinate plane.

We see that the graphs of the functions intersect at x=1. For the extraneous solution of x=4, the graphs do not intersect. In fact, x=4 is a vertical asymptote for both functions. There is no asymptote at x=3.

We can conclude that I, II, and IV are the correct ones among the given statements. As a general result, if the graphs intersect, then we can say that the point of intersection is a solution. If the graphs do not intersect, then the possible solution is extraneous.

We will begin by graphing y= 3x-4+ 1x. Let's first simplify it!

Now, we can find the asymptotes of the function. To find the vertical asymptotes, we will solve x(x-4)=0 by using the Zero Product Property.

Therefore, the function has vertical asymptotes at x=0 and x=4. Next, we will find the horizontal asymptote of the function. To do so, let's pay close attention to the degrees of the numerator and denominator. y=4x^1-4/x^2-4x We see that the degree of the denominator is higher than the degree of the numerator. Therefore, the line y=0 is a horizontal asymptote. Now, we will plot the asymptotes.

Next, we will make a table of values. Recall that we are asked to graph the function for x-values greater than 0 and less than 7. Since the function has a vertical asymptote at x=4, this value will be excluded from the domain.

x	4x-4/x(x-4)	y
0.5	4( 0.5)-4/0.5( 0.5-4)	≈ 1.1
1	4( 1)-4/1( 1-4)	0
2	4( 2)-4/2( 2-4)	- 1
3	4( 3)-4/3( 3-4)	≈ -2.7
3.5	4( 3.5)-4/3.5( 3.5-4)	≈ -5.7
4.5	4( 4.5)-4/4.5( 4.5-4)	≈ 6.2
5	4( 5)-4/5( 5-4)	3.2
6	4( 6)-4/6( 6-4)	≈ 1.7

Finally, let's plot and connect the points.

We can graph y= x-1x-4 proceeding in the same way. The function has also a vertical asymptote at x=4. Let's pay close attention to the degrees of the numerator and denominator. y=x^1-1/x^1-4 We see that the degrees of the numerator and denominator are the same. To find the horizontal asymptote, we need to find the quotient between the leading coefficients. y=1x-1/1x-3 Since 11=1, there is a horizontal asymptote at y=1. By finding, plotting, and connecting points, we can graph the function as we did for the first one.

x	x-1/x-4	y
0.5	0.5-1/0.5-4	≈ 0.1
1	1-1/1-4	0
2	2-1/2-4	- 0.5
3	3-1/3-4	- 2
3.5	3.5-1/3.5-4	- 5
4.5	4.5-1/4.5-4	7
5	5-1/5-4	4
6	6-1/6-4	2.5

Let's draw it!

We can now plot the graphs on the same coordinate plane.

Consider the set of rational functions with the following form.

\frac{1}{x} = a x + b, where a \neq = 0

Which of the following can be the number of solutions to an equation in this form?

We want to determine the number of solutions of an equation of the following form. 1/x = ax+b, a ≠ 0 We can write two functions from the equation, one function for each side. y = 1/x and y = ax+b For simplicity, let's consider the graph of y= 1x and y=x.

We see that there are two points of intersection for these functions. We can also find lines that intersect the given function at one point or that do not intersect at any point.

Each of these intersection points is a solution of the equation. Let's now use an interactive graph of the two functions to check if there is another possible case. Change the values of a and b.

By observing the applet, we can see that there are indeed three cases. We can generalize these cases as follows.

There are no solutions when the slope a of the line is negative and b=0.
There is exactly one solution when the line is tangent to the graph of y= 1x.
There are exactly two solutions when the slope a of the line is positive.

Therefore, we can conclude that a rational equation in the form 1x=ax+b can have exactly one solution, exactly two solutions, or no solution.

Solving Rational Equations and Inequalities

Catch-Up and Review

Dominika's Winning Percentage in Basketball

Rational Equations and Finding Their Solutions

Solving a Rational Equation by Using the Cross Products Property

Creating a $45 %$ Acid Solution

Hint

Solution

Solving a Rational Equation by Using the Least Common Denominator

Canoe Trip

Hint

Solution

Identifying an Extraneous Solution

Answer

Hint

Solution

Extra

Finding the Time Spent Painting the Canoe

Hint

Solution

Rational Inequalities and Finding Their Solution Sets

Solving a Rational Inequality

The Average Cost of a Basketball

Hint

Solution

Excluded Values

Solve the Related Equation

Divide the Number Line Into Intervals

Test Values

Calculating Dominika’s Winning Percentages for Different Situations

Answer

Hint

Solution

Solving Rational Equations and Inequalities

Recommended exercises

	11 Theory slides
	8 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions