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

Solve each equation by cross multiplying for the given variable. Write the answer in the simplest form.

\frac{- 2}{x + 8} = \frac{7}{x - 1}

Solve for

x .

\frac{p - 8}{3} = \frac{p ^{2} - 4}{2 p + 5}

Solve for

p .

\frac{P}{n} - \frac{R T}{V} = 0

Solve for

V .

We will solve the rational equation by cross multiplying. If a numerator or denominator contains more than one term, make sure to treat them as a parenthetical factor in the cross multiplication process.

For rational equations, extraneous solutions are values that make any denominator in the original equation 0. When we have 0 in the denominator we have an undefined expression. We have to disregard the solutions that would cause any undefined expressions. Let's check if substituting x=- 6 into the original equation causes any problem.

Since we obtained a true statement, x=- 6 is a solution of the equation.

We will solve the rational equation by using the Cross Products Property..

We have obtained a quadratic equation in the standard form. Let's identify the values of a, b, and c. p^2+11p+28=0 ⇕ 1p^2+ 11p+ 28=0 We can see that a= 1, b= 11, and c= 28. Next, we will substitute these values into the Quadratic Formula.

We will find the values for p by using the positive and negative signs. p_1=-11 + 3/2 &⇒ p_1=- 4 [1em] p_2=-11 - 3/2 &⇒ p_2=- 7 We found that p_1=- 4 and p_2=- 7 are possible solutions. Now, we have to check them to see whether any of them are extraneous. To do so, we will substitute p_1=- 4 and p_2=- 7 into the given rational equation. Let's start with p_1=- 4.

Since we obtained a true statement, p_1=- 4 is a solution to the equation. Let's now substitute p_2=- 7 into the equation.

Since we obtained a true statement, p_2=- 7 is also a solution to the equation.

We want to solve the given equation for the variable V. P/n-RT/V=0 Recall that we can solve a rational equation by cross multiplying if it is expressed as a proportion. Therefore, we should first write the given equation as a proportion by adding RTV to both sides.

We can now apply the Cross Product Property and solve the equation for V.

Solve each equation by using the least common denominators (LCD) of the rational expressions.

\frac{3 x - 7}{x + 2} - \frac{5}{x} = 3

\frac{4}{x ^{2} - 2 x - 3} = \frac{8}{x - 3} + \frac{x}{x + 1}

We will begin by looking at the given rational equation. 3x-7/x+2-5/x=3 To solve this equation, we will first determine the least common denominator (LCD) of the rational expressions in the equation. In this case, the denominators are already factored. Let's highlight the denominators. We can rewrite 3 as 31. 3x-7/x+2-5/x= 3/1 The least common denominator is the product of these highlighted terms, x+2, x, and 1. Therefore, the least common denominator is (x+2) x. We will now multiply both sides of the equation by this common denominator and solve for x.

We will now check whether x=- 59 causes any undefined expressions in the original equation. If so, it is an extraneous solution. Let's substitute - 59 for x in the original equation.

Substituting - 59 into the original equation resulted in a true statement. Therefore, it is a solution to the equation.

We will solve the given rational equation by using the LCD. 4/x^2-2x-3=8/x-3+x/x+1 Let's start by factoring the denominators to find the least common denominator (LCD). Since the denominators on the right-hand side are already factored, we need to factor only the one on the left-hand side.

We can now rewrite the given expression. 4/x^2-2x-3=8/x-3+x/x+1 ⇕ 4/(x-3) (x+1)=8/x-3+x/x+1 In this case, the LCD is the product of (x-3) and (x+1). We will solve the equation by multiplying each side by the LCD to simplify the denominators.

Note that we have obtained 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 have that a= 1, b= 5, and c= 4. Let's substitute these values into the Quadratic Formula and solve for x.

We found that x_1= - 5 - 32 and x_2= - 5 + 32 are possible solutions to the given equation. x_1=- 5 - 3/2 &⇒ x_1=- 4 [0.75em] x_2=- 5 + 3/2 &⇒ x_2=- 1 Finally, we will check whether either x=- 4 or x=- 1 is an extraneous solution. To do so, we will substitute - 4 and - 1 for x in the original equation. Let's start with x= -4.

Since we obtained a true statement, x_1=- 4 is a solution of the equation. Let's now check our second solution, x_2= - 1.

Since the denominator cannot be 0, x_2=- 1 is an extraneous solution. Therefore, x=- 4 is the only solution to x

Solve each of the following inequalities and express the answer as an inequality.

\frac{1}{4 k} + \frac{2}{3 k} < \frac{1}{7}

If the answer is in the form

a < k < b,

write it as is in one box. If the answer is in the form

k < a

k > b,

write each individual inequality in a separate box.

\frac{3 x}{x + 3} + \frac{1}{x - 3} > 2

If the answer is in the form

a < x < b,

write it as is in one box. If the answer is in the form

x < a

x > b,

write each individual inequality in a separate box.

To solve the rational inequality, we will follow these four steps.

Identify the excluded values. These are the values for which any of the denominators are 0.
Solve the related rational equation.
Use the values determined in the previous steps to divide a number line into intervals.
Test a value in each interval to determine which intervals contain values that satisfy the inequality.

Let's apply these steps to the given rational inequality one at a time.

Excluded Values

Consider the given inequality. 1/4k + 2/3k < 1/7 Let's find the values of k for which any of the denominators is 0. c|c 4k=0 & 3k=0 ⇕ & ⇕ k=0 & k=0 The excluded value for this inequality is 0.

Solve the Related Equation

We obtain the related equation by replacing the inequality sign with an equals sign. c|c Inequality & Equation [0.8em] 1/4k + 2/3k < 1/7 & 1/4k + 2/3k = 1/7 Let's solve the equation. To do so, we will start by finding the least common denominator (LCD). 1/4k + 2/3k = 1/7 [0.4em] ⇕ [0.4em] 1/4 k + 2/3 k = 1/7 The least common denominator is the product of 4, k, 3, and 7. Least Common Denominator 4( k)( 3)( 7) ⇔ 84k To clear the denominators, we will solve the equation by multiplying each side by the least common denominator.

Divide the Number Line Into Intervals

We will divide the number line considering the excluded value k=0 and the solution to the related equation k= 7712.

After dividing the number line, we get three possible intervals.

Test the Values

We will now test a value from each interval. Let's start with the interval k < 0. Any value from this interval can be chosen. Try testing the value k=- 1. If this substitution produces a true statement, the values in this interval are solutions to the rational inequality. If not, this interval is not included in the solution set of the inequality.

Since we obtained a true statement, we can conclude that the numbers less than 0 are in the solution set of the inequality. By following the same reasoning, we can determine whether the remaining intervals are also solutions to the given inequality.

Interval	Test Value	Statement
k< 0	k=- 1	- 77 < 12 ✓
0< k < 77/12	k=1	77 ≮ 12 *
k> 77/12	k=7	11 < 12 ✓

We found that the solutions to the given inequality are all the numbers less than 0 or greater than 7712. Solution Set [0.6em] k<0 or k>77/12

To solve the given rational inequality, we will follow the same steps as we did in the previous part.

Excluded Values

Consider the given inequality. 3x/x+3 + 1/x-3 > 2 Let's find the values of x for which any of the denominators is 0. c|c x+3=0 & x-3=0 ⇕ & ⇕ x=- 3 & x=3 The excluded values for this inequality are - 3 and 3.

Solve the Related Equation

We obtain the related equation by replacing the inequality sign with an equals sign. c|c Inequality & Equation [0.8em] 3x/x+3 + 1/x-3 > 2 & 3x/x+3 + 1/x-3 = 2 Let's solve the equation. First, we will get rid of all of the denominators by using a combination of methods.

We obtained a quadratic equation in standard form. Let's identify the values of a, b, and c. x^2-8x+21=0 ⇕ 1x^2+( - 8)x+( 21)=0 We have that a= 1, b= - 8, and c= 21. We will solve this equation by substituting these values into the Quadratic Formula.

Since the radicand is a negative number, the equation does not have real solutions.

Divide the Number Line Into Intervals

Since the related equation does not have real solutions, we will divide the number line considering only the excluded values x=- 3 and x=3.

In this case, we get three possible intervals.

Test the Values

We will now test a value from each interval. Let's start with the interval x < - 3. We will test the point x=- 4. If substituting this point into the original inequality produces a true statement, then the numbers in this interval are solutions. If not, these numbers are not solutions.

Since we have obtained a true statement, we can conclude that the numbers less than - 3 are solutions to the inequality. Using the same process, we can determine whether the numbers that belong to the other intervals are solutions.

Interval	Test Value	Statement
x< - 3	x=- 4	11.86... > 2 ✓
- 3< x < 3	x=0	- 0.33... > 2 *
x>3	x=4	2.71... > 2 ✓

We found that the solutions to the given inequality are all the numbers less than - 3 or greater than 3. Solution Set x<- 3 or x>3

Below are the steps a student takes when solving for the solutions of the following rational equation.

Consider the following statements.

A . B . C . D . Each term should be multiplied by the same expression in Step I . The expressions were distributed incorrectly in Step III . The Properties of Equality were applied incorrectly in Step IV . The Quadratic Formula was applied incorrectly in Step V .

Which option describes the error made by the student when solving the rational equation?

Find the solutions to the rational equation.

We will examine each step performed to identify which one has the mistake.

When solving a rational equation, we begin by multiplying each term of the equation by an algebraic expression that clears the denominators. Note that in Step I, the student multiplied the terms 2x-1x+1 and x+1x-1 by (x-1)(x+1). However, the student forgot to multiply the term - 1 by this expression.

However, we can notice that the student forgot to multiply the term - 1 by this expression. Let's multiplied every term of the equation by the term (x-1)(x+1).

This means that the mistake is that each term should be multiplied by the same expression in Step I, which is given in option A.

We will solve the equation by multiplying each term in the equation by (x-1)(x+1). Next, we will simplify the equation by using the Properties of Equality and solve it for x.

Therefore, the solution to the equation is x= 15.

Let's substitute 15, or 0.2, for x in the original equation to check if it is an extraneous solution.

Since we obtained a true statement, x = 15 is a solution.

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

Solving Rational Equations and Inequalities

Catch-Up and Review

Hint

Solution

Hint

Solution

Answer

Hint

Solution

Extra

Hint

Solution

Hint

Solution

Excluded Values

Solve the Related Equation

Divide the Number Line Into Intervals

Test Values

Answer

Hint

Solution

Excluded Values

Solve the Related Equation

Divide the Number Line Into Intervals

Test the Values

Excluded Values

Solve the Related Equation

Divide the Number Line Into Intervals

Test the Values

Solving Rational Equations and Inequalities

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