rational inequalities assignment quizlet edgenuity

Module 1: Equations and Inequalities

▪ solving rational inequalities, learning outcomes.

• Solve rational inequalities using boundary value method.

Solving Rational Inequalities using Boundary Value Method

Any inequality that can be put into one of the following forms

[latex]f(x)>0, f(x) \geq 0, f(x)<0, or f(x) \leq 0[/latex], where [latex]f[/latex] is a rational function

is called rational inequality .

How To: Solve Rational Inequalities

• Find restriction(s) of the given rational expression(s) in the inequality. The restriction(s) is(are) the boundary point(s) .
• Rewrite the given rational inequality as an equation by replacing the inequality symbol with the equal sign.
• Solve the rational equation. The real solution(s) of the equation is(are) the boundary point(s) .
• Plot the boundary point(s) from Step 1 & 3 on a number line. [latex] \Rightarrow [/latex] Use an open circle ALL restrictions. [latex] \Rightarrow [/latex] Use an open circle when the given inequality has [latex]<[/latex] or [latex]>[/latex] [latex]\Rightarrow[/latex] Use a closed circle when the given inequality has [latex]\leq[/latex] or [latex]\geq[/latex].
• Choose one number, which is called a test value , from each interval and test the intervals by evaluating the given inequality at that number. [latex]\Rightarrow[/latex] If the inequality is TRUE, then the interval is a solution of the inequality. [latex]\Rightarrow[/latex] If the inequality is FALSE, then the interval is not a solution of the inequality.
• Write the solution set (usually in interval notation), selecting the interval(s) from Step 5.

Example: Solving Rational Inequality using Boundary Value Method

Solve the rational inequality using boundary value method. Graph the solution set and write the solution in interval notation.

[latex]\frac{2x-1}{x+2} > 1[/latex]

First, find restriction(s):

[latex]x+2 \ne 0[/latex]

[latex]x \ne -2[/latex]

Second, let’s rewrite the inequality as an equation and then solve it:

[latex]\frac{2x-1}{x+2} = 1[/latex]

[latex]\frac{2x-1}{x+2} = \frac{1}{1}[/latex]

[latex]2x-1=x+2[/latex]

[latex]x=3[/latex]

Third, plot the restriction and the solution on the number line. All restrictions should be an  open circle. Since the inequality has “[latex]>[/latex]” (greater than), which doesn’t have “[latex]=[/latex]” (equal sign), the solution should be an  open circle as well:

Note that now there are three intervals on the number line: [latex](-\infty, -2), (-2, 3),[/latex] and [latex](3, \infty)[/latex].

Fourth, choose your choice of test value from each interval:

I chose [latex]-3, 0,[/latex] and [latex]4[/latex] from those three intervals.

fifth, test each interval using the test value:

When [latex]x=-3[/latex], [latex]\frac{2(-3)-1}{(-3)+2} > 1 \rightarrow \frac{-7}{-1}>1 \rightarrow 7>1[/latex]  (True)   [latex]\therefore[/latex] solution

When [latex]x=-3[/latex], [latex]\frac{2(0)-1}{(0)+2} > 1 \rightarrow \frac{-1}{2}>1 \rightarrow -\frac{1}{2}>1[/latex]  (False)   [latex]\therefore[/latex] not a solution

When [latex]x=4[/latex], [latex]\frac{2(4)-1}{(4)+2} > 1 \rightarrow \frac{7}{6}>1[/latex]  (True)   [latex]\therefore[/latex] solution

Actually, we can check the results of test values from the graph of [latex]f(x)=\frac{2x-1}{x+2}-1[/latex] below:

Therefore, the solution of the rational inequality [latex]\frac{2x-1}{x+2} > 1[/latex] is [latex](-\infty, -2) \cup (3, \infty)[/latex].

[latex]\frac{5x}{x-3} \leq 2[/latex]

[latex]x-3 \ne 0[/latex]

[latex]x \ne 3[/latex]

[latex]\frac{5x}{x-3}=2[/latex]

[latex]\frac{5x}{x-3} = \frac{2}{1}[/latex]

[latex]5x=2(x-3)[/latex]

[latex]5x=2x-6[/latex]

[latex]3x=-6[/latex]

[latex]x=-2[/latex]

Third, plot the restriction and the solution on the number line. All restrictions should be an  open circle. Since the inequality has “[latex]\leq[/latex]” (less than or equal to), which has “[latex]=[/latex]” (equal sign), the solution should be a  closed circle:

Note that now there are three intervals on the number line: [latex](-\infty, -2], [-2, 3),[/latex] and [latex](3, \infty)[/latex].

When [latex]x=-3[/latex], [latex]\frac{5(-3)}{(-3)-3} \leq 2 \rightarrow \frac{-15}{-6} \leq 2 \rightarrow 2.5 \leq 2[/latex]  (False)   [latex]\therefore[/latex] not a solution

When [latex]x=0[/latex], [latex]\frac{5(0)}{(0)-3} \leq 2 \rightarrow \frac{0}{-3} \leq 2 \rightarrow 0 \leq 2[/latex]  (True)   [latex]\therefore[/latex] solution

When [latex]x=4[/latex], [latex]\frac{5(4)}{(4)-3} \leq 2 \rightarrow \frac{20}{1} \leq 2 \rightarrow 20 \leq 2[/latex]  (False)   [latex]\therefore[/latex] not a solution

Actually, we can check the results of test values from the graph of [latex]f(x)=\frac{5x}{x-3}-2[/latex] below:

Therefore, the solution of the rational inequality [latex]\frac{5x}{x-3} \leq 2[/latex] is [latex][-2, 3)[/latex].

[latex]\frac{4x-5}{x+1} \geq 0[/latex]

[latex]x+1 \ne 0[/latex]

[latex]x \ne -1[/latex]

[latex]\frac{4x-5}{x+1}=0[/latex]

[latex]\frac{4x-5}{x+1} = \frac{0}{1}[/latex]

[latex]4x-5=0[/latex]

[latex]4x=5[/latex]

[latex]x=\frac{5}{4}[/latex]

Third, plot the restriction and the solution on the number line. All restrictions should be an  open circle. Since the inequality has “[latex]\geq[/latex]” (greater than or equal to), which has “[latex]=[/latex]” (equal sign), the solution should be a  closed circle:

Note that now there are three intervals on the number line: [latex](-\infty, -1), (-1, \frac{5}{4}],[/latex] and [latex][\frac{5}{4}, \infty)[/latex].

I chose [latex]-2, 0,[/latex] and [latex]2[/latex] from those three intervals.

When [latex]x=-2[/latex], [latex]\frac{4(-2)-5}{(-2)+1} \geq 0 \rightarrow \frac{-13}{-1} \geq 0 \rightarrow 13 \geq 0[/latex]  (True)   [latex]\therefore[/latex] solution

When [latex]x=0[/latex], [latex]\frac{4(0)-5}{(0)+1} \geq 0 \rightarrow \frac{-5}{1} \geq 0 \rightarrow -5 \geq 0[/latex]  (False)   [latex]\therefore[/latex] not a solution

When [latex]x=2[/latex], [latex]\frac{4(2)-5}{(2)+1} \geq 0 \rightarrow \frac{3}{3} \geq 0 \rightarrow 1 \geq 0[/latex]  (True)   [latex]\therefore[/latex] solution

Actually, we can check the results of test values from the graph of [latex]f(x)=\frac{4x-5}{x+1}[/latex] below:

Therefore, the solution of the rational inequality [latex]\frac{4x-5}{x+1} \geq 0[/latex] is [latex](-\infty, -1) \cup [\frac{5}{4}, \infty)[/latex].

• Solving Rational Inequalities. Authored by : Michelle Eunhee Chung. Provided by : Georgia State University. License : CC BY: Attribution

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Solving Rational Inequalities

Linear Inequalities Quadratic Inequalities Polynomial Inequalities Rational Inequalities

Solving rational inequalities works much like solving polynomial inequalities , in that the rational expressions will often have many factors which will split the number line into many intervals. However, the important difference for rational inequalities is that, by definition, there will be factors in the denominator, which can cause division by zero.

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So the basic process will be the same, but you'll need to take extra care with the denominators.

How do you solve rational inequalities?

To solve rational inequalities, follow these steps:

• Move terms, as necessary, to isolate the rational expression on one side of the inequality symbol, with zero on the other side.
• Simplify the rational expression, as necessary, to get one fraction (rather than added terms).
• Factor the numerator and denominator completely.
• Solve each factor to find the zeroes.
• Use the zeroes to split the number line into intervals. (Remember that zeroes in the denominator cannot be included in any solution, even if the inequality is an "or equal to" inequality, because you can never divide by zero.)
• Find the solution intervals by using the Test Point Method or the Factor Method.

Using the Test Point Method to test the intervals is generally time-consuming and error-prone. So I would recommend using the Factor Method, since you'll have the factors already anyway.

First off, I have to remember that I can't begin solving until I have the inequality in " = 0 " format. So I'll subtract the 2 over to the left-hand side:

Now I need to convert to a common denominator:

...and then I can simplify:

So the two factors of this rational expression are − x  + 6 and x − 3 . Note that x cannot equal 3 , or else I would be dividing by zero, which is not allowed.

The numerator's factor, − x  + 6 , is positive for:

− x + 6 > 0

The other factor, x  − 3 , is positive for:

x − 3 > 0

I need to remember that x cannot actually equal 3 , so this endpoint can not be included in any solution interval — even though this is an "or equal to" inequality — because I can't divide by zero.

From my inequality endpoints, I see that my intervals are:

(−∞, 3), (3, 6], [6, +∞)

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I have used square brackets at the 6 to remind me that 6 can be included in the solution; the parenthesis at 3 remind me that 3 cannot be included.

Using the Test-Point Method, I would pick a point in each interval and test for the sign on the result. I could use, say, x  = 0 , x  = 4 , and x  = 7 .

Using the Factor Method, I do my grid. I draw the number line aross the bottom of my grid, draw rows for each of the factors, draw vertical lines at the interval endpoints which split the number line into its intervals, and leave the top empty for the signs of the rational expression on each interval. My grid looks like this:

In each factor's row, I fill in the signs of the factors on each of the intervals:

Then I count the "minus" signs in each column to find the sign of the rational expression on that column's interval. I put that interval's sign in the top row:

Checking the original exercise, I am reminded that I am needing the intervals where the rational expression is negative. So my solution is all x 's in the following intervals:

(−∞, 3), [6, +∞)

A quick graph can serve to confirm the algebra:

Everything is already over on one side of the inequality, combined, and factored, so this will be pretty easy to do. I'll start with solving the various factors:

x + 3 > 0  ⇒  x > −3

x − 4 > 0  ⇒  x > 4

x − 1 > 0  ⇒  x > 1

x + 2 > 0  ⇒  x > −2

Now I can draw my factor table:

Checking the original exercise, I see that I need the intervals where the rational expression is positive. I need to remember that, though this is an "or equal to" inequality, I *cannot* include the zeroes x  = −2 and x  = 1 , because doing so would cause division by zero.

So my solution intervals are:

(−∞, −3], (−2, 1), [4, +∞)

A quick graph confirms the solution:

URL: https://www.purplemath.com/modules/ineqsolv4.htm

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Rational Inequalities

Solving rational inequalities.

The key approach in solving rational inequalities relies on finding the critical values of the rational expression which divide the number line into distinct open intervals.

The critical values are simply the zeros of both the numerator and the denominator. You must remember that the zeros of the denominator make the rational expression undefined, so they must be immediately disregarded or excluded as a possible solution. However, zeros of the numerator also need to be checked for their possible inclusion in the overall solution.

In this lesson, I will go over five (5) worked examples with varying levels of difficulty to illustrate both the procedures and concepts.

Examples of How to Solve Rational Inequalities

Example 1: Solve the rational inequality below.

I begin solving this rational inequality by writing it in general form. The general form implies that the rational expression is located on the left side of the inequality while the zero stays on the right.

The general form has four (4) types.

It’s good to know that this problem is already in the general form. My next step is to find the zeros of both numerator and denominator .

I can find the zeros of the numerator by factoring it out completely and then separately set each factor equal to zero and solve for [latex]x[/latex]. Likewise, finding the zeros of the denominator is done the same way.

• Zeros of numerator
• Zeros of denominator

Now, I will use the zeros to separate or partition the number line into intervals. The zeros of the numerator and denominator are also known as the critical numbers . In this case, the two critical numbers divide the number line into three distinct intervals.

The next step is to pick or select a number in every interval and evaluate it back into the original rational inequality; to determine if it is a true or false statement. A true statement means that an interval is part of the solution, otherwise, it is not.

As you can see, the numbers I picked for each interval are highlighted in yellow.

Notice that the open interval between [latex]−1[/latex] and [latex]3[/latex], written as [latex]\left( { – 1,3} \right)[/latex], yields a true statement which implies that it is part of the solution.

So, where else do we look for possible solutions to finish this off?

Check the zeros or critical numbers of the numerators only into the original equation. If it gives a true statement then include that critical number as part of the overall solution.

The zeros of the numerator is [latex]3[/latex]. Now I will verify it.

The use of a square bracket  indicates that it is part of the solution, while an open bracket (parenthesis) denotes that it’s not. I will write my final answer as [latex]\left( { – 1,\left. 3 \right]} \right.[/latex].

Example 2: Solve the rational inequality below.

First off, the given rational inequality is in general form because the rational expression is on the left while the zero is on the right side. That’s good!

Next, I will factor out the numerator and the denominator. After doing so, you should have something like this.

I can now find the zeros of the numerator and denominator.

These zeros or critical numbers divide the number line into distinct intervals or partitions.

Select a test number for each interval and substitute back to the original rational inequality.

  Use the factored form of the original rational inequality to evaluate test numbers for ease of calculation.

The numbers in yellow are the ones I chose to test the validity of each interval.

The intervals yielding true statements are

To find the rest of the solution, check the validity of the zeros of the numerator only into the original rational inequality.

If you have done it correctly, you should agree that [latex]−4 [/latex] and [latex]2[/latex] are not valid answers because they don’t give true statements after checking.

The final answer to this problem in interval notation is

Example 3: Solve the rational inequality below.

I would factor out the numerator and denominator first to find their zeros. In factored form, I got

Next, determine the zeros of the rational inequality by setting each factor equal to zero then solving for [latex]x[/latex].

• Zeros of the numerator: [latex]–1[/latex] and [latex]4[/latex]
• Zeros of the denominator: [latex]4[/latex]

Use the zeros as critical numbers to divide the number line into distinct intervals. I start testing the validity of each interval by selecting test value and evaluating them into the original rational inequality. The ones in yellow are the numbers I picked.

Notice that the only interval giving a true statement is [latex]\left( { – 1,4} \right)[/latex].

More so, the zeros of the numerator don’t check with the original rational inequality so I must disregard them.

The final answer is just [latex]\left( { – 1,4} \right)[/latex].

Example 4: Solve the rational inequality below.

This rational inequality is not in general form . The right-hand side must be zero. The first step is to get rid of the constant on that side by subtracting both sides by [latex]1[/latex]. After that, simplify into a single rational expression. You should have a similar preliminary step just like this.

Next, find the zeros of the numerator and denominator.

• Zeros of the numerator: [latex]-7[/latex]
• Zeros of the denominator: [latex]-3[/latex]

Use the zeros as critical numbers to partition the number line into sections or intervals.

Then pick test numbers for each interval and evaluate them into the general form to determine their truth values. The ones in yellow are the selected values. You may choose other numbers as long as they are in the interval being tested.

The intervals giving true statements are

Meanwhile, after checking the zero of the numerator at [latex]x = – \,7[/latex], it also results in a true statement. Use the square bracket for that to indicate it’s being included as a solution.

The final answer in interval notation should be

Example 5: Solve the rational inequality below.

I need to make the right side of the rational inequality zero. To do that, I will simultaneously add [latex]x[/latex] and subtract [latex]5[/latex] on both sides. However, my ultimate goal is to express it in a single rational expression. This is where your skills on how to add and subtract rational expressions will be useful. You should have similar steps below.

• Zeros of the numerator: [latex]-3[/latex] and [latex]5[/latex]
• Zeros of the denominator: [latex]0[/latex]

Make use of the zeros to divide the number line into distinct intervals. Choose test numbers for each interval to check if it results in true statements. The selected test values for [latex]x[/latex] are in yellow.

The “true” intervals are [latex]\left( { – \,\infty , – 3} \right)[/latex] and [latex]\left( {0,5} \right)[/latex]. More so, the zeros of the numerator also check with the general form of the given rational inequality. Consequently, I have to include [latex]-3[/latex] and [latex]5[/latex] as part of the solution with the use of square brackets.

The final answer now becomes

You might also like these tutorials:

• Solving Rational Equations
• Adding and Subtracting Rational Expressions
• Multiplying Rational Expressions

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looks like a and has a in the denominator. A rational just means we now have a combined with a might look like.

Write the inequality so that there is a single rational expression on the left side of the inequality and zero on the right side of the inequality. Determine the changing points by setting the numerator equal to zero and setting the denominator equal to zero. Use the changing points to separate the number line into intervals. Select test values in each interval and substitute those values into the inequality. If the test value makes the inequality true, then the entire interval is a solution to the inequality. If the test value makes the inequality false, then the entire interval is not a solution to the inequality.

• Express your answer in interval notation.

Example
		on the left and zero on the right? up the into intervals. A number between 0 and 2: x =1 A number larger than 2: x = 3 What interval would make the true?
	What is your answer?

is undefined.
	What is your answer?

is
		on the left and zero on the right? Now you can continue your by first determining the changing points and test three values.
	What is your answer?

in interval notation is is
	What is your answer?

   Catharine H. Colwell
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