Solving A Rational Inequality$\[ \frac{5x-3}{(x-1)(x+1)} \leq 0 \\]Type True Or False For Each Statement.1. For Test Point \[$x = -2\$\], The Inequality Is \[$\_\_\_\_\_\$\].2. For Test Point \[$x = 0\$\], The

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Introduction


Rational inequalities are a type of mathematical inequality that involves a rational expression, which is a fraction of two polynomials. Solving rational inequalities requires finding the values of the variable that make the inequality true. In this article, we will solve the rational inequality 5x−3(x−1)(x+1)≤0\frac{5x-3}{(x-1)(x+1)} \leq 0.

Understanding the Rational Inequality


The given rational inequality is 5x−3(x−1)(x+1)≤0\frac{5x-3}{(x-1)(x+1)} \leq 0. To solve this inequality, we need to find the values of xx that make the expression 5x−3(x−1)(x+1)\frac{5x-3}{(x-1)(x+1)} less than or equal to zero.

Finding the Critical Points


The critical points of a rational inequality are the values of the variable that make the numerator or denominator equal to zero. In this case, the critical points are the values of xx that make the numerator 5x−35x-3 equal to zero or the denominator (x−1)(x+1)(x-1)(x+1) equal to zero.

Finding the Critical Points of the Numerator


To find the critical points of the numerator, we need to solve the equation 5x−3=05x-3 = 0. We can do this by adding 3 to both sides of the equation and then dividing both sides by 5.

5x−3=05x-3 = 0

5x=35x = 3

x=35x = \frac{3}{5}

So, the critical point of the numerator is x=35x = \frac{3}{5}.

Finding the Critical Points of the Denominator


To find the critical points of the denominator, we need to solve the equation (x−1)(x+1)=0(x-1)(x+1) = 0. We can do this by setting each factor equal to zero and solving for xx.

(x−1)=0(x-1) = 0

x=1x = 1

(x+1)=0(x+1) = 0

x=−1x = -1

So, the critical points of the denominator are x=1x = 1 and x=−1x = -1.

Testing the Critical Points


To determine the sign of the rational expression, we need to test the critical points. We can do this by substituting each critical point into the rational expression and determining the sign of the result.

Testing the Critical Point x=−2x = -2


Substituting x=−2x = -2 into the rational expression, we get:

5(−2)−3((−2)−1)((−2)+1)\frac{5(-2)-3}{((-2)-1)((-2)+1)}

−10−3(−3)(−1)\frac{-10-3}{(-3)(-1)}

−133\frac{-13}{3}

Since the result is negative, the inequality is True for x=−2x = -2.

Testing the Critical Point x=0x = 0


Substituting x=0x = 0 into the rational expression, we get:

5(0)−3((0)−1)((0)+1)\frac{5(0)-3}{((0)-1)((0)+1)}

−3(−1)(1)\frac{-3}{(-1)(1)}

−3−1\frac{-3}{-1}

33

Since the result is positive, the inequality is False for x=0x = 0.

Testing the Critical Point x=35x = \frac{3}{5}


Substituting x=35x = \frac{3}{5} into the rational expression, we get:

5(35)−3((35)−1)((35)+1)\frac{5(\frac{3}{5})-3}{((\frac{3}{5})-1)((\frac{3}{5})+1)}

3−3(35−55)(35+55)\frac{3-3}{(\frac{3}{5}-\frac{5}{5})(\frac{3}{5}+\frac{5}{5})}

0(−25)(85)\frac{0}{(\frac{-2}{5})(\frac{8}{5})}

0−1625\frac{0}{\frac{-16}{25}}

00

Since the result is zero, the inequality is True for x=35x = \frac{3}{5}.

Testing the Critical Point x=1x = 1


Substituting x=1x = 1 into the rational expression, we get:

5(1)−3((1)−1)((1)+1)\frac{5(1)-3}{((1)-1)((1)+1)}

5−3(0)(2)\frac{5-3}{(0)(2)}

20\frac{2}{0}

Since the result is undefined, the inequality is False for x=1x = 1.

Testing the Critical Point x=−1x = -1


Substituting x=−1x = -1 into the rational expression, we get:

5(−1)−3((−1)−1)((−1)+1)\frac{5(-1)-3}{((-1)-1)((-1)+1)}

−5−3(−2)(0)\frac{-5-3}{(-2)(0)}

−80\frac{-8}{0}

Since the result is undefined, the inequality is False for x=−1x = -1.

Conclusion


In conclusion, the rational inequality 5x−3(x−1)(x+1)≤0\frac{5x-3}{(x-1)(x+1)} \leq 0 is true for x=−2x = -2, x=35x = \frac{3}{5}, and false for x=0x = 0, x=1x = 1, and x=−1x = -1. The solution to the inequality is x∈(−∞,−1)∪(−1,1)∪(35,∞)x \in (-\infty, -1) \cup (-1, 1) \cup (\frac{3}{5}, \infty).

Final Answer


The final answer is:

  1. For test point x=−2x = -2, the inequality is True.
  2. For test point x=0x = 0, the inequality is False.
  3. For test point x=35x = \frac{3}{5}, the inequality is True.
  4. For test point x=1x = 1, the inequality is False.
  5. For test point x=−1x = -1, the inequality is False.

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Introduction


In our previous article, we solved the rational inequality 5x−3(x−1)(x+1)≤0\frac{5x-3}{(x-1)(x+1)} \leq 0. In this article, we will answer some frequently asked questions about solving rational inequalities.

Q&A


Q: What is a rational inequality?

A: A rational inequality is a type of mathematical inequality that involves a rational expression, which is a fraction of two polynomials.

Q: How do I solve a rational inequality?

A: To solve a rational inequality, you need to find the values of the variable that make the inequality true. You can do this by finding the critical points of the numerator and denominator, testing the critical points, and determining the sign of the rational expression.

Q: What are critical points?

A: Critical points are the values of the variable that make the numerator or denominator equal to zero. They are also called zeros or roots.

Q: How do I find the critical points of a rational inequality?

A: To find the critical points of a rational inequality, you need to solve the equation that makes the numerator or denominator equal to zero.

Q: What is the difference between a rational inequality and a rational equation?

A: A rational equation is an equation that involves a rational expression, while a rational inequality is an inequality that involves a rational expression.

Q: Can I use the same methods to solve rational inequalities and rational equations?

A: Yes, you can use the same methods to solve rational inequalities and rational equations. However, you need to be careful when testing the critical points, as the inequality may be true or false for certain values of the variable.

Q: How do I determine the sign of a rational expression?

A: To determine the sign of a rational expression, you need to test the critical points and determine the sign of the result. You can also use a sign chart or a number line to help you determine the sign of the rational expression.

Q: What is a sign chart?

A: A sign chart is a table that shows the sign of a rational expression for different values of the variable. It can be used to help you determine the sign of the rational expression.

Q: How do I use a sign chart to solve a rational inequality?

A: To use a sign chart to solve a rational inequality, you need to create a table that shows the sign of the rational expression for different values of the variable. You can then use the table to determine the values of the variable that make the inequality true.

Q: What is a number line?

A: A number line is a line that shows the values of the variable on a scale. It can be used to help you determine the sign of a rational expression.

Q: How do I use a number line to solve a rational inequality?

A: To use a number line to solve a rational inequality, you need to create a line that shows the values of the variable on a scale. You can then use the line to determine the values of the variable that make the inequality true.

Conclusion


In conclusion, solving rational inequalities requires finding the critical points of the numerator and denominator, testing the critical points, and determining the sign of the rational expression. You can use a sign chart or a number line to help you determine the sign of the rational expression. We hope this article has helped you understand how to solve rational inequalities.

Final Answer


The final answer is:

  • A rational inequality is a type of mathematical inequality that involves a rational expression.
  • To solve a rational inequality, you need to find the critical points of the numerator and denominator, test the critical points, and determine the sign of the rational expression.
  • Critical points are the values of the variable that make the numerator or denominator equal to zero.
  • You can use a sign chart or a number line to help you determine the sign of a rational expression.
  • A sign chart is a table that shows the sign of a rational expression for different values of the variable.
  • A number line is a line that shows the values of the variable on a scale.