Problem 46
Question
Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{1}{x+1}>\frac{2}{x-1} $$
Step-by-Step Solution
Verified Answer
The solution in interval notation is \((-\infty, -3)\).
1Step 1: Find common denominator
Common denominator of \(x+1\) and \(x-1\) is \((x+1)(x-1)\), which simplifies to \(x^2-1\). Multiply each fraction by \(x^2-1\) to remove the denominators. This gives: \((x-1) > 2(x+1)\)
2Step 2: Simplify and solve the inequality
Expanding the brackets we get \(x-1 > 2x+2\). Simplifying this further gives: \(-x > 3\). To keep the inequality sign the same, we multiply each side by -1 and switch the inequality sign to '<'. This gives: \(x < -3\)
3Step 3: Find the restriction
To find the restriction in x, we set denominators equal to zero. This gives us \(x=-1, x=1\). Since the given inequality excluded these points, our solution for the inequality \(x < -3\) holds.
4Step 4: Interval notation and number line representation
Express the solution as an interval. Our solution \(x < -3\) can be written in interval notation as \((-\infty, -3)\). The graph on the number line will be a ray pointing to the left from the point -3, excluding the point itself.
Key Concepts
Common DenominatorInequality SolutionInterval NotationNumber Line Representation
Common Denominator
In rational inequalities, finding a common denominator is a crucial step. It allows us to compare fractions without denominators. Given an inequality such as \(\frac{1}{x+1} > \frac{2}{x-1}\), we first find a common denominator for the two fractions. For the terms \(x+1\) and \(x-1\), a common denominator is their product, which is \((x+1)(x-1)\). Simplifying the product, we get \(x^2 - 1\). This common denominator turns our fractions into expressions that can be compared directly.
\[\frac{(x-1)}{x^2 - 1} > \frac{2(x+1)}{x^2 - 1}\]By multiplying both sides by \(x^2 - 1\), we clear the denominators, simplifying the inequality to an easier form: \(x-1 > 2(x+1)\).
This process transforms the problem from a complex inequality involving fractions into a simpler, more straightforward algebraic inequality.
\[\frac{(x-1)}{x^2 - 1} > \frac{2(x+1)}{x^2 - 1}\]By multiplying both sides by \(x^2 - 1\), we clear the denominators, simplifying the inequality to an easier form: \(x-1 > 2(x+1)\).
This process transforms the problem from a complex inequality involving fractions into a simpler, more straightforward algebraic inequality.
Inequality Solution
After finding a common denominator and clearing it, the next step is solving the resulting inequality. From the transformation made in the previous step, we have the inequality \(x-1 > 2(x+1)\).
Expanding and simplifying this, we get: \[x - 1 > 2x + 2\]
Rearrange terms to isolate the variable \(x\):
Expanding and simplifying this, we get: \[x - 1 > 2x + 2\]
Rearrange terms to isolate the variable \(x\):
- Subtract \(2x\) from both sides: \(-x - 1 > 2\)
- Add \(1\) to both sides: \(-x > 3\)
Interval Notation
Interval notation is a concise way to describe sets of solutions. It harnesses brackets and parenthesis to denote the start and end of solution intervals on the number line. In our current example, the solution to the inequality is \(x < -3\). In interval notation, this is written as \((-\infty, -3)\).
- Round or open parenthesis, \((\), indicates that the endpoint is not included in the solution set.
- An interval starting with negative infinity, \(\ (-\infty\)\, indicates that the range extends infinitely in the negative direction.
- The number at the end, in this case \(-3\), represents the boundary point that is not part of the solution set.
Number Line Representation
A number line representation visually displays the solutions of inequalities, making it easier to understand at a glance. For the solution \(x < -3\), draw a number line and locate the point \(-3\). Indicate an open circle at \(-3\) to show that \(-3\) itself is not included in the solution.
- Draw a ray or an arrow starting from this point and extending to the left, depicting all numbers less than \(-3\).
- The open circle clearly shows exclusion of the boundary point.
- The arrow signifies the continuation in the negative direction.
Other exercises in this chapter
Problem 45
Exercises \(31-50\) contain equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator ze
View solution Problem 46
Explain how to add complex numbers. Provide an example with your explanation.
View solution Problem 46
Solve each equation in Exercises \(39-54\) by completing the square. $$ x^{2}+6 x-5=0 $$
View solution Problem 46
Solve each linear inequality in Exercises 27-48 and graph the solution set on a number line. Express the solution set using interval notation. $$\frac{4 x-3}{6}
View solution