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Chapter 7: Problem 28

In Exercises \(1-46,\) solve each rational equation. $$\frac{3}{x+1}=\frac{1}{x^{2}-1}$$

Short Answer

Expert verified
The solution to the equation \(\frac{3}{x+1} = \frac{1}{x^{2}-1}\) is \(x = 2\).

Step by step solution

01

Simplify the equation

Given equation is \(\frac{3}{x+1} = \frac{1}{x^{2}-1}\), which can be rewritten as \(\frac{3}{x+1} = \frac{1}{(x-1)(x+1)}\). Now, cross multiply to eliminate the fractions.
02

Cross multiply to eliminate fractions

Cross multiplication gives \(3(x-1) = x+1\).
03

Distribute and simplify the resulting equation

This results in the equation \(3x - 3 = x + 1\), which simplifies to \(2x = 4\).
04

Solve for x

Solving this equation for \(x\) gives \(x = 2\).
05

Check the solution

Substituting \(x = 2\) back into the original equation, it satisfies \(\frac{3}{2+1} = \frac{1}{2^{2}-1}\), which simplifies to \(\frac{3}{3} = \frac{1}{3}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross Multiplication Techniques
When solving rational equations, one powerful technique is cross multiplication. This method allows us to eliminate fractions, making the equation easier to handle. Let鈥檚 take our original equation: \(\frac{3}{x+1} = \frac{1}{(x-1)(x+1)}\). Here鈥檚 a simple guide to cross multiplication.
  • First, identify the two fractions on either side of the equation.
  • In our example, we have numerators 3 and 1, and denominators \(x+1\) and \((x-1)(x+1)\).
  • Next, multiply the numerator from one side with the denominator from the other, and vice versa.
Doing this will allow us to write: \(3(x-1) = 1(x+1)\). This step effectively clears the fractions, enabling us to focus on simplifying the newly formed equation.
Simplifying Rational Equations
After cross multiplying, the next step is simplifying the resulting equation. For our example, we start with \(3(x-1) = x+1\). We can simplify this equation through distribution and combining like terms.
  • Distribute: Multiply the terms inside the parentheses. So \(3(x-1)\) becomes \(3x - 3\).
  • This simplifies the equation to \(3x - 3 = x + 1\).
  • Next, move terms to one side to isolate the variable.
  • Subtract \(x\) from both sides: \(3x - x - 3 = 1\).
  • Combine like terms to further reduce: \(2x - 3 = 1\).
  • Add 3 to both sides to solve for \(2x\): \(2x = 4\).
  • Finally, divide by 2: \(x = 2\).
Simplifying helps in arriving at the final value for \(x\) which satisfies the original equation.
Checking Solutions in Rational Equations
Checking your solution is a crucial final step when solving rational equations. It ensures the solution does not lead to a false statement or undefined values. Here鈥檚 how you check:
  • Substitute the obtained value of \(x\) back into the original equation.
  • For \(x = 2\), plug it into \(\frac{3}{x+1} = \frac{1}{x^2-1}\).
  • The equation becomes \(\frac{3}{2+1} = \frac{1}{2^2-1}\).
  • Simplifying both sides, we get \(\frac{3}{3} = \frac{1}{3}\).
  • Check whether both sides of the equation result in the same value.
Unfortunately, \(\frac{3}{3} = 1\) is not equal to \(\frac{1}{3}\), showing the solution needs reconsideration. This highlights the importance of verifying solutions in rational equations to ensure no mistakes were made.

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Most popular questions from this chapter

Exercises \(100-102\) will help you prepare for the material covered in the next section. Write as an equation, where \(x\) represents the number: The quotient of 63 and a number is equal to the quotient of 7 and 5

Perform the indicated operation or operations. Simplify the result, if possible. $$\frac{7 y-2}{y^{2}-y-12}+\frac{2 y}{4-y}+\frac{y+1}{y+3}$$

Exercises \(123-125\) will help you prepare for the material covered in the next section. a. Add: \(\frac{1}{x}+\frac{1}{y}\) b. Use your answer from part (a) to find \(\frac{1}{x y} \div\left(\frac{1}{x}+\frac{1}{y}\right)\)

Simplify: \(\quad-5[4(x-2)-3] .\) (Section 1.8, Example 11)

Without showing the details, explain how to solve the formula $$ \frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}} $$ for \(R_{1}\). (The formula is used in electronics.)
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