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Chapter 7: Problem 14

Solve each rational equation. $$\frac{7}{x+1}=\frac{5}{x-3}$$

Short Answer

Expert verified
The solution to the equation \(\frac{7}{x+1}=\frac{5}{x-3}\) is \(x = 13\).

Step by step solution

01

Cross Multiply

Cross multiply the fractions to get rid of the denominators. We get: \(7*(x-3)=5*(x+1)\)
02

Expand Both Sides

Multiply out both sides of the equation. This gives: \(7x - 21 = 5x + 5\)
03

Rearrange and Solve for x

Now, rearrange the terms to get all the variables on one side and the numbers on the other side of the equation. Subtract \(5x\) from both sides of the equation, this gives us: \(7x-5x = 21 + 5\). Simplifying this gives us: \(2x = 26\). Finally, divide by 2 on both sides of the equation to find the value of \(x\), which gives: \(x=13\)

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Key Concepts

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

Cross-Multiplication
Cross-multiplication is a fundamental technique used in solving rational equations. It involves multiplying the numerator of one fraction by the denominator of the other to eliminate the fractions altogether. In our example, the fractions \( \frac{7}{x+1} \) and \( \frac{5}{x-3} \) are given, and cross-multiplying them results in the equation \( 7 \times (x - 3) = 5 \times (x + 1) \).
  • Determine each cross product: one side becomes the numerator of one fraction times the denominator of the other.
  • Cross-multiplying clears the denominators, simplifying the equation to a linear form.
  • Remember: It's akin to seeing a balance 鈥� what you do to one fraction, you must do to the other.
This step is powerful because it transforms a comparison of two rational expressions into a more straightforward algebraic equation. This makes it easier to solve, especially by novice learners.
Solving Rational Equations
Solving rational equations involves more than just solving typical linear equations. The goal is to identify values of the variable that make the equation true, while eliminating any potential restrictions caused by the denominators. After cross-multiplying in our example, we obtain a simpler equation \( 7x - 21 = 5x + 5 \). This is straightforward to solve, but keep in mind a few key points:
  • First, ensure that the values found do not make any denominator zero, as this would make the original equation undefined.
  • Isolate the variable by getting terms involving the variable on one side and constants on the other.
  • Check your solution back in the original equation, ensuring it doesn't lead to an impossible situation (like dividing by zero).
This ensures that you not only solve the equation but also understand any limitations posed by the original rational expressions.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging and simplifying equations to isolate variables and solve them. This involves systematic steps to ensure that equations remain balanced until the solution is found. In our equation, once we cross-multiply, we distribute and rearrange terms effectively.
  • Expand each side after cross-multiplication: multiply each term within the parentheses by the external factor (e.g., \( 7 \times (x-3) = 7x - 21 \)).
  • Rearrange by moving terms: ensure variable terms are on one side, constants on the other by adding or subtracting terms from both sides.
  • Simplify and solve: once isolated, solve for the unknown variable by performing arithmetic operations (adding, subtracting, multiplying, or dividing).
Through proper manipulation, we solve for \( x \), giving us the final result: \( x = 13 \). This example shows how rational equations can be tackled systematically using algebraic principles.

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Most popular questions from this chapter

Use a proportion to solve each problem. According to the authors of Number Freaking, in a global village of 200 people, 9 get drunk every day. How many of the world's 6.9 billion people \((2010\) population) get drunk everyday? Round to the nearest hundredth of a billion.

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{x-8}{x^{2}-16}-\frac{x-8}{16-x^{2}}$$

add or subtract as indicated. Simplify the result, if possible. $$\frac{6 y^{2}+y}{2 y^{2}-9 y+9}-\frac{2 y+9}{2 y^{2}-9 y+9}-\frac{4 y-3}{2 y^{2}-9 y+9}$$

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{6 x+7}{x-6}+\frac{3 x}{6-x}$$

perform the indicated operations. Simplify the result, if possible. $$\left(\frac{3 x-1}{x^{2}+5 x-6}-\frac{2 x-7}{x^{2}+5 x-6}\right) \div \frac{x+2}{x^{2}-1}$$
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