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Chapter 1: Problem 32

\(9- 46\) The given equation is either linear or equivalent to a linear equation. Solve the equation. $$ \frac{6}{x-3}=\frac{5}{x+4} $$

Short Answer

Expert verified
The solution is \(x = -39\).

Step by step solution

01

Cross-Multiply the Fractions

Start by cross-multiplying the two fractions to eliminate the denominators. This means multiplying the numerator of each fraction by the denominator of the other:\[ 6(x + 4) = 5(x - 3) \]
02

Distribute the Terms

Distribute the terms on both sides of the equation:\[ 6x + 24 = 5x - 15 \]
03

Move Variables to One Side

Subtract \(5x\) from both sides to get the variables on one side of the equation:\[ 6x - 5x + 24 = -15 \]This simplifies to:\[ x + 24 = -15 \]
04

Solve for x

Subtract 24 from both sides to solve for \(x\):\[ x = -15 - 24 \]\[ x = -39 \]
05

Verify the Solution

Substitute \(x = -39\) back into the original equation to verify it's correct:\[ \frac{6}{-39-3} = \frac{5}{-39+4} \]Both simplify to:\[ \frac{6}{-42} = \frac{5}{-35} \]Simplifying gives approximately -0.143 (both sides), which verifies the solution.

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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 powerful technique used to solve equations involving fractions. It helps us eliminate the denominators, making the equation easier to handle.
  • Consider the equation with two fractions on either side of an equal sign: \(\frac{a}{b} = \frac{c}{d}\).
  • To cross-multiply, multiply the numerator of the first fraction by the denominator of the second, and vice versa. This process turns the equation into: \(a \, \cdot \, d = c \, \cdot \, b\).
  • The resulting equation no longer has any fractions, allowing us to focus on solving a simpler linear equation.
Cross-multiplication is especially useful when the equation consists of one fraction on each side. Carefully performing this step will set the stage for solving the equation with minimal complexity.
Solving Equations
Solving linear equations involves finding the value of a variable that makes the equation true. After cross-multiplying, you're typically left with a linear equation where the variable appears in one or both terms.
  • First, look at the distributed terms on both sides of the equation. Distributing means multiplying out any parentheses. For example, \(6(x + 4)\) becomes \(6x + 24\).
  • Gather all terms involving the variable on one side by adding or subtracting terms appropriately. Here, subtract \(5x\) from both sides to get all \(x\) terms on the left side.
  • Simplify the equation step-by-step: \(6x - 5x + 24 = -15\) simplifies to \(x + 24 = -15\).
  • Finally, isolate the variable by performing inverse operations. In this case, subtract 24 from both sides to find that \(x = -39\).
When solving, it's crucial to maintain balance in the equation by performing the same operation on both sides. This ensures the equation stays true throughout the process.
Verification of Solutions
Verification of solutions is a crucial step in the problem-solving process. It helps confirm that the solution found is correct.
  • After calculating the variable, substitute it back into the original equation. For example, substitute \(x = -39\) back into \(\frac{6}{x-3} = \frac{5}{x+4}\).
  • Simplify both sides of the equation with the substituted value: \(\frac{6}{-42} = \frac{5}{-35}\).
  • Check if both sides are numerically equal once simplified. In this instance, both sides equal approximately -0.143.
If the sides are equal, this confirms the accuracy of your solution. Verification not only confirms correctness but also improves understanding and confidence in solving similar equations.

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Most popular questions from this chapter

Solve the inequality. Express the answer using interval notation. $$ 3-|2 x+4| \leq 1 $$

\(33-66\) . Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ \frac{6}{x-1}-\frac{6}{x} \geq 1 $$

Falling Ball Using calculus it can be shown that if a ball is thrown upward with an initial velocity of 16 \(\mathrm{ft} / \mathrm{s}\) from the top of a building 128 \(\mathrm{ft}\) high, then its height \(h\) above the ground \(t\) seconds later will be $$ h=128+16 t-16 t^{2} $$ During what time interval will the ball be at least 32 \(\mathrm{ft}\) above the ground?

\(33-66\) . Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ -2 x^{2} \leq 4 $$

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