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Chapter 6: Problem 35

Find the solution set to each equation. $$\frac{x}{6}=\frac{5}{x-1}$$

Short Answer

Expert verified
The solution set is \( x = 6 \) and \( x = -5 \).

Step by step solution

01

Cross-multiply the fractions

To eliminate the fractions, cross-multiply both sides of the equation. This means multiplying the numerator of one fraction by the denominator of the other fraction and setting the products equal to each other. \frac{x}{6}=\frac{5}{x-1} becomes \( x(x-1) = 5 \times 6 \)
02

Expand and simplify

Expand the equation obtained from cross-multiplication. \( x(x-1) = 5 \times 6 \) This simplifies to: \( x^2 - x = 30 \)
03

Form a quadratic equation

To form a standard quadratic equation, move all terms to one side of the equation. \( x^2 - x - 30 = 0 \)
04

Factor the quadratic equation

Factor the quadratic equation \( x^2 - x - 30 \) to find the values of x. \( x^2 - x - 30 = (x - 6)(x + 5) = 0 \)
05

Solve for x

Set each factor equal to zero to solve for x. \( x - 6 = 0 \) or \( x + 5 = 0 \) So, \( x = 6 \) or \( x = -5 \)
06

Verify the solutions

Substitute \( x = 6 \) and \( x = -5 \) back into the original equation to verify they are valid solutions. For \( x = 6 \): \( \frac{6}{6} = 1 \) and \( \frac{5}{6-1} = 1 \) which is true. For \( x = -5 \): \( \frac{-5}{6} = -\frac{5}{6} \) and \( \frac{5}{-5-1} = -\frac{5}{6} \) which is also true.

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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 method used to eliminate fractions from an equation. When dealing with an equation like \(\frac{x}{6} = \frac{5}{x-1}\), you can get rid of the fractions by multiplying the numerator of each fraction by the denominator of the other fraction. This process forms a new equation where the fractions are gone.
For instance, from \(\frac{x}{6} = \frac{5}{x-1}\), you multiply across to get \(x \times (x-1) = 6 \times 5\).
Hence, this becomes \(x(x-1) = 30\). Now, you have an equation that's easier to work with, without the fractions.
Factoring quadratic equations
Factoring is a key method for solving quadratic equations. Once you have the quadratic equation in the form \(ax^2 + bx + c = 0\), you can start to factor it if possible.
Let's continue with our example. From cross-multiplication, we obtained \(x^2 - x = 30\).
To get a standard quadratic form, rearrange it to \(x^2 - x - 30 = 0\).
Now, we need to factor this quadratic equation.
We look for two numbers that multiply to \-30\ and add to \-1\. These numbers are \-6\ and \+5\. So, \(x^2 - x - 30\) factors into \((x - 6)(x + 5) = 0\).
Splitting the equation, we get two possible solutions: \(x - 6 = 0\) and \(x + 5 = 0\).
Thus, \(x = 6\) and \(x = -5\) are the solutions.
Verifying solutions
It is essential to verify your solutions to ensure they are correct, particularly in quadratic equations.
Let's go back to our solutions \(x = 6\) and \(x = -5\) and substitute them back into the original equation \(\frac{x}{6} = \frac{5}{x-1}\).
For \(x = 6\), substitute to get \(\frac{6}{6} = \frac{5}{6-1}\), which simplifies to \(1 = 1\), confirming \(x = 6\) is correct.
For \(x = -5\), substitute to get \(\frac{-5}{6} = \frac{5}{-5-1}\), which simplifies to \(-\frac{5}{6} = -\frac{5}{6}\), confirming \(x = -5\) is also correct.
By verifying both solutions, you ensure that you have accurate answers to the problem.

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Most popular questions from this chapter

Find the solution set to each equation. $$\frac{x+1}{x-5}=\frac{x+2}{x-4}$$

Perform the indicated operations. Variables in exponents represent integers. $$\frac{w^{2 b}+2 w^{b}-8}{w^{2 b}+3 w^{b}-4} \div \frac{w^{2 b}-w^{b}-2}{w^{2 b}-1}$$

Perform the indicated operations. Variables in exponents represent integers. $$\frac{x^{2 a}+x^{a}-6}{x^{2 a}+6 x^{a}+9} \div \frac{x^{2 a}-4}{x^{2 a}+2 x^{a}-3}$$

Perform the indicated operations. $$\frac{3 x^{2}+13 x-10}{x} \cdot \frac{x^{3}}{9 x^{2}-4} \cdot \frac{7 x-35}{x^{2}-25}$$

Perform the indicated operations. When possible write down only the answer. $$\frac{\frac{3 a}{5 b}}{2}$$
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