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Chapter 9: Problem 4

Solve each equation. Check each solution. $$ \frac{2}{x-1}=\frac{x+4}{3} $$

Short Answer

Expert verified
The solutions to the equation \(\frac{2}{x-1}=\frac{x+4}{3}\) are \(x=1\) and \(x=2\). However, after checking, only \(x=2\) is a valid solution as \(x=1\) makes the denominator of the original equation zero.

Step by step solution

01

Finding the Least Common Multiple (LCM)

The denominators of the given rational equation are \(x-1\) and \(3\). The least common multiple of \(x-1\) and \(3\) is \(3(x-1)\). This will be used to eliminate the denominator.
02

Applying the LCM to remove the denominators

Multiply the entire equation by \(3(x-1)\) to get rid of the denominators:\n \[3(x-1)*\frac{2}{x-1} = 3(x-1)*\frac{x+4}{3}\] This simplifies to \(2*3(x-1)= (x-1)(x+4)\). Distributing the multiplication further simplifies the equation to \(6x-6= x^2+x-4\).
03

Solve the quadratic equation

Like any quadratic equation, it needs to be rearranged in the form of \(ax^2+bx+c=0\). Thus, the equation becomes: \(x^2-5x+2=0\). With this equation, we can solve for x using the quadratic formula \(x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
04

Calculate the solutions

Using the quadratic formula, the solutions are \(x=\frac{5 \pm \sqrt{25-8}}{2}\).This simplifies to \(x=1\) or \(x=2\).
05

Check the solutions

Substitute \(x=1\) and \(x=2\) back into the original equation to see if both of them are valid solutions.

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Key Concepts

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

Least Common Multiple (LCM)
When solving rational equations, finding the Least Common Multiple (LCM) is a crucial step. The LCM helps us combine terms and eliminate fractions, making the equation easier to manage. In this exercise, the denominators are \(x-1\) and \(3\). By finding the LCM, which is \(3(x-1)\), we can multiply the entire equation by this term to remove the denominators.

This results in both sides of the equation being free of fractions, allowing us to simplify and solve the equation more easily. Remember, multiplying by the LCM ensures the equation stays balanced, as long as we apply it to both sides.
Quadratic Equation
A quadratic equation is any equation that can be rearranged into the standard form \(ax^2 + bx + c = 0\). The standard form helps us in identifying the nature of the equation and solving it efficiently. After removing the denominators from the rational equation, the result is simplified to a quadratic equation: \(6x - 6 = x^2 + x - 4\).

We can then further rearrange it to form \(x^2 - 5x + 2 = 0\). This is now set to be solved using methods designated for quadratic equations, such as factoring, completing the square, or the quadratic formula.
Quadratic Formula
The Quadratic Formula is a dependable method for finding the roots of a quadratic equation. It is expressed as \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients from the equation \(ax^2 + bx + c = 0\).

In our example, we have \(a = 1\), \(b = -5\), and \(c = 2\). Substituting these values into the formula provides the solutions for \(x\). This method is especially helpful when the quadratic cannot be easily factored. Simply calculate under the square root \(b^2 - 4ac\) to ensure it's a valid input before proceeding with the solution.
Checking Solutions
Once we've calculated solutions using the quadratic formula, it's important to ensure that each solution is valid for the original equation. This step involves substituting back the solutions into the original rational equation.

In our task, after solving for \(x\), the possible solutions were \(x = 1\) and \(x = 2\). By substituting these values into \(\frac{2}{x-1} = \frac{x+4}{3}\), we can verify which values satisfy the equation. In practice, solutions that result in division by zero or false statements must be discarded. Always remember: the goal is to ensure solutions are consistent with the original equation.

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Most popular questions from this chapter

Physics The acceleration of an object is a measure of how much its velocity changes in a given period of time. acceleration \(=\frac{\text { final velocity }-\text { initial velocity }}{\text { time }}\) Suppose you are riding a bicycle at 6 \(\mathrm{m} / \mathrm{s}\) . You step hard on the pedals and increase your speed to 12 \(\mathrm{m} / \mathrm{s}\) in about 5 \(\mathrm{s}\) . a. Find your acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) . b. A sedan can go from 0 to 60 \(\mathrm{mi} / \mathrm{h}\) in about 10 s. What is the acceleration in \(\mathrm{m} / \mathrm{s}^{2} ?(\text { Hints: } 1 \mathrm{mi} \approx 1609 \mathrm{m} ; 1 \mathrm{h}=3600 \mathrm{s} .)\)

A jar contains four blue marbles and two red marbles. Suppose you choose a marble at random, and do not replace it. Then you choose a second marble. Find the probability of each event. You select two red marbles or two blue marbles.

Solve each equation. $$ \ln x-\ln 4=5 $$

Solve each equation. Check your answer. $$ \frac{1}{2}-x=\frac{x}{6} $$

Use this information for Exercises \(53-58\) . Bag 1 contains 5 red marbles, 1 blue marble, 3 yellow marbles, and 2 green marbles. Bag 2 contains 1 red pencil, 3 red pens, 2 blue pencils, and 5 blue pens. One marble is drawn from bag \(1 .\) What is the probability that the marble is blue or not green?
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