/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 Solve the equation and check you... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 0: Problem 43

Solve the equation and check your solution. (If not possible, explain why.) $$ \frac{3}{x^{2}-3 x}+\frac{4}{x}=\frac{1}{x-3} $$

Short Answer

Expert verified
The equation has no solution, as substitution of the obtained potential solutions (\(x = 0\) and \(x = 6\)) back into the original equation renders the equation undefined.

Step by step solution

01

Fractions elimination

First step is to eliminate the fractions. We do this by multiplying every term in the equation by the least common multiple (LCM) of the denominators. So, the LCM of \(x^{2}-3 x\), \(x\) and \(x-3\) is \(x(x-3)(x^{2}-3 x)\). Using this, the equation becomes: \[ 3x(x-3)+ 4(x-3)(x^{2}-3 x) = (x)(x^{2}-3 x) \] Simplify this equation to get: \[ 3x^2 - 3x^3 + 4x^{3} - 12x^{2} = x^{3} - 3x^{2} \]
02

Solve the equation

Next, we simplify and solve the algebraic equation. By collecting all the terms to one side, we will have: \[3x^3 - 3x^2- x^{3} + 3x^{2}- 4x^{3} + 12x^{2} = 0 \] This simplifies to: \[-2x^3 + 12x^2 = 0\] Factor out the equation to get: \[ -2x^2(x - 6) = 0 \] Solving the equation, we will have \(x = 0\) or \(x = 6\) as potential solutions.
03

Check the solutions

We need to confirm whether these solutions are valid by substituting them back into the original equation. If substituting a value results in the denominator being zero (thus undefined), then it is not a solution. When substituting \(x= 6\), we have: \[\frac{3}{(6)^{2}-3(6)}+\frac{4}{6}=\frac{1}{6-3}\] Which results in \[\frac{3}{0}+\frac{4}{6}=\frac{1}{3}\]which is not possible since we have a zero at the denominator. Thus, \(x = 6\) is not a solution to the original equation. When substituting \(x=0\), we get: \[\frac{3}{(0)^{2}-3(0)}+\frac{4}{0}=\frac{1}{0-3}\] here, we have 4 divided by zero in the second term, meaning \(x = 0\) is not a solution either. Therefore, the original equation has no solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the average rate of change of the function from \(x_{1}\) to \(x_{2}\). $$ \begin{array}{cc} \text { Function } & x \text {-Values } \\ f(x)=x^{2}+12 x-4 & \quad x_{1}=1, x_{2}=5 \end{array} $$

You need a total of 50 pounds of two types of ground beef costing \(\$ 1.25\) and \(\$ 1.60\) per pound, respectively. A model for the total cost \(y\) of the two types of beef is $$ y=1.25 x+1.60(50-x) $$ where \(x\) is the number of pounds of the less expensive ground beef. (a) Find the inverse function of the cost function. What does each variable represent in the inverse function? (b) Use the context of the problem to determine the domain of the inverse function. (c) Determine the number of pounds of the less expensive ground beef purchased when the total cost is \(\$ 73\).

Find the average rate of change of the function from \(x_{1}\) to \(x_{2}\). $$ \begin{array}{cc} \text { Function } & x \text {-Values } \\ f(x)=-\sqrt{x+1}+3 &\quad x_{1}=3, x_{2}=8 \end{array} $$

In Exercises 39-54, (a) find the inverse function of \(f\), (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\), and (d) state the domain and range of \(f\) and \(f^{-1}\). $$ f(x)=\frac{x+1}{x-2} $$

True or False? In Exercises 85 and 86, determine whether the statement is true or false. Justify your answer. Proof Prove that if \(f\) is a one-to-one odd function, then \(f^{-1}\) is an odd function.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.