/*! 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 99 Solve the equation. $$ \frac... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 99

Solve the equation. $$ \frac{1}{3} x+\frac{1}{2}=\frac{1}{2}(x+1)-\frac{1}{6} x $$

Short Answer

Expert verified
All real numbers are solutions.

Step by step solution

01

Distribute and simplify

Distribute the \(\frac{1}{2}\) on the right side of the equation: \(\frac{1}{2}(x+1) = \frac{1}{2}x + \frac{1}{2}\). The equation becomes: \[\frac{1}{3}x + \frac{1}{2} = \frac{1}{2}x + \frac{1}{2} - \frac{1}{6}x\]
02

Combine like terms on the right side

Combine the terms involving \(x\) on the right side: \(\frac{1}{2}x - \frac{1}{6}x = \frac{3}{6}x - \frac{1}{6}x = \frac{2}{6}x = \frac{1}{3}x\). The equation now reads: \[\frac{1}{3}x + \frac{1}{2} = \frac{1}{3}x + \frac{1}{2}\]
03

Eliminate x from both sides

Subtract \(\frac{1}{3}x\) from both sides to eliminate \(x\): \[\frac{1}{3}x + \frac{1}{2} - \frac{1}{3}x = \frac{1}{3}x + \frac{1}{2} - \frac{1}{3}x\]This simplifies to: \[\frac{1}{2} = \frac{1}{2}\]
04

Analyze the result

The equation \(\frac{1}{2} = \frac{1}{2}\) is always true, indicating that any \(x\) value will satisfy the equation.

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!

Key Concepts

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

Distribution Property
In the equation given, the first step is to use the distribution property. This property allows you to multiply a term outside the parentheses with every term inside the parentheses. For example, in our equation, we apply \(\frac{1}{2} (x + 1)\) to get \(\frac{1}{2} x + \frac{1}{2} \).By distributing correctly, equations can be simplified, making it easier to see what steps to take next. Breaking down complicated parts by distributing helps in combining and eliminating terms later on.
Combining Like Terms
After distribution, we combine like terms. Like terms are terms that have the same variable raised to the same power.For example, \(\frac{1}{2} x \) and \(\frac{1}{6} x \) are like terms. By combining these, we get:\[ \frac{1}{2} x - \frac{1}{6} x = \frac{3}{6} x - \frac{1}{6} x = \frac{2}{6} x = \frac{1}{3} x \]Combining like terms simplifies the equation, making it clearer which operations need to be carried out next.
Identity Equation
An identity equation is one that is true for all values of the variable involved. In simpler terms, it鈥檚 an equation where both sides are identical if we follow algebra rules correctly. In our case, after simplifying and eliminating x, we ended up with \[ \frac{1}{2} = \frac{1}{2} \] which is always true regardless of the value of x. This means any value of xdoes not change the truth of the equation, indicating it as an identity equation.
Eliminating Variables
Eliminating variables means removing the variable terms to see if an equation holds true without them. When we subtracted \(\frac{1}{3} x\)from both sides, we aimed to simplify the equation further. This step revealed that the variable was not crucial to the equation's truth, simplifying it down to \(\frac{1}{2} = \frac{1}{2} \).By successfully eliminating the variable, we confirmed that the equation holds for any value of x,indicating it as an identity equation.

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

A police officer uses a radar detector to determine that a motorist is traveling \(34 \mathrm{mph}\) in a \(25 \mathrm{mph}\) school zone. The driver goes to court and argues that the radar detector is not accurate. The manufacturer claims that the radar detector is calibrated to be in error by no more than 3 mph. a. If \(x\) represents the motorist's actual speed, write an inequality that represents an interval in which to estimate \(x\). b. Solve the inequality and interpret the answer. Should the motorist receive a ticket?

Solve the inequality and write the solution set in interval notation. \(5 \leq|2 x+1| \leq 7\)

Donovan has offers for two sales jobs. Job A pays a base salary of \(\$ 25,000\) plus a \(10 \%\) commission on sales. Job B pays a base salary of \(\$ 30,000\) plus \(8 \%\) commission on sales. a. How much would Donovan have to sell for the salary from Job A to exceed the salary from Job B? b. If Donovan routinely sells more than \(\$ 500,000\) in merchandise, which job would result in a higher salary?

Solve the inequality, and write the solution set in interval notation. \(5|x+1|-9 \geq-4\)

Solve the inequality. Write the solution set in interval notation. $$2 x+1<-3\( or \)2 x+3>3$$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Calculus

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.