/*! 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 6 Solve and check linear equation.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 6

Solve and check linear equation. \(3 x+5=2 x+13\)

Short Answer

Expert verified
The solution for the linear equation \(3x+ 5 = 2x + 13\) is \(x = 8\).

Step by step solution

01

- Isolate variable terms on one side

Subtract \(2x\) from both sides of the equation. This gives \(3x - 2x = 13 - 5\), which simplifies to \(x = 13 - 5\)
02

- Finding the Variable

Now do the subtraction on the right-hand side. This gives \(x = 8\)
03

- Checking Solution

Substitute x=8 into the original equation \(3 x+5=2 x+13\), to check solution, if left hand side equals right hand side, the solution is valid.

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.

Solving Equations
Solving equations is a fundamental concept in algebra that involves finding the value of a variable that makes an equation true. The process is like solving a puzzle, where each step brings you closer to the solution. You must perform the same operation on both sides of an equation to keep it balanced. Think of it as a seesaw, where both sides need to be equal for it to be in balance.

Steps to solving an equation usually include:
  • Isolating the variable term to one side of the equation.
  • Simplifying both sides of the equation if necessary.
  • Solving the simplified equation for the unknown variable.
  • Checking your solution by substituting back into the original equation.


When you subtract, add, multiply, or divide, remember to do so to both sides of the equation! This ensures that the equation stays valid throughout the process.
Algebraic Equations
Algebraic equations are mathematical statements indicating that two expressions are equal. These equations can range from simple equations with one variable, like the one we are solving, to more complex systems involving several variables. The particular equation we are considering is a linear equation, which means it forms a straight line when graphed.

An essential characteristic of linear equations is that they contain no exponents higher than one, such as in our example: \[3x + 5 = 2x + 13.\]

Understanding how to manipulate these equations is crucial. Here are some key points to remember:
  • Linear equations have one solution when they involve one variable.
  • They form the backbone of algebra and are foundational for understanding more intricate algebraic concepts.
  • They always graph as straight lines on a coordinate plane, which can make visualization easier.
Mathematical Solutions
In mathematics, finding a solution refers to determining the value(s) of variable(s) that satisfy the equation. The solution provides the answer to the posed problem and confirms that the left and right sides of the equation are indeed equal.

For the equation \[3x + 5 = 2x + 13,\]
the solution can be found by first simplifying the terms and isolating the variable. Once \(x = 8\) is obtained, it is important to verify this solution. Verification reassures that no mistake was made during calculation.

Here鈥檚 how you check your solution:
  • Substitute the value back into the original equation.
  • Simplify both sides of the equation to see if they balance.
  • If both sides are equal, your solution is correct!


Always double-checking helps deepen understanding and builds confidence for tackling more challenging equations.

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

$$\text { Solve for } C: \quad V=C-\frac{C-S}{L} N$$

In Exercises \(127-130,\) solve each equation by the method of your choice. $$ \frac{1}{x^{2}-3 x+2}=\frac{1}{x+2}+\frac{5}{x^{2}-4} $$

Things did not go quite as planned. You invested \(\$ 8000\), part of it in stock that paid \(12 \%\) annual interest. However, the rest of the money suffered a \(5 \%\) loss. If the total annual income from both investments was \(\$ 620,\) how much was invested at each rate?

Exercises \(177-179\) will help you prepare for the material covered in the next section. If \(-8\) is substituted for \(x\) in the equation \(5 x^{\frac{2}{3}}+11 x^{\frac{1}{3}}+2=0\) is the resulting statement true or false?

Solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe? $$A=\frac{1}{2} h(a+b) \text { for } a$$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

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.