If $\frac{x}{3}+\frac{x}{4}+15=x$, Then $x$ Equals:A. 18 B. 24 C. 36 D. 48 E. 60

Introduction

Linear equations are a fundamental concept in mathematics, and solving them is a crucial skill for students to master. In this article, we will focus on solving a specific type of linear equation, which involves fractions and a constant term. We will use the given equation $\frac{x}{3}+\frac{x}{4}+15=x$ to demonstrate the step-by-step process of solving linear equations.

Understanding the Equation

The given equation is $\frac{x}{3}+\frac{x}{4}+15=x$. To solve this equation, we need to isolate the variable $x$ on one side of the equation. The equation involves fractions, which can be challenging to work with. However, with the right approach, we can simplify the equation and solve for $x$.

Step 1: Simplify the Equation

The first step in solving the equation is to simplify it by combining the fractions. We can do this by finding a common denominator for the fractions. In this case, the common denominator is 12.

\frac{x}{3}+\frac{x}{4}+15=x
\implies \frac{4x}{12}+\frac{3x}{12}+15=x
\implies \frac{7x}{12}+15=x

Step 2: Isolate the Variable

Now that we have simplified the equation, we can isolate the variable $x$ by subtracting 15 from both sides of the equation.

\frac{7x}{12}+15=x
\implies \frac{7x}{12}=x-15
\implies \frac{7x}{12}-x=-15
\implies \frac{7x-12x}{12}=-15
\implies \frac{-5x}{12}=-15

Step 3: Solve for $x$

Now that we have isolated the variable $x$, we can solve for $x$ by multiplying both sides of the equation by 12.

\frac{-5x}{12}=-15
\implies -5x=-15\times 12
\implies -5x=-180
\implies x=\frac{-180}{-5}
\implies x=36

Conclusion

In this article, we have demonstrated the step-by-step process of solving a linear equation involving fractions and a constant term. We have used the given equation $\frac{x}{3}+\frac{x}{4}+15=x$ to illustrate the process of simplifying the equation, isolating the variable, and solving for $x$. By following these steps, we have arrived at the solution $x=36$.

Answer

The correct answer is C. 36.

Discussion

This problem is a great example of how to solve linear equations involving fractions and a constant term. The key steps involved in solving this equation are simplifying the equation, isolating the variable, and solving for $x$. By following these steps, we can solve a wide range of linear equations, including those involving fractions and a constant term.

Tips and Tricks

• When simplifying an equation involving fractions, find a common denominator to combine the fractions.
• When isolating a variable, subtract the constant term from both sides of the equation.
• When solving for a variable, multiply both sides of the equation by the reciprocal of the coefficient of the variable.

Practice Problems

• Solve the equation $\frac{x}{2}+\frac{x}{3}+10=x$.
• Solve the equation $\frac{x}{4}+\frac{x}{5}+12=x$.

Conclusion

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Introduction

In our previous article, we demonstrated the step-by-step process of solving a linear equation involving fractions and a constant term. In this article, we will provide a Q&A guide to help students better understand the concept of solving linear equations.

Q: What is a linear equation?

A: A linear equation is an equation in which the highest power of the variable is 1. In other words, it is an equation that can be written in the form ax + b = c, where a, b, and c are constants.

Q: What are the steps to solve a linear equation?

A: The steps to solve a linear equation are:

1. Simplify the equation by combining like terms.
2. Isolate the variable by subtracting the constant term from both sides of the equation.
3. Solve for the variable by multiplying both sides of the equation by the reciprocal of the coefficient of the variable.

Q: How do I simplify a linear equation involving fractions?

A: To simplify a linear equation involving fractions, find a common denominator to combine the fractions. For example, if the equation is $\frac{x}{3}+\frac{x}{4}+15=x$, the common denominator is 12. You can then rewrite the equation as $\frac{4x}{12}+\frac{3x}{12}+15=x$.

Q: How do I isolate the variable in a linear equation?

A: To isolate the variable in a linear equation, subtract the constant term from both sides of the equation. For example, if the equation is $\frac{x}{3}+\frac{x}{4}+15=x$, you can subtract 15 from both sides of the equation to get $\frac{x}{3}+\frac{x}{4}=-15$.

Q: How do I solve for the variable in a linear equation?

A: To solve for the variable in a linear equation, multiply both sides of the equation by the reciprocal of the coefficient of the variable. For example, if the equation is $\frac{x}{3}+\frac{x}{4}=-15$, you can multiply both sides of the equation by 12 to get $4x+3x=-180$.

Q: What are some common mistakes to avoid when solving linear equations?

A: Some common mistakes to avoid when solving linear equations include:

• Not simplifying the equation before isolating the variable.
• Not isolating the variable before solving for it.
• Not multiplying both sides of the equation by the reciprocal of the coefficient of the variable.

Q: How can I practice solving linear equations?

A: You can practice solving linear equations by working through a series of problems, such as:

• Solving equations involving fractions and a constant term.
• Solving equations involving decimals and a constant term.
• Solving equations involving variables and a constant term.

Q: What are some real-world applications of solving linear equations?

A: Solving linear equations has many real-world applications, including:

• Calculating the cost of goods and services.
• Determining the amount of time it takes to complete a task.
• Finding the area and perimeter of a shape.

Conclusion

In conclusion, solving linear equations is a crucial skill for students to master. By understanding the steps to solve a linear equation, students can apply this knowledge to a wide range of mathematical problems. With practice and patience, students can become proficient in solving linear equations and apply it to real-world situations.

Practice Problems

• Solve the equation $\frac{x}{2}+\frac{x}{3}+10=x$.
• Solve the equation $\frac{x}{4}+\frac{x}{5}+12=x$.
• Solve the equation $2x+5=15$.

Answer Key

• The solution to the equation $\frac{x}{2}+\frac{x}{3}+10=x$ is $x=18$.
• The solution to the equation $\frac{x}{4}+\frac{x}{5}+12=x$ is $x=24$.
• The solution to the equation $2x+5=15$ is $x=5$.