Solve The Equation:${ \frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4} }$

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Introduction

In this article, we will delve into the world of mathematics and focus on solving a linear equation. The equation we will be working with is x2βˆ’15=x3+14\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}. This equation may seem daunting at first, but with the right approach and techniques, we can easily solve for the variable xx. In this discussion, we will break down the solution into manageable steps, making it easy to follow and understand.

Understanding the Equation

Before we dive into solving the equation, let's take a closer look at what we're dealing with. The equation is a linear equation, which means it can be written in the form of ax+b=cax + b = c. In this case, the equation is x2βˆ’15=x3+14\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}. Our goal is to isolate the variable xx and find its value.

Step 1: Multiply Both Sides by the Least Common Multiple (LCM)

To eliminate the fractions, we need to multiply both sides of the equation by the least common multiple (LCM) of the denominators. In this case, the LCM of 2, 5, and 3 is 30. Multiplying both sides by 30, we get:

30(x2βˆ’15)=30(x3+14)30\left(\frac{x}{2} - \frac{1}{5}\right) = 30\left(\frac{x}{3} + \frac{1}{4}\right)

Step 3: Distribute the Multiplication

Now, let's distribute the multiplication to both sides of the equation:

15xβˆ’6=10x+7.515x - 6 = 10x + 7.5

Step 4: Isolate the Variable xx

Our goal is to isolate the variable xx. To do this, we need to get all the terms with xx on one side of the equation and the constant terms on the other side. Let's subtract 10x10x from both sides of the equation:

5xβˆ’6=7.55x - 6 = 7.5

Step 5: Add 6 to Both Sides

Next, let's add 6 to both sides of the equation to get rid of the negative term:

5x=13.55x = 13.5

Step 6: Divide Both Sides by 5

Finally, let's divide both sides of the equation by 5 to solve for xx:

x=13.55x = \frac{13.5}{5}

Conclusion

And there you have it! We have successfully solved the equation x2βˆ’15=x3+14\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}. By following the steps outlined above, we were able to isolate the variable xx and find its value. This equation may have seemed daunting at first, but with the right approach and techniques, we were able to solve it with ease.

Tips and Tricks

  • When dealing with fractions, it's often helpful to multiply both sides of the equation by the least common multiple (LCM) of the denominators.
  • To isolate the variable, try to get all the terms with the variable on one side of the equation and the constant terms on the other side.
  • Don't be afraid to add or subtract the same value from both sides of the equation to get rid of negative terms.

Real-World Applications

Solving linear equations is a fundamental skill that has many real-world applications. Here are a few examples:

  • In physics, linear equations are used to describe the motion of objects.
  • In finance, linear equations are used to calculate interest rates and investment returns.
  • In engineering, linear equations are used to design and optimize systems.

Final Thoughts

Solving linear equations may seem like a daunting task, but with the right approach and techniques, it can be done with ease. By following the steps outlined above, we were able to solve the equation x2βˆ’15=x3+14\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}. Whether you're a student or a professional, mastering the art of solving linear equations is an essential skill that will serve you well in many areas of life.

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Introduction

In our previous article, we delved into the world of mathematics and focused on solving a linear equation. The equation we worked with was x2βˆ’15=x3+14\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}. In this article, we will continue to explore the topic of solving linear equations and answer some of the most frequently asked questions.

Q&A

Q: What is a linear equation?

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

Q: How do I know if an equation is linear?

A: To determine if an equation is linear, look for the highest power of the variable(s). If the highest power is 1, then the equation is linear.

Q: What is the least common multiple (LCM)?

A: The least common multiple (LCM) is the smallest multiple that two or more numbers have in common. In the context of solving linear equations, the LCM is used to eliminate fractions.

Q: How do I find the LCM of two or more numbers?

A: To find the LCM of two or more numbers, list the multiples of each number and find the smallest multiple that they all have in common.

Q: What is the difference between a linear equation and a quadratic equation?

A: A linear equation is an equation in which the highest power of the variable(s) is 1, while a quadratic equation is an equation in which the highest power of the variable(s) is 2.

Q: How do I solve a quadratic equation?

A: To solve a quadratic equation, use the quadratic formula: x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

Q: What is the quadratic formula?

A: The quadratic formula is a formula used to solve quadratic equations. It is: x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

Q: How do I use the quadratic formula?

A: To use the quadratic formula, plug in the values of aa, bb, and cc into the formula and simplify.

Q: What is the difference between a linear equation and a system of linear equations?

A: A linear equation is a single equation, while a system of linear equations is a set of two or more equations.

Q: How do I solve a system of linear equations?

A: To solve a system of linear equations, use the method of substitution or elimination.

Q: What is the method of substitution?

A: The method of substitution is a technique used to solve a system of linear equations. It involves substituting the expression for one variable into the other equation.

Q: What is the method of elimination?

A: The method of elimination is a technique used to solve a system of linear equations. It involves adding or subtracting the equations to eliminate one of the variables.

Conclusion

Solving linear equations is a fundamental skill that has many real-world applications. By understanding the concepts and techniques outlined in this article, you will be able to solve linear equations with ease. Whether you're a student or a professional, mastering the art of solving linear equations is an essential skill that will serve you well in many areas of life.

Tips and Tricks

  • When dealing with fractions, it's often helpful to multiply both sides of the equation by the least common multiple (LCM) of the denominators.
  • To isolate the variable, try to get all the terms with the variable on one side of the equation and the constant terms on the other side.
  • Don't be afraid to add or subtract the same value from both sides of the equation to get rid of negative terms.

Real-World Applications

Solving linear equations has many real-world applications. Here are a few examples:

  • In physics, linear equations are used to describe the motion of objects.
  • In finance, linear equations are used to calculate interest rates and investment returns.
  • In engineering, linear equations are used to design and optimize systems.

Final Thoughts

Solving linear equations may seem like a daunting task, but with the right approach and techniques, it can be done with ease. By following the steps outlined in this article, you will be able to solve linear equations with confidence. Whether you're a student or a professional, mastering the art of solving linear equations is an essential skill that will serve you well in many areas of life.