Not multiplying both sides of the equation by the LCD

Oct 13, 2025 by thesweetdesigns.com 109 views

Introduction

In this article, we will delve into solving a complex equation involving fractions and variables. The equation is $\frac{t}{6} + \frac{1}{2-t} = \frac{1}{3}$ , and our goal is to isolate the variable $t$ and find its value. We will break down the solution into manageable steps, making it easier to understand and follow along.

Step 1: Multiply Both Sides by the Least Common Denominator (LCD)

To simplify the equation, we need to eliminate the fractions. The least common denominator (LCD) of the fractions is $6(2-t)$ . We will multiply both sides of the equation by the LCD to get rid of the fractions.

$\frac{t}{6} + \frac{1}{2-t} = \frac{1}{3}$

Multiply both sides by $6(2-t)$ :

$6(2-t)\left(\frac{t}{6}\right) + 6(2-t)\left(\frac{1}{2-t}\right) = 6(2-t)\left(\frac{1}{3}\right)$

Simplify the equation:

$t(2-t) + 6 = 2(2-t)$

Expand and simplify:

$2t - t^2 + 6 = 4 - 2t$

Step 2: Move All Terms to One Side of the Equation

Now that we have simplified the equation, we need to move all the terms to one side of the equation to set it equal to zero. This will make it easier to solve for $t$ .

$2t - t^2 + 6 = 4 - 2t$

Subtract $4$ from both sides:

$-t^2 + 4t + 2 = 0$

Step 3: Factor the Quadratic Equation

The equation is now a quadratic equation in the form of $ax^2 + bx + c = 0$ . We can try to factor the equation to find the values of $t$ .

$-t^2 + 4t + 2 = 0$

Unfortunately, this equation does not factor easily. We will need to use other methods to solve for $t$ .

Step 4: Use the Quadratic Formula

Since the equation does not factor easily, we can use the quadratic formula to find the values of $t$ . The quadratic formula is:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = -1$ , $b = 4$ , and $c = 2$ . Plugging these values into the formula, we get:

$t = \frac{-4 \pm \sqrt{4^2 - 4(-1)(2)}}{2(-1)}$

Simplify the equation:

$t = \frac{-4 \pm \sqrt{16 + 8}}{-2}$

$t = \frac{-4 \pm \sqrt{24}}{-2}$

$t = \frac{-4 \pm 2\sqrt{6}}{-2}$

$t = 2 \pm \sqrt{6}$

Conclusion

We have successfully solved the equation for $t$ . The values of $t$ are $2 + \sqrt{6}$ and $2 - \sqrt{6}$ . These values satisfy the original equation, and we can verify this by plugging them back into the equation.

Final Answer

Introduction

In our previous article, we solved the equation $\frac{t}{6} + \frac{1}{2-t} = \frac{1}{3}$ for the variable $t$ . We broke down the solution into manageable steps, making it easier to understand and follow along. In this article, we will provide a Q&A guide to help you better understand the solution and answer any questions you may have.

Q: What is the least common denominator (LCD) of the fractions in the equation?

A: The least common denominator (LCD) of the fractions in the equation is $6(2-t)$ .

Q: Why do we need to multiply both sides of the equation by the LCD?

A: We need to multiply both sides of the equation by the LCD to eliminate the fractions. This makes it easier to simplify the equation and solve for $t$ .

Q: What is the simplified equation after multiplying both sides by the LCD?

A: The simplified equation is $t(2-t) + 6 = 2(2-t)$ .

Q: How do we move all the terms to one side of the equation?

A: We move all the terms to one side of the equation by subtracting $4$ from both sides. This gives us the equation $-t^2 + 4t + 2 = 0$ .

Q: Why can't we factor the quadratic equation easily?

A: We can't factor the quadratic equation easily because it does not have any obvious factors. This means we need to use other methods to solve for $t$ .

Q: What is the quadratic formula, and how do we use it to solve for $t$ ?

A: The quadratic formula is $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . We use this formula to solve for $t$ when the equation does not factor easily. In this case, $a = -1$ , $b = 4$ , and $c = 2$ . Plugging these values into the formula, we get $t = 2 \pm \sqrt{6}$ .

Q: What are the values of $t$ that satisfy the original equation?

A: The values of $t$ that satisfy the original equation are $2 + \sqrt{6}$ and $2 - \sqrt{6}$ .

Q: How do we verify that these values satisfy the original equation?

A: We can verify that these values satisfy the original equation by plugging them back into the equation. If the equation holds true, then we know that these values are the correct solutions.

Conclusion

We hope this Q&A guide has helped you better understand the solution to the equation $\frac{t}{6} + \frac{1}{2-t} = \frac{1}{3}$ . If you have any further questions or need clarification on any of the steps, please don't hesitate to ask.

Common Mistakes to Avoid

Not multiplying both sides of the equation by the LCD: This can lead to incorrect simplification of the equation.
Not moving all the terms to one side of the equation: This can make it difficult to solve for $t$ .
Not using the quadratic formula correctly: This can lead to incorrect solutions for $t$ .

Tips and Tricks

Make sure to simplify the equation correctly: This will make it easier to solve for $t$ .
Use the quadratic formula when the equation does not factor easily: This will give you the correct solutions for $t$ .
Verify your solutions by plugging them back into the equation: This will ensure that you have the correct solutions.

Final Answer

The final answer is $\boxed{2 + \sqrt{6}, 2 - \sqrt{6}}$ .