Solve The Equation 1/(x-4) = (x-3)/6 Step-by-Step Guide

In this article, we will walk through the step-by-step process of solving the equation $\frac{1}{x-4} = \frac{x-3}{6}$. This type of equation involves fractions and requires careful manipulation to arrive at the correct solution(s). We will explore the algebraic techniques used to clear the fractions, simplify the equation, and ultimately find the value(s) of $x$ that satisfy the given condition. If multiple solutions exist, they will be listed separated by commas. In the event that there is no solution, we will denote it with the symbol $\varnothing$.

Step-by-Step Solution

To solve the equation $\frac{1}{x-4} = \frac{x-3}{6}$, our initial goal is to eliminate the fractions. This can be achieved by multiplying both sides of the equation by the least common multiple (LCM) of the denominators, which in this case is $6(x-4)$.

Clearing the Fractions

Multiplying both sides by $6(x-4)$, we get:

$6(x-4) \cdot \frac{1}{x-4} = 6(x-4) \cdot \frac{x-3}{6}$

This simplifies to:

$6 = (x-4)(x-3)$

Expanding and Simplifying

Next, we expand the right side of the equation by multiplying the binomials:

$6 = x^2 - 3x - 4x + 12$

Combining like terms, we have:

$6 = x^2 - 7x + 12$

To solve for $x$, we need to set the equation to zero. Subtracting 6 from both sides gives us a quadratic equation:

$0 = x^2 - 7x + 6$

Solving the Quadratic Equation

We now have a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a = 1$, $b = -7$, and $c = 6$. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, factoring is the most straightforward approach.

We are looking for two numbers that multiply to 6 and add to -7. These numbers are -1 and -6. Thus, we can factor the quadratic equation as follows:

$0 = (x - 1)(x - 6)$

Setting each factor equal to zero gives us the possible solutions for $x$:

$x - 1 = 0$ or $x - 6 = 0$

Solving these linear equations, we find:

$x = 1$ or $x = 6$

Checking for Extraneous Solutions

It is crucial to check these solutions in the original equation to ensure they are valid. We need to make sure that the solutions do not make the denominator of any fraction equal to zero, as this would result in an undefined expression.

Checking $x = 1$

Plugging $x = 1$ into the original equation:

$\frac{1}{1-4} = \frac{1-3}{6}$

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

$-\frac{1}{3} = -\frac{1}{3}$

This solution is valid.

Checking $x = 6$

Plugging $x = 6$ into the original equation:

$\frac{1}{6-4} = \frac{6-3}{6}$

$\frac{1}{2} = \frac{3}{6}$

$\frac{1}{2} = \frac{1}{2}$

This solution is also valid.

Final Answer

Both $x = 1$ and $x = 6$ are valid solutions to the equation. Therefore, the solutions are $x = 1, 6$.

Alternative Methods for Solving Quadratic Equations

While factoring was the most efficient method in this case, it's worth exploring other techniques for solving quadratic equations. The quadratic formula and completing the square are two such methods that can be used when factoring is not straightforward.

The Quadratic Formula

The quadratic formula is a general method for solving equations of the form $ax^2 + bx + c = 0$. The formula is given by:

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

For our equation $x^2 - 7x + 6 = 0$, we have $a = 1$, $b = -7$, and $c = 6$. Plugging these values into the quadratic formula, we get:

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(6)}}{2(1)}$

$x = \frac{7 \pm \sqrt{49 - 24}}{2}$

$x = \frac{7 \pm \sqrt{25}}{2}$

$x = \frac{7 \pm 5}{2}$

This gives us two solutions:

$x = \frac{7 + 5}{2} = \frac{12}{2} = 6$

$x = \frac{7 - 5}{2} = \frac{2}{2} = 1$

As we found through factoring, the solutions are $x = 1$ and $x = 6$.

Completing the Square

Completing the square is another method for solving quadratic equations. It involves transforming the equation into the form $(x - h)^2 = k$, where $h$ and $k$ are constants. This method can be particularly useful when the quadratic equation cannot be easily factored.

To complete the square for the equation $x^2 - 7x + 6 = 0$, we first isolate the terms with $x$:

$x^2 - 7x = -6$

Next, we add $(\frac{b}{2})^2$ to both sides of the equation. In this case, $b = -7$, so we add $(\frac{-7}{2})^2 = \frac{49}{4}$ to both sides:

$x^2 - 7x + \frac{49}{4} = -6 + \frac{49}{4}$

The left side of the equation is now a perfect square:

$(x - \frac{7}{2})^2 = -\frac{24}{4} + \frac{49}{4}$

$(x - \frac{7}{2})^2 = \frac{25}{4}$

Taking the square root of both sides:

$x - \frac{7}{2} = \pm \sqrt{\frac{25}{4}}$

$x - \frac{7}{2} = \pm \frac{5}{2}$

Now, we solve for $x$:

$x = \frac{7}{2} \pm \frac{5}{2}$

This gives us two solutions:

$x = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6$

$x = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1$

Again, we find the solutions to be $x = 1$ and $x = 6$.

Common Mistakes and How to Avoid Them

When solving equations, especially those involving fractions and quadratic expressions, there are several common mistakes that students often make. Being aware of these pitfalls can help you avoid them and arrive at the correct solutions.

Forgetting to Check for Extraneous Solutions

One of the most critical steps in solving equations with fractions is to check for extraneous solutions. These are solutions that satisfy the transformed equation but not the original equation. Extraneous solutions typically arise when multiplying both sides of an equation by an expression that could be zero.

In our example, we multiplied both sides of the equation by $6(x-4)$. If either $x$ were equal to 4, the denominator of the original equation would be zero, making the expression undefined. Thus, we must check that our solutions $x = 1$ and $x = 6$ do not make the denominator zero. Fortunately, neither of these values does, so they are both valid solutions.

To avoid this mistake, always check your solutions in the original equation, especially when dealing with rational expressions, radicals, or logarithms.

Incorrectly Expanding Binomials

Another common mistake is incorrectly expanding binomials, such as $(x-4)(x-3)$. This can lead to an incorrect quadratic equation and, consequently, wrong solutions. Remember to use the distributive property (also known as the FOIL method) correctly:

$(x-4)(x-3) = x(x) + x(-3) - 4(x) - 4(-3) = x^2 - 3x - 4x + 12$

A common error is to forget the middle terms or to get the signs wrong. Always double-check your expansion to ensure accuracy.

Sign Errors

Sign errors are frequent culprits in algebraic mistakes. Whether it's distributing a negative sign or combining like terms, a simple sign error can throw off the entire solution. For example, when rearranging the equation $6 = x^2 - 7x + 12$ to $0 = x^2 - 7x + 6$, it's crucial to subtract 6 correctly from both sides.

To minimize sign errors, write each step clearly and double-check your work. Pay close attention to the signs when distributing, combining terms, and moving terms across the equals sign.

Incorrect Factoring or Applying the Quadratic Formula

When solving quadratic equations, errors can occur during factoring or when applying the quadratic formula. Factoring requires identifying the correct pair of numbers that multiply to the constant term and add up to the coefficient of the linear term. If you're unsure about factoring, the quadratic formula is a reliable alternative.

When using the quadratic formula, ensure that you substitute the values of $a$, $b$, and $c$ correctly. A common mistake is to misidentify the signs or mix up the values. Write out the formula and the values separately before plugging them in to reduce the risk of error.

Not Setting the Equation to Zero Before Solving

For quadratic equations, it's essential to set the equation to zero before attempting to solve it. The quadratic formula and factoring methods are designed for equations in the form $ax^2 + bx + c = 0$. If you try to factor or apply the quadratic formula without setting the equation to zero, you will likely obtain incorrect solutions.

Making Arithmetic Errors

Simple arithmetic errors can derail the entire solution process. Whether it's adding, subtracting, multiplying, or dividing, a small mistake can lead to an incorrect answer. Use a calculator if necessary, and double-check your arithmetic to ensure accuracy.

Skipping Steps

While it may be tempting to skip steps to save time, this can often lead to errors. Writing out each step clearly makes it easier to track your work and identify any mistakes. It also helps in understanding the logic behind the solution process.

Misunderstanding the Problem

Finally, one of the most fundamental errors is misunderstanding the problem itself. Before you start solving, make sure you understand what the question is asking and what you are trying to find. Read the problem carefully and identify any key information or constraints.

By being mindful of these common mistakes and taking steps to avoid them, you can improve your accuracy and confidence in solving equations.

Conclusion

In conclusion, solving the equation $\frac{1}{x-4} = \frac{x-3}{6}$ involves several key steps: clearing fractions, expanding and simplifying, solving the resulting quadratic equation, and checking for extraneous solutions. We found that the solutions to the equation are $x = 1$ and $x = 6$. By understanding the underlying principles and practicing problem-solving techniques, you can confidently tackle similar algebraic challenges. Remember to always check your solutions and be mindful of common mistakes to ensure accuracy. Whether you use factoring, the quadratic formula, or completing the square, the ability to solve quadratic equations is a fundamental skill in mathematics and has applications in various fields.