Problem 39 - Prealgebra and Introductory Algebra

Question

Solve. $$x-\frac{6}{x-3}=\frac{2 x}{x-3}$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = -1\).

Step 1: Identify the common denominator

The common denominator for both fractions in the equation \(x-\frac{6}{x-3}=\frac{2 x}{x-3}\) is \(x-3\).

Step 2: Multiply through by the common denominator

Multiply every term in the equation by \(x-3\) to clear the fractions. This will result in the new equation: \((x)(x-3)-6=2x\)

Step 3: Expand and simplify the equation

When expanded, the equation becomes \(x^2-3x-6=2x\). Transpose all terms to one side and simplify, we have \(x^2-5x-6=0\).

Step 4: Factor the equation

Factor the equation, this results in \((x-6)(x+1)=0\).

Step 5: Solve for the variable x

By setting both factors equal to zero, the solutions emerge to be \(x = 6\) and \(x = -1\). However, looking back at the original equation, we noticed that \(x = 3\) makes the denominator zero, therefore, the fraction is undefined at \(x = 3\). Since 6 is the solution we found, the only valid solution is \(x = -1\).

Key Concepts

Common Denominator

Understanding the concept of a common denominator is crucial when dealing with algebraic fractions. It refers to a shared denominator that two or more fractions have, which makes addition, subtraction, and comparison of fractions possible. In algebra, finding the common denominator can simplify equations and make further manipulation easier.

For instance, in the equation \(x-\frac{6}{x-3}=\frac{2x}{x-3}\), the term \(x-3\) is the common denominator. It's the shared bottom part of both fractions. This common denominator is a critical starting point for solving the equation, as it allows you to manipulate the equation in a way that eliminates the fractions altogether. The strategy here is to multiply every term by the common denominator, which leads us to the next concept—clearing fractions.

Clearing Fractions

Clearing fractions from an equation can transform a potentially complex problem into a much more manageable one. This technique involves multiplying every term in the equation by the common denominator. The purpose is to get rid of the fractions, leaving behind a simpler, whole-number equation that is easier to solve.

By using this method on our exercise, \(x-\frac{6}{x-3}=\frac{2x}{x-3}\), when we multiply each term by the common denominator \(x-3\), we eliminate the denominators, forcing the x terms out of the fraction and simplifying the equation to \(x(x-3)-6=2x\), without any division remaining. It's akin to reducing the fraction to its most basic form before addressing the equation as a whole. This method streamlines the equation, preparing it for the next operation: factoring.

Factoring Quadratic Equations

Once we have cleared the fractions and simplified the equation, we often end up with a quadratic equation. Factoring is a technique used to solve these equations, which typically take the form \(ax^2 + bx + c = 0\). This method involves breaking down the quadratic equation into a product of two binomial expressions that set to zero and solving for the variable x.

In our exercise example, once we've cleared fractions and simplified, we obtain \(x^2-5x-6=0\). Factoring this equation means finding two numbers that multiply to give -6 (the constant term) and add to give -5 (the coefficient of x). These numbers are -6 and +1, giving us the factors \(x-6\) and \(x+1\). Therefore, our factored equation is \(x-6)(x+1)=0\). The solutions are the values of x that make each factor equal to zero, so we set each binomial expression to zero and solve for x. We must also remember to check for any values that would make the original denominators undefined, as these cannot be solutions.