Problem 67 - College Algebra

Question

The equations in Exercises \(59-70\) combine the types of equations we have discussed in this section. Solve each equation or state that it is true for all real numbers or no real numbers. $$ \frac{4}{x-2}+\frac{3}{x+5}=\frac{7}{(x+5)(x-2)} $$

Step-by-Step Solution

Answer

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

Step 1 - Identify common denominator

The equation given: \[\frac{4}{x-2}+\frac{3}{x+5}=\frac{7}{(x+5)(x-2)}\]has the common denominator of \((x-2)(x+5)\). The common denominator is the product of all the distinct denominator factors from the original equation.

Step 2 - Rewrite the equation

Multiply every term by the common denominator, this removes the fractions. This results in:\[4(x+5) + 3(x-2) = 7\]We then simplify this to:\[4x + 20 + 3x - 6 = 7\]

Step 3 - Simplify and solve

Simplify the equation further to obtain: \[7x + 14 = 7\]Subtract 14 from both sides of the equation to isolate the variable terms. \[7x = -7\]Then, divide both sides by 7 to solve for x:\[x = -1\]

Step 4 - Check the solution

We always need to check that the solution we found doesn't make the denominator of the original equation zero. This would result in an undefined solution. If we plug in -1 for x in the original equation, we see that the denominator does not become zero, thus our solution \(x = -1\) is valid.

Key Concepts

Understanding Common Denominators

A crucial step in solving equations with fractions, or algebraic fractions, is finding a common denominator. This is key to combining fractions effectively. In our exercise, we encountered fractions with different denominators: one with an \(x-2\) and another with \(x+5\). The least common denominator (LCD) here is the product of these two unique factors, giving us \( (x-2)(x+5) \).

Why do we need a common denominator? To perform addition or subtraction between fractions, they must have the same bottom number, their denominator. This allows us to combine the numerators directly on top. Think of it as trying to add slices of two different pies; to combine them sensibly, you need to consider the whole pies the same size. Thus, identifying the LCD is like figuring out the size of the pie each slice came from, making it possible to put those slices together in a meaningful way.

Simplifying Equations: An Essential Skill

Once we have a common denominator, simplifying equations is the next pivotal step. In our example, after multiplying each term by the common denominator, we obtained a simpler equation without fractions. We then combined like terms and constants, resulting in \(7x + 14 = 7\). Simplifying an equation makes it more manageable and prepares it for solving.

Think of simplification as tidying up a messy room. Initially, you have items scattered everywhere (the original equation with fractions), but after organizing and putting things together (combining like terms), you end up with a neat and orderly space (a linear equation). This clean-up is essential for two reasons: it helps you to spot the solution clearly, and it reduces the likelihood of making mistakes that can come from handling complex expressions.

Dealing with Algebraic Fractions

Working with algebraic fractions requires careful attention to avoid errors. As we saw in the exercise, when dealing with algebraic fractions, we first identify the LCD to combine them into a single fraction, or eliminate the fractions entirely to simplify the equation. It's important to remember that these fractions are not much different from regular fractions: they adhere to the same rules but involve variables.

In our exercise, we started with the fraction equation and ended with a linear equation by clearing the fractions. But be cautious, algebraic fractions also bring in the possibility of restrictions on the variable—values that could make the denominator zero, leading to undefined fractions. Always substitute your solution back into the original equation to ensure you're not hitting any of these restrictions. The cleared equation should represent equivalent relationships to the original, but in a much simpler form, allowing you to solve for the variable with ease.