Problem 148 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve: $$\frac{3}{x+1}-\frac{5}{x}=\frac{19}{x^{2}+x}$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = -12\)

Step 1: Identify the Least Common Denominator (LCD)

The LCD of \(x\), \(x+1\), and \(x^{2}+x\) is \(x(x+1)\), because each denominator is a factor of this expression.

Step 2: Multiply each term by LCD

Multiplying each term in the equation by \(x(x+1)\) gives us: \(3x - 5(x+1) = 19\). Simplify this to: \(3x - 5x - 5 = 19\).

Step 3: Solve the resulting equation

The resulting equation, \(-2x - 5 =19\), is a simple linear equation. Add 5 to both sides to isolate the terms involving \(x\): \(-2x = 24\). Finally, divide by -2 to solve for \(x\): \(x = -12\)

Key Concepts

Least Common Denominator

Understanding the least common denominator (LCD) is crucial when working with rational equations that include algebraic fractions (fractions with variables). The LCD is the smallest expression that all denominators in the equation can divide into. This is akin to the least common multiple for integers, but for algebraic expressions.

In the given exercise, the denominators are different, so it's not immediately clear how to combine the fractions. Identifying the LCD allows to transform these fractions with unlike denominators into equivalent fractions with a common denominator, making it possible to combine them just as you would with simple numerical fractions.

To find the LCD, consider the factors that make up each denominator. In this case, the denominators are x, x+1, and x^2+x. The latter factors as x(x+1) which already includes the other two denominators. Therefore, x(x+1) is the LCD here. Multiplying each term in the equation by the LCD eliminates the denominators and reduces the rational equation to a simpler linear equation.

Linear Equations

Once the LCD is utilized to eliminate the fractions, we are left with a linear equation. A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. Linear equations are of the form ax + b = 0, where a and b are constants. The solution to a linear equation is the value of x that makes the equation true.

In our exercise, after clearing the denominators using the LCD, we end up with the linear equation 3x - 5x - 5 = 19. Linear equations are solved by isolating the variable on one side of the equation. By combining like terms and performing basic arithmetic operations—addition, subtraction, multiplication, division—we can find the value of x that satisfies the equation. The simplicity of linear equations makes them a fundamental concept in algebra, serving as the foundation for solving more complex equations.

Algebraic Fractions

Algebraic fractions are simply fractions where the numerator, the denominator, or both contain algebraic expressions. Just like numeric fractions, they can represent parts of wholes, ratios, or divisions. The presence of variables, however, means that simplifying and solving algebraic fractions often involves additional steps beyond those used for numeric fractions.

In the context of the exercise, we're given algebraic fractions that need to be combined and solved for a particular value of x. To do this effectively, one has to manipulate the expressions to achieve a common denominator across all terms, thus allowing for the combination and reduction of these fractions into a single linear equation. Algebraic fractions are excellent exercises in algebraic manipulation and understanding their behavior is key in advancing algebraic problem-solving skills.