Problem 23 - Introductory Algebra for College Students

Question

Solve each rational equation. $$\frac{4}{y}-\frac{y}{2}=\frac{7}{2}$$

Step-by-Step Solution

Answer

The solutions to the equation are \(y = 1\) and \(y = -8\).

Step 1: Clearing the Fractions

To eliminate the fractions, first find the least common denominator (LCD), which in this case is \(2y\). Multiply both sides of the equation by the LCD to get rid of the fractions: \(2y(\frac{4}{y}-\frac{y}{2})=2y*\frac{7}{2}\). Then simplify each term by cancelling out where possible.

Step 2: Simplifying and Rearranging

The equation now becomes: \(8 - y^2 = 7y\). Move \(7y\) to the left side of the equation by subtracting \(7y\) from both sides, resulting in: \(y^2 + 7y - 8 = 0\). This is a quadratic equation.

Step 3: Solving the Quadratic Equation

Factor the equation into: (y - 1)(y + 8) = 0. Then, set each factor equal to 0 and solve for y to get \(y = 1\) and \(y = -8\). These are the solutions to the equation.

Step 4: Checking the Solutions

Always check your answers back into the original equation. If a number for y makes the denominator zero, discard that solution. Here, neither 1 nor -8 lead to a zero denominator, so both are valid solutions.

Key Concepts

Least Common Denominator

In order to solve rational equations, a key step is to eliminate the fractions involved, and the best way to do this is by finding the Least Common Denominator (LCD). The LCD is the smallest multiple that two or more denominators share. This is crucial because it allows you to transform a fractional equation into a simpler, easier-to-solve equation without changing the solutions.

To determine the LCD in any given rational equation, one must:

• Identify all the denominators in the equation.
• Compute the smallest common multiple for these denominators.

In our exercise, the denominators are simply 2 and y, making the LCD equal to 2y. By multiplying every term in the equation by this LCD, the fractions are effectively cleared.

This results in a new equation, free from fractions, which is much simpler to handle. It translates the problem so that it's easier to apply algebraic operations and solve for the unknown variable.

Quadratic Equations

Once the fractions are cleared in a rational equation, the result is often a form of a quadratic equation. Quadratic equations are polynomials of degree two, usually expressed in the form:\[ax^2 + bx + c = 0\]where a, b, and c are constants.

In the case of our given problem after clearing the fractions, the equation becomes:\[y^2 + 7y - 8 = 0\]This equation is typical of what we recognize as a standard quadratic form, where:

• a = 1 (the coefficient of \(y^2\))
• b = 7 (the coefficient of y)
• c = -8 (the constant term)

The aim is to find solutions for y that satisfy this equation. Quadratic equations can be solved by factoring, using the quadratic formula, or completing the square, each designed to offer solutions based on the nature of the quadratic.

Factoring Quadratics

Factoring is one of the simplest yet powerful methods for solving quadratic equations. The goal is to express the quadratic equation in the form of factors that multiply to zero, according to the zero-product property. When you factor a quadratic equation, you rewrite it as a product of binomials.

Consider our quadratic equation:\[y^2 + 7y - 8 = 0\]To factor this, we look for two numbers that multiply to the constant term, c (-8), and add to the linear coefficient, b (7). These numbers happen to be 8 and -1 because:\[8 \times (-1) = -8 \text{ and } 8 + (-1) = 7\]
Thus, the equation can be factored into:\[(y + 8)(y - 1) = 0\]
Solve each binomial equation separately, setting each factor equal to zero. This gives us the solutions:

• \(y + 8 = 0 \rightarrow y = -8\)
• \(y - 1 = 0 \rightarrow y = 1\)

Factoring is efficient and provides real solutions quickly when the quadratic is easily factored into binomials.