Problem 48 - College Algebra and Calculus: An Applied Approach

Question

Solve the equation and check your solution. (Some equations have no solution.) $$ 3=2+\frac{2}{z+2} $$

Step-by-Step Solution

Answer

The solution to the equation is \(z = 0\).

Step 1: Rearrange the Equation

The first step is to isolate the fraction on one side of the equation. Subtract 2 from both sides to achieve this: \(3 - 2 = 2 + \frac{2}{z+2} - 2\), which simplifies to \(1 = \frac{2}{z+2}\).

Step 2: Solve for z

In the equation \(1= \frac{2}{z+2}\), cross multiply to eliminate the denominator on the right side: \(1 * (z+2) = 2*1\), which simplifies to \(z + 2 = 2\).

Step 3: Isolate the Variable z

Subtract 2 from both sides of the equation to isolate \(z\): \(z + 2 - 2 = 2 - 2\), which simplifies to \(z = 0\).

Step 4: Check the Solution

Substitute the solution back into the original equation to check: \(3 = 2 + \frac{2}{0+2}\), this simplifies to \(3 = 2 + 1 = 3\), confirming that the solution is correct.

Key Concepts

Understanding Algebra

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating these symbols. In algebra, those symbols often represent numbers in equations or expressions. For instance, in the equation \(3 = 2 + \frac{2}{z+2}\), \(z\) represents an unknown variable that we need to find.
One of the primary goals in algebra is to solve these equations by determining the values of unknown variables. Solving equations involves understanding and applying various mathematical operations such as addition, subtraction, multiplication, and division in a structured way.
When working on algebraic equations, you often encounter situations where you need to isolate the variable. This means rearranging the equation so that the unknown variable you want to solve for is on one side of the equation, and all other terms are on the other side.

• Identify the variable to solve for.
• Use inverse operations to isolate the variable.
• Ensure that any operations done on one side are applied to the other side.

Algebra is like solving a puzzle in which you find the missing piece that completes the whole picture, such as finding the value of \(z\) in our equation.

Exploring Rational Equations

Rational equations involve ratios or fractions within the equation that must be solved. These equations often contain a variable in the denominator and can sometimes seem more complex due to the fractions involved.
In our example, the equation \(3 = 2 + \frac{2}{z+2}\) is a rational equation. Here, the fraction \(\frac{2}{z+2}\) involves the variable \(z\) as part of its denominator.
To work with rational equations, the main strategy is eliminating the denominators. This usually involves manipulating the equation so that the variable is no longer part of the denominator. Solving rational equations typically follows these steps:

• Move terms around to isolate the fraction on one side of the equation.
• Clear the fractions by cross-multiplying or multiplying through by the least common denominator.
• Simplify the equation until you isolate the variable.

Rational equations require careful handling to avoid potential pitfalls such as dividing by zero, which is why checking solutions is essential.

Applying Cross-Multiplication

Cross-multiplication is a technique used to solve equations that have fractions. It's particularly useful in rational equations to eliminate the fractions and simplify the process of solving for the variable.
For the equation \(1 = \frac{2}{z+2}\), cross-multiplication helps by removing the denominator \(z+2\) from the fraction. Here's how it works:1. Consider each fraction as a ratio, \(a/b = c/d\).2. Cross-multiplication means you multiply \(a\) with \(d\) and \(b\) with \(c\), giving you the equation \(a \cdot d = b \cdot c\).
In our specific example, \(1 \cdot (z+2) = 2 \cdot 1\), leads to the equation \(z + 2 = 2\). By using cross-multiplication, we easily clear the fraction and are left with a clearer path to solve for \(z\).
This technique is powerful, especially in rational equations, as it allows you to work with simplified linear equations, making it easier to isolate and determine the value of the variable.