Problem 48 - Elementary Algebra

Question

For Problems \(33-50\), set up an equation and solve the problem. (Objective 2 ) It takes two pipes 3 hours to fill a water tank. Pipe \(B\) can fill the tank alone in 8 hours more than it takes pipe A to fill the tank alone. How long would it take each pipe to fill the tank by itself?

Step-by-Step Solution

Answer

Pipe A takes 4 hours, and Pipe B takes 12 hours.

Step 1: Understand the Problem

We have two pipes filling a tank. Together, they fill the tank in 3 hours. Pipe B takes 8 hours more than Pipe A alone to fill the tank. We need to find out how long each pipe takes individually.

Step 2: Define the Variables

Let \( x \) be the time in hours it takes for Pipe A to fill the tank alone. Then Pipe B takes \( x + 8 \) hours.

Step 3: Set Up the Equation

Since Pipe A can fill \( \frac{1}{x} \) of the tank in 1 hour and Pipe B can fill \( \frac{1}{x+8} \) of the tank in 1 hour, together they fill \( \frac{1}{3} \) of the tank in one hour. Thus, the equation is: \[ \frac{1}{x} + \frac{1}{x+8} = \frac{1}{3} \]

Step 4: Solve the Equation

First, find a common denominator and combine the fractions: \[ \frac{x+8 + x}{x(x+8)} = \frac{2x + 8}{x(x+8)} = \frac{1}{3} \]Cross-multiply to solve: \[ 3(2x + 8) = x(x+8) \] \[ 6x + 24 = x^2 + 8x \]Move all terms to one side to form a quadratic equation: \[ x^2 + 2x - 24 = 0 \]Factor the quadratic equation: \[ (x + 6)(x - 4) = 0 \] The solutions are \( x = -6 \) or \( x = 4 \). Since time cannot be negative, \( x = 4 \).

Step 5: Interpret the Solution

Pipe A takes 4 hours to fill the tank by itself, and Pipe B takes \( 4 + 8 = 12 \) hours to fill the tank by itself.

Key Concepts

Quadratic Equations

Quadratic equations are a crucial part of elementary algebra. They are equations of the form \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Solving these equations helps uncover the values of \( x \) that satisfy the equation.
In the exercise example, we arrived at the quadratic equation: \[ x^2 + 2x - 24 = 0 \]. To solve it, we applied factoring. This means finding two binomials that multiply to give the original quadratic equation. In this case, \[ (x + 6)(x - 4) = 0 \]. This leads us to the solutions \( x = -6 \) or \( x = 4 \).Since time cannot be negative in our real-world problem, we dismiss \( x = -6 \) and accept \( x = 4 \). Knowing how to identify and solve quadratics is useful, as it finds applications in various problem-solving contexts.

Problem Solving

Problem-solving is an essential skill in mathematics that involves interpreting a problem, defining the necessary variables, and creating an equation to find solutions. The goal is to convert a real-world situation into a mathematical problem that we can solve using algebraic techniques.
In our exercise, the problem tells us about two pipes that fill a water tank in 3 hours when working together. We needed to find out how long each could do this task alone. The strategy included:

• Understanding that each pipe's contribution adds up to the total fill rate
• Defining variables to represent unknown quantities, such as \( x \) for the time taken by pipe A
• Formulating an equation that encapsulates the rates of both pipes

With the equation set up, the solution followed logically by combining terms and solving the quadratic equation. Effective problem-solving in algebra requires careful thought and analytical skills.

Algebraic Fractions

Algebraic fractions involve expressions with fractions where the numerator, the denominator, or both contain algebraic expressions. Understanding them is vital in algebra as they appear frequently in problems involving rates and ratios, like our exercise with pipes.
In the example, each pipe's rate of filling the tank per hour translates into an algebraic fraction:

• Pipe A fills \( \frac{1}{x} \) of the tank each hour
• Pipe B fills \( \frac{1}{x+8} \)

Together, their combined rate is \( \frac{1}{3} \) of the tank per hour, leading to the equation \[ \frac{1}{x} + \frac{1}{x+8} = \frac{1}{3} \]. Solving such equations often involves finding a common denominator and combining the fractions. Here, we simplify the equation and apply known methods for solving quadratics to find values for \( x \).Mastery of algebraic fractions helps students in tackling various algebraic problems, enhancing their mathematical proficiency.