Problem 87 - Elementary and Intermediate Algebra

Question

A repairman purchased some washing machine motors for a total of 224 dollar. When the unit cost decreased by 4 dollar, he was able to buy one extra motor for the same total price. How many motors did he buy originally?

Step-by-Step Solution

Answer

The repairman originally bought 7 motors.

Step 1: Define Variables

Let the original number of motors be \( x \), and the original cost per motor be \( y \, \text{dollars} \). Thus, the equation for the total cost is \( xy = 224 \).

Step 2: Set Up Equation for Decreased Price

When the cost per motor is decreased by 4 dollars, the new cost per motor is \( y - 4 \). Now, he can purchase one additional motor, which is \( x + 1 \). Therefore, the new equation becomes \((x + 1)(y - 4) = 224\).

Step 3: Expand and Simplify Second Equation

Expand the second equation: \( (x + 1)(y - 4) = xy - 4x + y - 4 = 224 \). We know from the first equation that \( xy = 224 \), so substitute \( xy \) into the expanded equation to get:\( 224 - 4x + y - 4 = 224 \).

Step 4: Simplify and Solve for y

Cancel 224 from both sides, which results in:\(-4x + y - 4 = 0 \). Rearranging gives:\( y = 4x + 4 \).

Step 5: Substitute and Solve for x

Substitute the expression for \( y \) into the first equation \( xy = 224 \):\( x(4x + 4) = 224 \).Expand: \( 4x^2 + 4x = 224 \).Simplify to find: \( x^2 + x = 56 \).Transform to: \( x^2 + x - 56 = 0 \), a quadratic equation.

Step 6: Solve Quadratic Equation

To solve \( x^2 + x - 56 = 0 \), use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).Substitute \( a = 1, \ b = 1, \ c = -56 \):\( x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-56)}}{2 \times 1} \).\( x = \frac{-1 \pm \sqrt{1 + 224}}{2} \).\( x = \frac{-1 \pm \sqrt{225}}{2} \).\( x = \frac{-1 \pm 15}{2} \).This gives \( x = 7 \) or \( x = -8 \), but \( x \) must be positive, so \( x = 7 \).

Key Concepts

Algebraic Word Problems

Algebraic word problems, like the one involving the repairman, require translating real-life situations into mathematical expressions. This process involves understanding which quantities to represent as variables and then using logical relationships to form equations.
In this problem, two main variables were defined.

• The original number of motors purchased, represented as \( x \).
• The original cost per motor, represented as \( y \).

The total expenditure for the motors, given as 224 dollars, suggests a multiplication relationship, forming the equation \( xy = 224 \). Understanding that word problems describe mathematical mechanisms is key. The decrease in unit cost by 4 dollars allows for an extra motor, highlighting a second condition used to form another equation. Solving these equations requires a systematic approach to unravel the number of motors originally bought, showing the practical application of algebra in real-world scenarios.

Variable Substitution

Variable substitution is a crucial step in simplifying complex equations, especially in word problems. Once we have defined our variables and formulated the equations, substitution helps simplify or solve these equations.
In our exercise, the repairman’s situation resulted in two equations:

• The original equation: \( xy = 224 \), representing the total cost of the motors.
• The modified equation: \( (x + 1)(y - 4) = 224 \), taking into account the cost change leading to more motors.

By expanding and rearranging the second equation, we derived an expression for \( y \), namely \( y = 4x + 4 \). Substituting this into the original equation enabled us to express everything in terms of \( x \), leading to a single-variable quadratic equation. This streamlining through substitution is what allows us to use mathematical techniques effectively for problem resolution.

Step-by-Step Problem Solving

Step-by-step problem solving in algebra involves breaking down complex problems into manageable actions. This ensures clarity and helps in systematically finding solutions.
In our example, the problem was tackled by:

• Defining the variables \( x \) and \( y \), leading to our fundamental relationship \( xy = 224 \).
• Developing a second condition reflecting the changes in cost and quantity, \((x+1)(y-4) = 224\).
• Expanding the second equation into \( xy - 4x + y - 4 = 224 \) and strategically utilizing the known \( xy = 224 \) for simplifying the equation.
• Expressing \( y \) in terms of \( x \) and substituting it back, resulting in \( x^2 + x - 56 = 0 \), a quadratic equation.
• Using the quadratic formula to solve for \( x \), ensuring logical checking by eliminating negative values to find \( x = 7 \).

Each step in this methodical approach serves to transform the problem into a solvable format using clear algebraic techniques, highlighting the importance of organization and clarity in keeping complex problems manageable.