Problem 54 - Elementary and Intermediate Algebra

Question

Furnace Repairs. A repairman purchased several furnaceblower motors for a total cost of $$ 210 .\( If his cost per motor had been $$ 5\) less, he could have purchased one additional motor. How many motors did he buy at the regular rate?

Step-by-Step Solution

Answer

He bought 6 motors at the regular rate.

Step 1: Define Variables

Let the number of motors he bought be \( n \) and the cost per motor be \( p \). So, the total cost can be expressed as \( n \times p = 210 \).

Step 2: Express the Problem Mathematically

If he could buy one more motor at \( p - 5 \) dollars each, the equation becomes \((n + 1) \times (p - 5) = 210\).

Step 3: Expand the Modified Equation

Expand \((n + 1) \times (p - 5) = 210\) to obtain: \( np - 5n + p - 5 = 210 \).

Step 4: Substitute Original Total Cost

Substitute \( np = 210 \) from the first equation into the expanded modified equation: \( 210 - 5n + p - 5 = 210 \).

Step 5: Simplify the Equation

Cancel \(210\) from both sides to get: \(-5n + p - 5 = 0\). Rearrange to find \( p = 5n + 5 \).

Step 6: Substitute to Find n

Substitute \( p = 5n + 5 \) into the original total cost equation \( n(5n + 5) = 210 \).

Step 7: Solve the Quadratic Equation

Simplify to get \( 5n^2 + 5n - 210 = 0 \). Divide through by 5 to simplify further: \( n^2 + n - 42 = 0 \).

Step 8: Factor the Quadratic Equation

Factor the quadratic equation \( n^2 + n - 42 = 0 \) as \( (n + 7)(n - 6) = 0 \).

Step 9: Determine the Solution for n

The solutions to the equation \( (n + 7)(n - 6) = 0 \) are \( n = -7 \) and \( n = 6 \). Since a negative number of motors doesn't make sense, \( n = 6 \).

Key Concepts

Quadratic Equations

Quadratic equations are a cornerstone of algebra and refer to any equation that can be rearranged to the standard form: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable.
In many word problems like the 'Furnace Repairs', the formation of a quadratic equation emerges naturally. Here, we had a cost problem involving motors, and by expressing the relationships in mathematical terms, it led us to this form.
When solving quadratic equations, there are several methods available:

• Factoring - This method is used if the quadratic can be easily broken down into two binomial expressions. It's quick but only works if the equation is factorable.
• Quadratic Formula - The formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \) works for any quadratic equation, providing a fair and comprehensive approach.
• Completing the Square - An algebraic technique that converts a quadratic equation into a perfect square trinomial, making it possible to solve by taking square roots.

This problem employs the factoring method, as the quadratic was simplified to \( n^2 + n - 42 = 0 \). Understanding these techniques not only helps solve problems but also deepens comprehension of how different elements in an equation interact.

Factoring

Factoring is a technique used to simplify quadratic equations, turning them into products of two simpler expressions. In the equation \( n^2 + n - 42 = 0 \), factoring involves finding two numbers that multiply to give \(-42\) (the constant term, "c") and add up to \(1\) (the coefficient of \(n\), the middle term).
The process usually starts by listing factor pairs of the constant term to find a pair that fits both criteria. For \(42\), potential factors include:

• 1 and 42
• 2 and 21
• 3 and 14
• 6 and 7

Through trial and error, the pair \(7\) and \(-6\) provides the correct sum and product, leading to the factorization: \((n + 7)(n - 6) = 0\).
Factoring can quickly provide the roots of a quadratic equation, essential for solving algebra word problems. Once factored correctly, setting each expression to zero efficiently identifies possible solutions, here leading to \(n = 6\) as the valid number of motors.

Linear Equations

Linear equations are fundamental to algebra and consist of terms that are linear in nature, meaning they involve no exponents higher than one. Expressions like \(-5n + p - 5 = 0\) represent linear equations and are straightforwardly solved for a specific variable.
These equations play well into systems involving multiple relationships. During the furnace motor problem, converting the quadratic context into linear relationships helped simplify one part of a broader complex equation.
Two common techniques for solving linear equations include:

• Substitution - Solving one equation for one variable and substituting it into another, helping isolate and determine each variable.
• Elimination - Adding or subtracting equations to eliminate a variable, simplifying the process of reaching a solution.

In the example provided, by rearranging the equation \(-5n + p - 5 = 0\) to \(p = 5n + 5\), it was directly substituted back into the initial cost equation, showcasing how converting to and from linear forms is valuable in solving algebraic word problems.