Problem 24 - Contemporary Precalculus

Question

A group of homeowners are to share equally in the $$ 210\( cost of repairing a bus-stop shelter near their homes. At the last moment, two members of the group decide not to participate, and this raises the share of each remaining person by \)\$ 28 .$ How many people were in the group at the beginning?

Step-by-Step Solution

Answer

Answer: There were initially 5 people in the group.

Step 1: Write down the initial equation

Initially, all x members of the group were to share the cost equally, so each person's share would be: $$\frac{210}{x}$$.

Step 2: Write down the new equation after two members dropped out

Since two members left the group, the new number of people in the group is x-2. Now, the each person's share is increased by \(\$28\), so for the remaining members:$$\frac{210}{x-2}= \frac{210}{x} + 28$$

Step 3: Remove the denominators by multiplying both sides by x(x-2)

To solve the equation easier, we eliminate the denominators by multiplying both sides by x(x-2). So, $$210(x-2) = 210x + 28x(x-2)$$

Step 4: Simplify and solve the equation

Now, we will simplify the equation and solve for x. $$210x - 420 = 210x + 28x^2 - 56x$$ $$0 = 28x^2 - 56x + 420$$ Divide the equation by 28 to simplify further: $$0 = x^2 - 2x + 15$$ Now, factor the equation: $$(x-5)(x+3) = 0$$

Step 5: Get the final answer

We see that two x values satisfy this equation, so x can be either 5 or -3. However, since we are talking about a group of people, only the positive value of x makes sense. So the initial number of people in the group was 5.

Key Concepts

Equation Solving

Equation solving is the process of finding the values that satisfy a given equation. In the context of algebra word problems, it is a matter of translating a real-world situation into a mathematical equation and then solving for the unknown variable.

For example, let's say that a group of people need to share a cost. If the total cost and the change in cost per person after some people opt-out are known, we can set up an equation to represent this situation. In the provided exercise, the cost shared was \(210 and the cost per person increased by \)28 after two members dropped out. By writing an equation for the initial state and another for the altered state, we linked the real-world problem to algebra.

It's crucial to model the problem accurately with the correct equations. Once the problem is properly set up, you can use different techniques like cross multiplying if crucial to eliminate denominator complexities or combining like terms to simplify the equations before solving for the variable that represents the unknown quantity, in this case, the original number of people in the group.

Rational Expressions

A rational expression is a fraction in which both the numerator and the denominator are polynomials. In our problem, we worked with the rational expression \(\frac{210}{x}\), where \(210\) is a constant and \(x\) is a variable representing the number of people.

When the number of participants changed, the rational expression became \(\frac{210}{x-2} + 28\), introducing two challenges: dealing with a new variable scenario and understanding how to solve equations with rational expressions. To tackle a rational expression equation, a common strategy is to find a common denominator in order to combine the terms which often involves eliminating the denominators to simplify the equation, as seen in the exercise.

Understanding and manipulating rational expressions are crucial for resolving algebra word problems accurately, especially when changes in a situation create new ratios or proportions to consider.

Factoring Quadratics

Factoring quadratics is a powerful technique used to solve equations where the highest power of the variable is 2. In the exercise, after applying equation solving and manipulation of rational expressions, we derived a quadratic equation in the form \(x^2 - 2x + 15 = 0\).

To solve this, factoring is often more efficient than other methods like completing the square or using the quadratic formula, especially when the equation simplifies to nice integer factors. Factoring involves rewriting the quadratic as a product of two binomials and finding the values of \(x\) that make the equation true.

In our example, we factored the quadratic to \(x-5)(x+3) = 0\), yielding two potential solutions. However, when applying factoring, it's important to consider the context; since we can't have a negative number of people, \(x = 5\) is the only viable solution. Factoring quadratics is thus a critical step in solving many algebra problems.