Question

A woman worked part-time a certain number of days, receiving for her pay \(\$ 1800 .\) If she had received \(\$ 10\) per day less than she did, she would have had to work 3 days longer to earn the same sum. How many days did she work?

Short answer

The woman worked for 20 days.

Step-by-step solution

Define Variables

Let the number of days the woman worked be represented by 'd', and the pay per day she received be 'p'. Therefore, her total earnings can be represented as \(1800 = d \times p\).

Set Up the Equation for the Adjusted Scenario

If the woman had earned \(10 less per day, she would have made \)(p - 10)\( per day. She would have had to work 'd + 3' days to earn the same amount of \)1800. So the equation for this scenario is \(1800 = (d + 3) \times (p - 10)\).

Express 'p' from the Initial Equation

We can express 'p' from the original earnings equation: \(p = \frac{1800}{d}\).

Substitute 'p' in the Adjusted Equation

Substitute the expression for 'p' into the adjusted scenario's equation: \(1800 = (d + 3) \times \bigg(\frac{1800}{d} - 10\bigg)\).

Solve the Equation

Multiply both sides of the equation to eliminate the fraction: \(1800d = (d + 3) \times (1800 - 10d)\). Expand the right side and simplify: \(1800d = 1800d + 5400 - 10d^2 - 30d\). Subtract \(1800d\) from both sides to get a quadratic equation: \(0 = -10d^2 - 30d + 5400\). Divide throughout by -10 to simplify further: \(0 = d^2 + 3d - 540\).

Factor the Quadratic Equation

Factor the quadratic equation: \(d^2 + 3d - 540 = (d + 27)(d - 20)\). When factored, we find two possible solutions for 'd': \(d = -27\) or \(d = 20\).

Find the Correct Solution

Since the number of days 'd' cannot be negative, the correct solution is \(d = 20\) days.

Key concepts

Quadratic Equations

Quadratic equations are fundamental to algebra and are characterized by the highest power of the variable being two, typically in the form of \(ax^2 + bx + c = 0\). In the given exercise, the quadratic equation is derived from a word problem involving earnings and the number of days worked. The equation \(d^2 + 3d - 540 = 0\) represents such a quadratic equation.

To solve quadratic equations, one can use several methods, such as factoring, applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), or completing the square. Factoring is especially handy when the equation breaks down nicely, as in our exercise: \(d^2 + 3d - 540 = (d + 27)(d - 20)\). Through factoring, the possible solutions for 'd' are obtained. Nevertheless, it is crucial to interpret the solutions in the context of the problem to determine which, if any, make sense. In this case, negative days worked are impossible, leaving the positive value as the sensible answer.

System of Equations

A system of equations consists of two or more equations that share variables, and the solution is the set of values that satisfies all equations simultaneously. In our word problem, we effectively have a system with two scenarios: one detailing the actual pay and days worked, and another hypothetical where the pay per day was less.

• The first scenario yields \(1800 = d \times p\)
• The second scenario provides \(1800 = (d + 3) \times (p - 10)\).

In solving this system, we isolate one variable, 'p' in this case, and substitute it into the other equation. This transformation yields a single equation in 'd', which turns out to be a quadratic equation. Understanding the system of equations concept and this particular method of substitution is essential, as it is commonly used in algebra to tackle a variety of problems.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and arithmetic operations. They represent values and can be as simple as a single term or as complex as a large polynomial. In our exercise, we deal with several algebraic expressions such as \(d \times p\) and \(d + 3) \times (p - 10)\). These expressions represent the total earnings under different conditions.

Understanding how to manipulate these expressions is vital in solving algebraic problems. In the step where we 'express 'p' from the Initial Equation,' we show one way of manipulating an expression to isolate a variable, which is crucial for solving equations and systems of equations. Mastering the manipulation of algebraic expressions allows students to transition from simply evaluating expressions to solving more complex equations.