Question

In a basketball game, where points are scored either by a 3 point shot, a 2 point shot or a 1 point free throw, 110 points were scored from 70 successful shots. Find all ways in which the points may have been scored in this game.

Short answer

Solve \( 2x + y = 40 \) and find all non-negative integer solutions for \( x, y, z \) with \( x + y + z = 70 \).

Step-by-step solution

Define Variables and Equations

Let \( x \) be the number of 3-point shots, \( y \) be the number of 2-point shots, and \( z \) be the number of 1-point free throws. Two equations can be formed based on the given information: one for the total number of shots, \( x + y + z = 70 \), and another for the total points scored, \( 3x + 2y + z = 110 \).

Formulate the System of Equations

We have a system of two equations: \( x + y + z = 70 \) (Equation 1) and \( 3x + 2y + z = 110 \) (Equation 2). We'll use these equations to find the values of \( x, y, \) and \( z \).

Subtract the Equations

Subtract Equation 1 from Equation 2 to eliminate \( z \): \( (3x + 2y + z) - (x + y + z) = 110 - 70 \), which simplifies to \( 2x + y = 40 \). This provides us a third equation.

Express Variables in Terms of One Variable

Using \( 2x + y = 40 \), we can express \( y \) in terms of \( x \): \( y = 40 - 2x \). Substituting this into \( x + y + z = 70 \), we get \( x + (40 - 2x) + z = 70 \), which simplifies to \( z = 30 - x \).

Solve for Integer Values of x, y, z

Both \( x \), \( y \), and \( z \) must be non-negative integers. From \( y = 40 - 2x \), \( x \) must satisfy \( 40 - 2x \geq 0 \), which gives \( x \leq 20 \). From \( z = 30 - x \), \( x \) must satisfy \( 30 - x \geq 0 \), which gives \( x \leq 30 \). Thus, \( x \) can range from 0 to 20.

List Possible Solutions

For each integer \( x \) from 0 to 20, compute \( y = 40 - 2x \) and \( z = 30 - x \) to find consistent solutions. Each consistent set of non-negative integers \((x, y, z)\) from these calculations fits the criteria.

Key concepts

Linear Algebra

Linear Algebra is a branch of mathematics that deals with vectors, vector spaces, and linear equations. In this exercise, we use it to set up and solve equations.
Imagine we have three types of basketball shots—3-point shots, 2-point shots, and 1-point free throws. Each type of shot can be thought of as a separate vector contributing a certain number of points.

• A successful 3-point shot adds 3 to the point total.
• A successful 2-point shot adds 2.
• A 1-point free throw adds 1.

To find how the points were scored, we form a system of linear equations that can be imagined as balancing scales. One equation sums up the total number of shots, with the weights being the number of successful shots of each type. The other equation sums up the total points scored. This creates a system of equations where each variable represents the count of a specific type of shot. The aim is to solve these equations to find all the possibilities or combinations that meet the given constraints.

Problem-Solving

When solving a system of equations, being strategic can simplify the process. Here, we are tasked with determining how 70 successful basketball shots can result in 110 points. To approach this:

• We define the variables: let \( x \) be the number of 3-point shots, \( y \) the number of 2-point shots, and \( z \) the number of 1-point free throws.
• Create two equations based on the information: \( x + y + z = 70 \) and \( 3x + 2y + z = 110 \).
• Subtract these equations to eliminate one variable getting \( 2x + y = 40 \), which simplifies the system.

The next step is to express some variables in terms of another, allowing us to plug back in these to one of the equations, reducing it further to solve for each variable. This method significantly narrows down the possibilities, providing a clear, efficient path to the solution.

Integer Solutions

In this scenario, the solutions must be plausible in real-world terms. This means each solution needs to be an integer. Each variable (\( x \), \( y \), and \( z \)) represents a count of shots taken and cannot be fractional or negative.
The exercise's approach involves iterating through potential values of \( x \) because both \( y \) and \( z \) are expressed in terms of \( x \). Given the constraints:

• \( y = 40 - 2x \)
• \( z = 30 - x \)

We determine the range for \( x \) that keeps \( y \) and \( z \) as non-negative integers. Once this range is established, testing each integer \( x \) within this range allows for exploration of all possible scoring combinations. It guarantees that each way to score is both logically valid and mathematically accurate.