Examples of Solving Systems of Equations

Understanding Systems of Equations

Solving systems of equations is a fundamental skill in algebra that allows us to find the values of variables that satisfy multiple equations simultaneously. This is particularly useful in real-life scenarios where various constraints must be considered. Let’s dive into three diverse examples to see how this works in practice.

Example 1: Planning a School Event

Context

Imagine you are part of a school committee planning a fundraising event. You need to find out how many tickets to sell for adults and children to meet your fundraising goal.

You know that adult tickets cost \(10 each and child tickets cost \)5 each. Your goal is to raise $200, and you plan to sell a total of 30 tickets. This scenario can be represented as a system of equations.

1. Let x be the number of adult tickets sold.
2. Let y be the number of child tickets sold.

The equations are:

• Equation 1: x + y = 30 (total tickets)
• Equation 2: 10x + 5y = 200 (total revenue)

To solve these equations, we can use the substitution or elimination method. Here, we will use substitution:

1. From Equation 1, express y in terms of x: y = 30 - x.
2. Substitute y into Equation 2: 10x + 5(30 - x) = 200.
3. Simplifying gives: 10x + 150 - 5x = 200, which simplifies to 5x = 50, leading to x = 10.
4. Then substitute x back to find y: y = 30 - 10 = 20.

Thus, you should sell 10 adult tickets and 20 child tickets to meet your fundraising goal.

Notes

• This example illustrates a practical application in event planning.
• Variations could include changing the ticket prices or the total revenue goal.

Example 2: Mixing Solutions in Chemistry

Context

In a chemistry lab, you are tasked with creating a solution with a specific concentration by mixing two solutions. Solution A has a concentration of 30% salt, and Solution B has a concentration of 10% salt. You need a total of 10 liters of a 20% salt solution.

Let x be the amount of Solution A and y be the amount of Solution B in liters. The system of equations is:

• Equation 1: x + y = 10 (total volume)
• Equation 2: 0.3x + 0.1y = 0.2(10) (total salt concentration)

To solve, we can again use substitution:

1. From Equation 1, express y in terms of x: y = 10 - x.
2. Substitute y into Equation 2: 0.3x + 0.1(10 - x) = 2.
3. Simplifying gives: 0.3x + 1 - 0.1x = 2, which leads to 0.2x = 1, giving x = 5.
4. Substitute x back to find y: y = 10 - 5 = 5.

So, you need to mix 5 liters of Solution A and 5 liters of Solution B to achieve the desired concentration.

Notes

• This example demonstrates a real-world application in chemistry.
• You could explore variations by changing the concentrations or the total volume required.

Example 3: Budgeting for a Trip

Context

You’re planning a trip and have a budget for accommodations and meals. You want to spend a total of \(800, and you know that hotel rooms cost \)120 per night while meals cost $40 per day. You plan to stay for 5 nights and want to determine how much you can spend on meals.

Let x be the number of nights at the hotel and y be the number of days for meals. The equations are:

• Equation 1: 120x + 40y = 800 (total budget)
• Equation 2: x = 5 (number of nights)

To solve this, substitute x into Equation 1:

1. Substitute x = 5 into Equation 1: 120(5) + 40y = 800.
2. This simplifies to 600 + 40y = 800.
3. Solving for y gives 40y = 200, leading to y = 5.

Thus, you can afford to spend on 5 days of meals while staying for 5 nights.

Notes

• This budgeting example is relatable for anyone planning a trip.
• Variations could include adjusting the number of nights or the costs of meals and hotels.