Sure! Let's create three different systems of linear equations: one with infinite solutions, one with no solutions, and one with exactly one solution. I'll also provide a real-world scenario for one of them.
System with Infinite Solutions
Equations:
- \(2x + 4y = 8\)
- \(x + 2y = 4\)
Solution: The second equation can be rewritten in terms of the first one: \[ 2x + 4y = 8 \quad \text{(Multiply the second equation by 2)} \] This means the second equation is simply a scalar multiple of the first, representing the same line in the coordinate plane.
- To prove that both equations represent the same line, we can express \(y\) in terms of \(x\): From the second equation: \[ x + 2y = 4 \implies 2y = 4 - x \implies y = 2 - \frac{1}{2}x \] From the first equation: \[ 2x + 4y = 8 \implies 4y = 8 - 2x \implies y = 2 - \frac{1}{2}x \] Since both equations yield the same expression for \(y\), there are infinite solutions along the line defined by either equation.
System with No Solutions
Equations:
- \(3x + 2y = 6\)
- \(3x + 2y = 10\)
Solution: Here, both equations describe parallel lines. They have the same slope but different intercepts.
- To show they are inconsistent, let’s look at the left-hand sides: \[ 3x + 2y = 6 \quad \text{(Equation 1)} \] \[ 3x + 2y = 10 \quad \text{(Equation 2)} \] Since these lines will never intersect, there are no solutions to this system.
System with One Solution
Equations:
- \(x + y = 10\)
- \(2x - y = 3\)
Solution: To solve this system, we can use the substitution or elimination method.
Step 1: Solve for one variable. From the first equation: \[ y = 10 - x \]
Step 2: Substitute into the second equation. \[ 2x - (10 - x) = 3 \] \[ 2x - 10 + x = 3 \] \[ 3x - 10 = 3 \] \[ 3x = 13 \implies x = \frac{13}{3} \approx 4.33 \]
Step 3: Substitute \(x\) back to find \(y\). \[ y = 10 - \frac{13}{3} = \frac{30}{3} - \frac{13}{3} = \frac{17}{3} \approx 5.67 \]
One solution: Thus, \(x = \frac{13}{3}\) and \(y = \frac{17}{3}\). The solution is \(\left( \frac{13}{3}, \frac{17}{3} \right)\).
Real-World Scenario
Scenario for the unique solution: Imagine you are planning a party. You have two constraints:
- The total number of guests (friends + family) should be 10.
- You want to buy 2 snacks for every friend you invite, and you have decided to spend $3 on each friend for snacks, while snacks for family members will make up the rest of the $10 total cost.
You can set up the equations based on this scenario:
- First equation: \(x\) (friends) + \(y\) (family) = 10
- Second equation: \(2x - y = 3\) (cost-related equation).
Here, the solution tells you how many friends and family members you can invite based on the defined constraints. Thus, with the solution \(x = \frac{13}{3}\) friends and \(y = \frac{17}{3}\) family, you can put together a plan that optimally includes both groups.