Sure! Let's create systems of linear equations to demonstrate different types of solutions.
### System of Linear Equations with One Solution
To find a system where the equations have exactly one solution, the lines must intersect at exactly one point. Here is how we can construct it:
- Equation 1: [tex]\( x + 2y = 4 \)[/tex]
- Equation 2: [tex]\( 2x - y = 1 \)[/tex]
This system has one unique solution because the lines intersect at exactly one point.
### System of Linear Equations with No Solution
For a system of linear equations with no solution, the lines must be parallel but not identical. This means they have the same slope but different y-intercepts. Here's how we can construct it:
- Equation 1: [tex]\( x + 2y = 4 \)[/tex]
- Equation 2: [tex]\( x + 2y = 6 \)[/tex]
In this case, the lines are parallel to each other (both have the slope [tex]\(-\frac{1}{2}\)[/tex]) but never intersect because they are shifted vertically apart, resulting in no solution.
### System of Linear Equations with Infinitely Many Solutions
To have infinitely many solutions, the two equations must represent the same line. This can be done by multiplying one equation by a scalar to get the other equation. Here's an example:
- Equation 1: [tex]\( x + 2y = 4 \)[/tex]
- Equation 2: [tex]\( 2x + 4y = 8 \)[/tex]
In this scenario, the second equation is just a multiple of the first equation (multiplied by 2), so they represent the same line and thus have infinitely many solutions.
Therefore, the systems of equations categorized by types of solutions are as follows:
### One Solution
[tex]\[
\begin{cases}
x + 2y = 4 \\
2x - y = 1
\end{cases}
\][/tex]
### No Solution
[tex]\[
\begin{cases}
x + 2y = 4 \\
x + 2y = 6
\end{cases}
\][/tex]
### Infinitely Many Solutions
[tex]\[
\begin{cases}
x + 2y = 4 \\
2x + 4y = 8
\end{cases}
\][/tex]
