Question

name: date: score: we do 1a.) \\(\

$$\begin{cases} y = 2x + 3 \\\\ y = 2x + 1 \\end{cases}$$

\\) 1b.) \\(\

$$\begin{cases} y = -\\frac{1}{2}x - 1 \\\\ y = -\\frac{1}{2}x - 4 \\end{cases}$$

\\) 2a.) \\(\

$$\begin{cases} x = 2 \\\\ x = 2 \\end{cases}$$

\\) 2b.) \\(\

$$\begin{cases} y = 2x - 5 \\\\ y = 2x - 5 \\end{cases}$$

\\) 3a.) \\(\

$$\begin{cases} 2x - 6 = y \\\\ 3 - x = y \\end{cases}$$

\\) 3b.) \\(\

$$\begin{cases} -\\frac{3}{2}x + 2 = y \\\\ -2 + \\frac{1}{2}x = y \\end{cases}$$

\\)

Explanation:

Let's solve one of the systems, say 3a: \(

$$\begin{cases} 2x - 6 = y \\ 3 - x = y \end{cases}$$

\)

Step1: Set the two equations equal

Since both equal \(y\), set \(2x - 6 = 3 - x\)

Step2: Solve for \(x\)

Add \(x\) to both sides: \(2x + x - 6 = 3\) → \(3x - 6 = 3\)
Add 6 to both sides: \(3x = 3 + 6\) → \(3x = 9\)
Divide by 3: \(x = \frac{9}{3} = 3\)

Step3: Find \(y\)

Substitute \(x = 3\) into \(3 - x = y\): \(y = 3 - 3 = 0\)

Step1: Analyze slopes

Both lines have slope \(m = 2\) (from \(y = mx + b\) form)

Step2: Check y-intercepts

Y-intercepts are \(3\) and \(1\), different. So parallel lines.

Step3: Conclusion

Parallel lines (same slope, different y-intercepts) have no solution.

Answer:

The solution is \(x = 3\), \(y = 0\) (for system 3a)

For another system, like 1a: \(

$$\begin{cases} y = 2x + 3 \\ y = 2x + 1 \end{cases}$$

\)