Systems of linear equations solved by graphing, with visual representations of solutions.

Let me solve Example 1 correctly. The system of equations is:
- y = x² - 2x - 3 (the blue parabola)
- y = 2x - 4 (the red line)

To find the intersection points, I'll set the equations equal to each other:
x² - 2x - 3 = 2x - 4
x² - 4x + 1 = 0

Using the quadratic formula: x = [4 ± √(16 - 4)]/2 = [4 ± √12]/2 = [4 ± 2√3]/2 = 2 ± √3

So x ≈ 2 + 1.732 = 3.732 or x ≈ 2 - 1.732 = 0.268

When x = 2 + √3: y = 2(2 + √3) - 4 = 4 + 2√3 - 4 = 2√3 ≈ 3.464
When x = 2 - √3: y = 2(2 - √3) - 4 = 4 - 2√3 - 4 = -2√3 ≈ -3.464

However, looking at the graph, the intersection points appear to be at integer coordinates. Let me check if I misread the equations.

Actually, let me re-examine the graph more carefully. The parabola appears to intersect the x-axis at x = -1 and x = 3, which would mean the equation is y = (x + 1)(x - 3) = x² - 2x - 3, which matches what's written.

The line appears to pass through (0, -4) and (2, 0), which gives a slope of (0 - (-4))/(2 - 0) = 4/2 = 2, so y = 2x - 4, which also matches.

Given that the algebraic solution gives irrational numbers but the graph appears to show integer intersections, there might be a discrepancy. However, based on the equations provided, the solutions are (2 + √3, 2√3) and (2 - √3, -2√3).

But let me check if there's an error in my interpretation. Looking at the graph again, the intersection points might be approximately (0.3, -3.5) and (3.7, 3.5), which aligns with my algebraic solution.

Now I'll summarize the solutions for all three examples:

Example 1: Solutions are (2 + √3, 2√3) and (2 - √3, -2√3), approximately (3.73, 3.46) and (0.27, -3.46).

Example 2: Solutions are (-2, -3) and (-6, -3).

Example 3: Solutions are (1, 2) and (4, 5).

This completes the analysis of the three examples shown in the image.