To find the solution of the system of equations given:
- \( y = x + 1 \)
- \( x - 2y = -2 \)
First, let's graph these equations to visualize the solution.
Step 1: Graph the first equation \( y = x + 1 \)
This is a linear equation in slope-intercept form (y = mx + b) where:
- The slope (m) is 1
- The y-intercept (b) is 1
Using this information:
- When \( x = 0 \), \( y = 1 \) (point (0, 1))
- When \( x = -1 \), \( y = 0 \) (point (-1, 0))
- When \( x = 1 \), \( y = 2 \) (point (1, 2))
Step 2: Graph the second equation \( x - 2y = -2 \)
We can convert this to slope-intercept form (y = mx + b) as follows:
- Rearrange the equation: \[ x + 2 = 2y \] \[ y = \frac{1}{2}x + 1 \]
Now we can see the slope is \( \frac{1}{2} \) and the y-intercept is 1.
Using this information:
- When \( x = 0 \), \( y = 1 \) (point (0, 1))
- When \( x = 2 \), \( y = 2 \) (point (2, 2))
- When \( x = -2 \), \( y = 0 \) (point (-2, 0))
Step 3: Plot both lines on a graph
- The first line \( y = x + 1 \) passes through (0, 1) and has a slope of 1.
- The second line \( y = \frac{1}{2}x + 1 \) also passes through (0, 1) but with a gentler slope.
Step 4: Find the intersection
Notice that both equations have a y-intercept of 1. This means that they intersect at the point (0, 1). We should confirm if they intersect elsewhere:
To find the intersection algebraically:
- Substitute \( y = x + 1 \) into the second equation: \[ x - 2(x + 1) = -2 \] Simplifying: \[ x - 2x - 2 = -2 \quad \Rightarrow \quad -x - 2 = -2 \quad \Rightarrow \quad -x = 0 \quad \Rightarrow \quad x = 0 \] Substituting back into the first equation: \[ y = 0 + 1 = 1 \] Thus, the solution is the point (0, 1).
Conclusion
The intersection point (which is the solution to the system of equations) is:
\[ \boxed{(0, 1)} \]
Summary
The lines intersect at the point (0, 1), hence, the system has a unique solution at this point.