To solve these systems of equations using the graphing method, follow these steps:
### System A
Given:
1. [tex]\( y = x + 3 \)[/tex]
2. [tex]\( y = -x + 3 \)[/tex]
Step 1: Graph the equations
1. For [tex]\( y = x + 3 \)[/tex]:
- This is a straight line with a slope of 1 and a y-intercept of 3.
- To graph, start at the point (0, 3) on the y-axis.
- Use the slope to find another point. From (0, 3), move up 1 unit and to the right 1 unit, reaching (1, 4).
- Connect these points with a straight line.
2. For [tex]\( y = -x + 3 \)[/tex]:
- This is a straight line with a slope of -1 and a y-intercept of 3.
- To graph, start at the point (0, 3) on the y-axis.
- Use the slope to find another point. From (0, 3), move down 1 unit and to the right 1 unit, reaching (1, 2).
- Connect these points with a straight line.
Step 2: Find the intersection
- The intersection point of the two lines [tex]\( y = x + 3 \)[/tex] and [tex]\( y = -x + 3 \)[/tex] is where they cross. The lines cross at (0, 3).
Solution for System A:
- The solution is [tex]\( x = 0 \)[/tex] and [tex]\( y = 3 \)[/tex].
- Thus, the point of intersection is (0, 3).
### System B
Given:
1. [tex]\( y = 3x + 6 \)[/tex]
2. [tex]\( y = x + 2 \)[/tex]
Step 1: Graph the equations
1. For [tex]\( y = 3x + 6 \)[/tex]:
- This is a straight line with a slope of 3 and a y-intercept of 6.
- To graph, start at the point (0, 6) on the y-axis.
- Use the slope to find another point. From (0, 6), move up 3 units and to the right 1 unit, reaching (1, 9).
- Connect these points with a straight line.
2. For [tex]\( y = x + 2 \)[/tex]:
- This is a straight line with a slope of 1 and a y-intercept of 2.
- To graph, start at the point (0, 2) on the y-axis.
- Use the slope to find another point. From (0, 2), move up 1 unit and to the right 1 unit, reaching (1, 3).
- Connect these points with a straight line.
Step 2: Find the intersection
- The intersection point of the two lines [tex]\( y = 3x + 6 \)[/tex] and [tex]\( y = x + 2 \)[/tex] is where they cross. To find this point algebraically:
- Set [tex]\( 3x + 6 = x + 2 \)[/tex].
- Simplify: [tex]\( 3x + 6 = x + 2 \)[/tex]
- [tex]\( 3x - x = 2 - 6 \)[/tex]
- [tex]\( 2x = -4 \)[/tex]
- [tex]\( x = -2 \)[/tex].
- Substitute [tex]\( x = -2 \)[/tex] back into one of the equations (e.g., [tex]\( y = x + 2 \)[/tex]):
- [tex]\( y = -2 + 2 \)[/tex]
- [tex]\( y = 0 \)[/tex].
Solution for System B:
- The solution is [tex]\( x = -2 \)[/tex] and [tex]\( y = 0 \)[/tex].
- Thus, the point of intersection is (-2, 0).
In conclusion:
- The solution for System A is the point (0, 3).
- The solution for System B is the point (-2, 0).
