solve system graphically. be sure to check your solution. if a system has infinite number of solutions, use set builder notation to write the solution det. if a system as no solution, state this. a=1+b b=5-2a
b = 5 - 2 (1+b)
b = 5 - 2 - 2 b
3 b = 3
b = 1
a = 1+b = 2
these are straight lines
choose some values for b and find a for each equation then find intersection at a = 2 and b = 1
To solve the system of equations graphically, we can start by representing each equation as a line on a graph.
First, let's rewrite the equations in the standard form (y = mx + b), where "y" represents "a" and "x" represents "b":
Equation 1: a = 1 + b --> a - b = 1 --> -b + a = 1 --> a = b + 1
Equation 2: b = 5 - 2a --> 2a + b = 5
Now, for each equation, let's generate some coordinate points and plot them on the graph:
For Equation 1 (a = b + 1):
- When b = 0, a = 1. Therefore, the point (0, 1) is on the line.
- When b = 2, a = 3. Therefore, the point (2, 3) is on the line.
For Equation 2 (2a + b = 5):
- When a = 0, b = 5. Therefore, the point (0, 5) is on the line.
- When a = 2, b = 1. Therefore, the point (2, 1) is on the line.
Now, we can plot these points on a graph:
|
4 |
|
3 | • (2, 3)
|
2 |
|
1 | (2, 1)
| •
0 |_________________
0 1 2 3 4 5
Now, we can draw the lines connecting the points for each equation:
|
4 |
|
3 |--------
| /
2 | /
| /
1 |/
| •
0 |_________________
0 1 2 3 4 5
As we can see from the graph, the lines intersect at the point (2, 3). This means that (b, a) = (2, 3) is the solution to the system of equations.
To check our solution, substitute the values of (b, a) = (2, 3) into both equations:
Equation 1: a = 1 + b
3 = 1 + 2 (True)
Equation 2: b = 5 - 2a
2 = 5 - 2(3) (True)
Therefore, (b, a) = (2, 3) is a valid solution to the system of equations.
In this case, the system has only one unique solution.