How do you solve #y = 4x - 3#, #y = 1# by graphing and classify the system?
1 Answer
See a solution process below:
Explanation:
Two solve this by graphing, for each equation:
- Plot two points for each equation
- Draw a line through the two points
- Identify where the lines cross
Equation 1:
Solve the equation for two points and plot the two points then draw a line through the two points:
- For
#x = 0#
- For
#x = 1#
graph{(y-4x+3)(x^2+(y+3)^2-0.04)((x-1)^2+(y-1)^2-0.04)=0 [-10, 10, -5, 5]
Equation 2:
-
For
#x = -2# ;#y = 1# or#(-2, 1)# -
For
#x = 2# ;#y = 1# or#(2, 1)#
graph{(y-1)(y-4x+3)((x+2)^2+(y-1)^2-0.04)((x-2)^2+(y-1)^2-0.04)=0 [-10, 10, -5, 5]
Identify where the lines cross:
graph{(y-1)(y-4x+3)((x-1)^2+(y-1)^2-0.04)=0 [-10, 10, -5, 5]
We can see from the graphs the lines cross at
Because there is at least one point in common the system of equations is considered to be Consistent
