How do you solve #y = 4x  3#, #y = 1# by graphing and classify the system?
1 Answer
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{(y4x+3)(x^2+(y+3)^20.04)((x1)^2+(y1)^20.04)=0 [10, 10, 5, 5]
Equation 2:

For
#x = 2# ;#y = 1# or#(2, 1)# 
For
#x = 2# ;#y = 1# or#(2, 1)#
graph{(y1)(y4x+3)((x+2)^2+(y1)^20.04)((x2)^2+(y1)^20.04)=0 [10, 10, 5, 5]
Identify where the lines cross:
graph{(y1)(y4x+3)((x1)^2+(y1)^20.04)=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