How do you solve #y = 4x - 3#, #y = 1# by graphing and classify the system?

1 Answer
Jul 23, 2018

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#

#y = (4 * 0) - 3#

#y = 0 - 3#

#y = -3# or #(0, -3)#

  • For #x = 1#

#y = (4 * 1) - 3#

#y = 4 - 3#

#y = 1# or #(1, 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:

#y = 1# is a horizontal line where for each and every value of #x#, #y# is equal to #1#

  • 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 #(1, 1)#

Because there is at least one point in common the system of equations is considered to be Consistent