How do you solve # 4x - y = 30#, # 4x + 5y = -6# by graphing and classify the system?

1 Answer
Jan 16, 2018

Answer:

See a solution process below:

Explanation:

To solve this problem, plot two points for each equation and draw a line through the points.

Equation 1:

For #x = 0#

#(4 * 0) - y = 30#

#0 - y = 30#

#-y = 30#

#-1 * -y = -1 * 30#

#y = -30# or #(0, -30)#

For #x = 5#

#(4 * 5) - y = 30#

#20 - y = 30#

#20 - color(red)(20) - y = 30 - color(red)(20)#

#0 - y = 10#

#-y = 10#

#-1 * -y = -1 * 10#

#y = -10# or #(5, -10)#

graph{(x^2 + (y + 30)^2 - 0.75)((x - 5)^2 + (y + 10)^2 - 0.75)(4x - y - 30) = 0 [-100, 100, -50, 50]}

Equation 2:

For #x = 0#

#(4 * 0) + 5y = -6#

#0 + 5y = -6#

#5y = -6#

#(5y)/color(red)(5) = -6/color(red)(5)#

#y = -6/5# or #(0, -6/5)#

For #y = 0#

#4x + (5 * 0) = -6#

#4x + 0 = -6#

#4x = -6#

#(4x)/color(red)(4) = -6/color(red)(4)#

#x = -3/2# or #(-3/2, 0)#

graph{(x^2 + (y + 6/5)^2 - 0.75)((x + 3/2)^2 + y^2 - 0.75)(4x + 5y + 6)(4x - y - 30) = 0 [-100, 100, -50, 50]}

We can see the lines cross at: #(6, -6)#

These two equations are consistent and independent because the have a solution and only one solution.

graph{((x-6)^2 + (y + 6)^2 - 0.05)(4x + 5y + 6)(4x - y - 30) = 0 [-2, 14, -7, 1]}