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

Question

Eva Adamson · Accepted Answer

See a solution process below:

Plotting two points for each equation and drawing a line through them will help you solve this problem. Formula 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)# Because there is only one solution to each of these two equations, they are independent and consistent. graph{[-2, 14, -7, 1]}((x-6)^2 + (y + 6)^2 - 0.05)(4x + 5y + 6)(4x - y - 30) = 0

Eva Abramson · Answer

undefined To solve the system of equations 4x - y = 30 and 4x + 5y = -6 by graphing:

1. Graph each equation on the same coordinate plane.
2. Find the point of intersection of the two lines.
3. Classify the system based on the number of solutions obtained from the intersection point.

If the lines intersect at one point, the system has one unique solution and is consistent and independent. If the lines are parallel and do not intersect, the system has no solution and is inconsistent. If the lines coincide (overlap), the system has infinitely many solutions and is consistent and dependent.