How do you solve the system of Equations #2x+8y=6, -5x-20y=-15#?

Question

Hazel Anglin · Accepted Answer

#y = 0# and #x = 3# To complete this problem, first solve the first equation for #x#: #2x + 8y - 8y = 6 - 8y# #2x = 6 - 8y# #(2x)/2 = (6 - 8y)/2# #x = 3 - 4y# Next, substitute #3 - 4y# for #x# in the second equation and solve the #y#: #-5(3 - 4y) - 20y = -15# #-15 + 60y - 20y = -15# #-15 + 40y = -15# #-15 + 15 + 40y = 15 + 15# #40y = 0# #(40y)/40 = 0/40# #y = 0# Finally, we substitute #0# for #y# into the solution for the first equation and calculate #x#: #x = 3 - 4*0# #x = 3 - 0# #x = 3#

Genesis Allen · Answer

undefined To solve the system of equations 2x + 8y = 6 and -5x - 20y = -15:

1. Multiply the first equation by 5 to eliminate the variable x:
   5(2x + 8y) = 5(6)
   10x + 40y = 30

2. Multiply the second equation by 2 to eliminate the variable x:
   2(-5x - 20y) = 2(-15)
   -10x - 40y = -30

3. Add the modified equations together:
   (10x + 40y) + (-10x - 40y) = 30 + (-30)
   0 = 0

4. Since 0 = 0 is always true, this means the system has infinitely many solutions.

Therefore, the system of equations is consistent and dependent. The solution set can be represented as any point on the line 2x + 8y = 6.