How do you solve #0.05x+0.25y=66# and #0.15x+0.05y=72# using substitution?

1 Answer
Aug 26, 2017

Answer:

See a solution process below: #(420, 180)#

Explanation:

Step 1) Solve the first equation for #y#:

#0.05x + 0.25y = 66#

#-color(red)(0.05x) + 0.05x + 0.25y = -color(red)(0.05x) + 66#

#0 + 0.25y = -0.05x + 66#

#0.25y = -0.05x + 66#

#color(red)(4) xx 0.25y = color(red)(4)(-0.05x + 66)#

#1y = (color(red)(4) xx -0.05x) + (color(red)(4) xx 66)#

#y = -0.2x + 264#

Step 2) Substitute #(-0.2x + 264)# for #y# in the second equation and solve for #x#:

#0.15x + 0.05y = 72# becomes:

#0.15x + 0.05(-0.2x + 264) = 72#

#0.15x + (0.05 xx -0.2x) + (0.05 xx 264) = 72#

#0.15x + (-0.01x) + 13.2 = 72#

#0.15x - 0.01x + 13.2 = 72#

#(0.15 - 0.01)x + 13.2 = 72#

#0.14x + 13.2 = 72#

#0.14x + 13.2 - color(red)(13.2) = 72 - color(red)(13.2)#

#0.14x + 0 = 58.8#

#0.14x = 58.8#

#(0.14x)/color(red)(0.14) = 58.8/color(red)(0.14)#

#(color(red)(cancel(color(black)(0.14)))x)/cancel(color(red)(0.14)) = 420#

#x = 420#

Step 3) Substitute #420# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = -0.2x + 264# becomes:

#y = (-0.2 xx 420) + 264#

#y = -84 + 264#

#y = 180#

The Solution Is: #x = 420# and #y = 180# or #(420, 180)#