How do you solve the system of equations: #4x - y = 15# and #2x + 3y = 25#?

1 Answer
Dec 14, 2017

See a solution process below:

Explanation:

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

#4x - y = 15#

#-color(blue)(15) + 4x - y + color(red)(y) = -color(blue)(15) + 15 + color(red)(y)#

#-15 + 4x - 0 = 0 + y#

#-15 + 4x = y#

#y = 4x - 15#

Step 2) Substitute #(4x - 15)# for #y# in the first equation and solve for #x#:

#2x + 3y = 25# becomes:

#2x + 3(4x - 15) = 25#

#2x + (3 xx 4x) - (3 xx 15) = 25#

#2x + 12x - 45 = 25#

#(2 + 12)x - 45 = 25#

#14x - 45 = 25#

#14x - 45 + color(red)(45) = 25 + color(red)(45)#

#14x - 0 = 70#

#14x = 70#

#(14x)/color(red)(14) = 70/color(red)(14)#

#(color(red)(cancel(color(black)(14)))x)/cancel(color(red)(14)) = 5#

#x = 5#

Step 3) Substitute #5# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = 4x - 15# becomes:

#y = (4 xx 5) - 15#

#y = 20 - 15#

#y = 5#

The Solution Is: #x = 5# and #y = 5# or #(5, 5)#