How do you solve #x+ y = 40# and #2x + .7y = 12#?

1 Answer
Feb 26, 2017

See the entire solution process below:

Explanation:

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

#x + y = 40#

#x + y - color(red)(y) = 40 - color(red)(y)#

#x + 0 = 40 - y#

#x = 40 - y#

Step 2) Substitute #40 - y# for #x# in the second equation and solve for #y#:

#2x + .7y = 12# becomes:

#2(40 - y) + .7y = 12#

#80 - 2y + .7y = 12#

#80 - 1.3y = 12#

#-color(red)(80) + 80 - 1.3y = -color(red)(80) + 12#

#0 - 1.3y = -68#

#-1.3y = -68#

#(-1.3y)/color(red)(-1.3) = (-68)/color(red)(-1.3)#

#(color(red)(cancel(color(black)(-1.3)))y)/cancel(color(red)(-1.3) )= 52.308#

#y = 52.308# rounded to the nearest thousandth.

Step 3) Substitute #52.308# for #x# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = 40 - y# becomes:

#x = 40 - 52.308#

#x = -12.308#

The solution is: #x = -12.308# and #y = 52.308# or #(-12.308, 52.308)# rounded to the nearest thousandth.