How do you solve #10x + 10y = 1# and #x = y - 3 # using substitution?

1 Answer
Feb 11, 2017

Answer:

See the entire solution process below:

Explanation:

Step 1) Because the second equation is already solved for #x# we can substitute #y - 3# for #x# in the first equation and solve for #y#:

#10x + 10y = 1# becomes:

#10(y - 3) + 10y = 1#

#10y - 30 + 10y = 1#

#10y + 10y - 30 = 1#

#20y - 30 = 1#

#20y - 30 + color(red)(30) = 1 + color(red)(30)#

#20y - 0 = 31#

#20y = 31#

#(20y)/color(red)(20) = 31/color(red)(20)#

#(color(red)(cancel(color(black)(20)))y)/cancel(color(red)(20)) = 31/20#

#y = 31/20#

Step 2) Substitute #31/20# for #y# in the second equation and calculate #x#:

#x = y - 3# becomes:

#x = 31/20 - 3#

#x = 31/20 - (20/20 xx 3)#

#x = 31/20 - 60/20#

#x = -29/20#

The solution is: #x = -29/20# and #y = 31/20# or #(-29/20, 31/20)#