How do you solve #3x+y=5# and #x-2y=11# using substitution?

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

#3x - color(red)(3x) + y = 5 - color(red)(3x)#

#0 + y = 5 - 3x#

#y = 5 - 3x#

Step 2) Substitute #color(red)(5 - 3x)# for #y# in the second equation and solve for #x#:

#x - 2(color(red)(5 - 3x))= 11#

#x - (2 xx color(red)(5)) + (2 xx color(red)(3x))= 11#

#x - 10 + 6x = 11#

#7x - 10 = 11#

#7x - 10 + color(red)(10) = 11 + color(red)(10)#

#7x - 0 = 21#

#7x = 21#

#(7x)/color(red)(7) = 21/color(red)(7)#

#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 3#

#x = 3#

Step 3) Substitute #color(red)(3)# for #x# in the solution to the first equation at the end of Step 1

#y = 5 - (3 xx color(red)(3))#

#y = 5 - 9#

#y = -4#

The solution is #x = 3# and #y = -4#

Using the second equation, we can express #y# in terms of #x# as such:

#-2y = 11 - x => y = -11/2+x/2#.

Substituting this #y# value in the first equation:

#3x + (-11/2 + x/2) = 5 => (7x)/2 = -1/2 =>#

#=>7x = -1 => x = -1/7#.

Now, we can substitute this #x# value in the second equation to find #y# (any equation will do, but #x# is easier to evaluate than #3x#)

#-1/7 - 2y = 11 => 2y = -78/7 => y = -78/14 = -39/7#