How do you solve #2x + y = 3#, #x - 3y = 5#?

1 Answer
May 18, 2018

Answer:

See a solution process below:

Explanation:

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

#2x + y = 3#

#2x - color(red)(2x) + y = 3 - color(red)(2x)#

#0 + y = 3 - 2x#

#y = 3 - 2x#

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

#x - 3y = 5# becomes:

#x - 3(3 - 2x) = 5#

#x - (3 * 3) + (3 * 2x) = 5#

#x - 9 + 6x = 5#

#1x + 6x - 9 = 5#

#(1 + 6)x - 9 = 5#

#7x - 9 = 5#

#7x - 9 + color(red)(9) = 5 + color(red)(9)#

#7x - 0 = 14#

#7x = 14#

#(7x)/color(red)(7) = 14/color(red)(7)#

#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 2#

#x = 2#

Step 3) Substitute #2# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = 3 - 2x# becomes:

#y = 3 - (2 * 2)#

#y = 3 - 4#

#y = -1#

The Solution Is:

#x = 2# and #y = -1#

Or

#(2, -1)#