How do you solve the system of equations #y= 2x - 5# and #2y - 3x = 6#?

1 Answer
Mar 25, 2017

See the entire solution process below:

Explanation:

Step 1) Because the first equation is already solved for #y#, substitute #2x - 5# for #y# in the second equation and solve for #x#:

#2y - 3x = 6# becomes:

#2(2x - 5) - 3x = 6#

#(2 xx 2x) - (2 xx 5) - 3x = 6#

#4x - 10 - 3x = 6#

#4x - 3x - 10 = 6#

#(4 - 3)x - 10 = 6#

#1x - 10 = 6#

#x - 10 = 6#

#x - 10 + color(red)(10) = 6 + color(red)(10)#

#x - 0 = 16#

#x = 16#

Step 2) Substitute #16# for #x# in the first equation and calculate #y#:

#y = 2x - 5# becomes:

#y = (2 xx 16) - 5#

#y = 32 - 5#

#y = 27#

The solution is: #x = 16# and #y = 27#