How do you solve the system of equations #2x+3y=16# and #5x-2y=21#?

1 Answer
Feb 5, 2017

See the entire solution process below:

Explanation:

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

#2x + 3y = 16#

#2x + 3y - color(red)(3y) = 16 - color(red)(3y)#

#2x + 0 = 16 - 3y#

#2x = 16 - 3y#

#(2x)/color(red)(2) = (16 - 3y)//color(red)(2)#

#x = 8 - 3/2y#

Step 2) Substitute #8 - 3/2y# for #x# in the second equation and solve for #y#:

5(8 - 3/2y) - 2y = 21#

#40 - 15/2y - 2y = 21#

#40 - 15/2y - (2/2 xx 2)y = 21#

#40 - 15/2y - 4/2y = 21#

#40 - 19/2y = 21#

#40 - color(red)(40) - 19/2y = 21 - color(red)(40)#

#0 - 19/2y = -19#

#-19/2y = -19#

#-2/19 xx -19/2y = -2/19 xx -19#

#38/38y = 2#

#y = 2#

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

#x = 8 - (3/2 xx 2)#

#x = 8 - 3#

#x = 5#

The solution is #x = 5# and #y = 2#