How do you solve #x-2y<8# and #2x-3y<=6#?

1 Answer
Feb 19, 2018

See below.

Explanation:

We have

#{(x-2y=8-lambda_1^2),(2x-3y=6-lambda_2^2):}#

with #{lambda_1 ne 0, lambda_2} in RR#

Now solving for #{x,y}# we get at

#{(x=3lambda_1^2-2lambda_2^2-12),(y=2lambda_1^2-lambda_2^2-10):}#

which means that the solution is the union set for all #{lambda_1 ne 0, lambda_2} in RR#