How do you solve #x-5y = -3# and #3x+8y = -9# by substitution?

Question

739 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-27T17:59:06

I got:
#x=-3#
#y=0#

From the first equation we get:

#x=5y-3#

and substitute for #x# into the second:

#3(color(red)(5y-3))+8y=-9#

solve for #y#:

#15ycancel(-9)+8y=cancel(-9)#

and: #y=0#

giving:

#x=-3#