How do you solve the system using elimination #3r - 2s = 1# and #2r - 3s = 9#?

1 Answer
Jul 4, 2015

Answer:

Subtract different multiples of the two equations from one another to eliminate #r# or #s# and find #s = -5# and #r = -3#

Explanation:

First subtract #3# times the second equation from twice the first equation to get:

#-25 = 2(1) - 3(9) = 2(3r-2s)-3(2r-3s) = 5s#

Divide both ends by #5# to get #s = -5#

Secondly, subtract twice the second equation from #3# times the first equation to get:

#-15 = 3(1) - 2(9) = 3(3r-2s)-2(2r-3s) = 5r#

Divide both ends by #5# to get #r = -3#