How do you solve the system #x/2-y/3=5/6# and #x/5-y/4=71/10#?

1 Answer
Mar 21, 2016

Answer:

Scale and combine the equations to eliminate #y# and solve for #x#, then substitute that value of #x# back in to solve for #y# and find:

#{ (x = -37), (y = -58) :}#

Explanation:

Given:

#{ (x/2-y/3=5/6), (x/5-y/4=71/10) :}#

Multiply the first equation by #6# and the second by #20# to get:

#{ (3x-2y=5), (4x-5y=142) :}#

To eliminate #y#, multiply the first of these by #5#, the second by #2# and subtract the second from the first...

#{(15x-10y=25),(8x-10y=284):}#

#=> 7x = 25-284 = -259#

Then divide both ends by #7# to find:

#x = -259/7 = -37#

Substitute this value of #x# into the equation #3x-2y=5# to get:

#-111-2y=5#

Add #111# to both sides to get:

#-2y=116#

Divide both sides by #-2# to get:

#y = -58#