How do you solve the following system: #x-y=3 , 4x-5y-23=0 #?

1 Answer
Mar 15, 2016

Answer:

#-4*(x - y) = -4*3#
+#4x - 5y = 23#


#-y = 11 or y = -11#

#-5(x - y) = -5*3#
+#4x - 5y = 23#


#-x = 8 or x = -8#

Therefore, #x = -8# and #y = - 11#

Explanation:

To solve this problem, you must first solve for one variable (#x#) and then the other (#y#). To solve for #y#, we eliminate the #x# variable by multiplying the first equation by -4 on both sides:

#-4(x - y) = -4*3# ----> #-4x +4y = -12#

Then we add the two equations:

#-4x + 4y = -12#
+#4x - 5y = 23#

This gives us #(-4x + 4x) + (4y - 5y) = (23-12)# = #-y = 11# or #y = -11#

To solve for #x# we then eliminate the #y# variable by multiplying the first equation by -5 on both sides:

#-5(x - y) = -5*3# ----> #-5x + 5y = -15#

Then we add the two equations:

#-5x + 5y = -15#
+ #4x - 5y = 23#

This gives us #(-5x + 4x) + (5y - 5y) = 8# = #-x = 8# or #x = -8#

You can check the answer by substituting -8 for #x# and -11 for #y#, and you will find that both equations are satisfied.