How do you solve the following system: #x-4y=2, x-2y=4 #?

3 Answers
Mar 4, 2018

Answer:

#x = 6#
#y = 1#

Explanation:

We set up the two equations with 'matching' components and then use algebra to reduce the terms to a solution (if there is one).

1) #x − 4y = 2#
2) #x − 2y = 4# Subtract 1) from 2)

#2y = 2# ; #y = 1# Put this into 1) to find x:

#x − 4(1) = 2# ; # x = 6# Put this back into 2) to check for validity.

#6 − 2(1) = 4# ; #6 = 6# CORRECT!

Answer:

#x=6, y=1#

Explanation:

#x-2y=4#

#x=2y+4#...........................eq.(1)

#x-4y=2#

#x=4y+2#...........................eq.(2)

#x=x#

eq(1)=eq(2)

#2y+4=4y+2#

#2y=2#

#y=1#

#x-2(1)=4#

#x=6#

Mar 4, 2018

Answer:

#x = 6#, #y = 1#

Explanation:

#x - 4y = 2#

solve for #x#

#x = 2 + 4y#

*substitute #x# in the second equation *

#2 +4y -2y = 4#

#2 + 2y = 4#

#2y = 2#

#y = 1#

Now we solve for #x#

#x = 2 + 4(1)#

#x = 2+4#

#x = 6#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * #

To double-check, let's plug in our values to the second equation

#x−2y=4#

#6 - 2(1) = 4#

#6 - 2 = 4#

#4 = 4#

We were right!