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