How do you solve the following system?: # 2x - 4y = 5 , x-4y=3 #

2 Answers
Gió
Nov 13, 2015

I found:
#x=2#
#y=-1/4#

Explanation:

I would isolate #x# from the second equation and substitute it into the first:
#x=3+4y#
and so:
#2(color(red)(3+4y))-4y=5# solve for #y#:
#6+8y-4y=5#
#4y=-1#
#y=-1/4#
Substitute this back into: #x=3+4y#
#x=3+4(color(red)(-1/4))=3-1=2#

mason m
Nov 13, 2015

Use substitution for #x# then plug back in with #y#.

Explanation:

If #x - 4y = 3#, then add #4y# to both sides to get #x = 4y + 3#.

Knowing this, you can plug in #4y + 3# in the other equation in place of #x# since they are equivalent.

This results in: #2(4y + 3) - 4y = 5#
Distribute the 2 to get: #8y + 6 - 4y = 5#
Combine like terms on the left: #4y + 6 = 5#
Subtract 6 from both sides: #4y = -1#
Divide both sides by 4: #color(red)(y = -1/4)#

With this knowledge, plug #y# into either equation to determine the value of #x#, as such: #x - 4(-1/4) = 3#
Multiply (remember that a negative time a negative is positive): #x + 1 = 3#
Add 1 to both sides: #color(red)(x = 2)#

Thus, our ordered pair is #(2, -1/4)#.

Related questions
See all questions in Systems Using Substitution
Impact of this question
5317 views around the world
You can reuse this answer
Creative Commons License