How do you solve #x + 2y = 6# and #x - 4y = 8# using substitution?

2 Answers
Apr 10, 2016

If #x-4y=8#
then #x=4y+8#

Therefore we can substitute #color(blue)(4y+8)# for #x#
in the other given equation: #x+2y=6#
giving:
#color(white)("XXX")color(blue)(4y+8)+2y=6#
which simplifies as:
#color(white)("XXX")6y=-2#
or
#color(white)("XXX")y=color(red)(-1/3)#

We can then substitute #color(red)(-1/3)# for #y#
in one of the given equations, say: #x-4y=8#
giving:
#color(white)("XXX")x-4(-1/3)=8#

#color(white)("XXX")x+4/3=8#

#color(white)("XXX")x=6 2/3#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Verification is always a good idea.
So testing the above solution #(x,y)=(6 2/3,-1/3)# in the given equation not used to evaluate #x#
#color(white)("XXX")x+2y=6# (????)

#color(white)("XXX")6 2/3+2xx(-1/3) = 6# (????)

#color(white)("XXX")6 2/3 - 2/3 = 6# (Correct!)

Apr 10, 2016

Answer:

substitute one of the following equations

Explanation:

x+2y=6 -----------------(1) and x-4y=8------------------(2)

Steps:
~>. Substitute one of the following equations, either (1) OR (2)

substituting (1)

Put x subject of formula,

x= 6-2y--------------(*)

~> Replace the equation (*) in equation (2)

x-4y=8
replacing.....

(6-2y)-4y=8
6-2y-4y=8
6-6y=8

Transpose "+6"on the other side of the equation
-6y=8-6
-6y=2

y= #2/-6#

y=#-(1/3)#

Hence, Replace the value obtained for y in equation (*)

x=6-2y
x=6-(#2*-1/3#)
x=6-(#-2/3#)
x=6+(#2/3#)
x=#20/3#

Therefore: x=#20/3# and y=#-(1/3)#