How do you solve #x-2y=4# and #-x+6y=12#?

1 Answer
Dec 1, 2015

Answer:

#color(green)(x = 12#
#color(green)(y = 4#

Explanation:

#x-2y=4# -----(1)
#-x+6y=12# ------(2)

The question is asking us the find the values of x and y which satisfy both these equations

To find the value of #x#, we just need to eliminate #y#

We can do that by adding the two equations. We get:

#(x - 2y) + (-x + 6y) = 4+12#

#->cancel(x) - 2y - cancel(x) + 6y = 16#

#4y = 16#

#color(green)(y = 4#

To get the value of #x#, we can easily substitute this value of #y# in any of the two equations

Substituting it in the first equation, we get

#x - 2(4) = 4#

#->x - 8 = 4#

Transposing #8# to the other side, we get

#-> x = 4 + 8 #

#color(green)(x = 12#