How do you solve this system of equation: #y= x - 4 and - 2x + y = 6#?

1 Answer
Nov 20, 2017

#x=-10, y=-14#

Explanation:

There are multiple ways to solve equation systems. The most straight forward way in my opinion is to solve for one variable in one equation and plugging it into the other equation. This is how to do that for #y=x-4# and #-2x+y=6#:

Let's call #y=x-4# equation 1, or #E_1# for short, and let's call #-2x+y=6# equation 2, or #E_2#. I usually write the name of the equation on the row I'm writing so I can keep track of which equation I'm manipulating.

We will start by isolating #y# in #E_2#:

#E_2#: #-2x+y=6#

#E_2#: #cancel(+2x-2x)+y=6+2x# (add 2x to both sides)

#E_2#: #y=6+2x#

#E_1#: #6+2x=x-4# (we use the definition of y from #E_2# in #E_1#)

#E_1#: #cancel(-6+6)+2x=x-4-6# (subtract 6 from both sides)

#E_1#: #2x=x-10#

#E_1#: #cancel(2)x-x=-10cancel(+x-x)# (subtract x from both sides)

#E_1#: #x=-10#

We now know that #x=-10#, so let's plug this into the y solution in #E_2# to get #y=6+2(-10)#, which simplifies to #y=6-20=-14#.

So, #x=-10# and #y=-14#