What are x and y if #4x-4y=-16# and #x-2y=-12#?

1 Answer
Nov 10, 2015

#x = 4#, #y = 8#

Explanation:

There are many ways to solve a system of linear equations.
One of those goes like this:

Take the equation that looks easier to you and solve it for #x# or #y#, whichever is easier.
In this case, if I were you, I would definitely take #x - 2y = -12# and solve it for #x#:

#x - 2y = - 12#
#<=> x = 2y - 12#

Now, plug #2y - 12# for #x# in the other equation:

#4*(2y-12) - 4y = -16#

...simplify the left side:
#<=> 8y - 48 - 4y = -16#
#<=> 4y - 48 = -16#

... add #48# on both sides:
#<=> 4y = 48 - 16#
#<=> 4y = 32#

... divide by 4 on both sides:
#<=> y = 8#

Now that you have the solution for #y#, you just need to plug this value into one of the two equations (again, whichever is easier) and compute #x#:

#x - 2* 8 = - 12#
#<=> x = 16 - 12#
#<=> x = 4#