How do solve the following linear system?: # x + y = 2, 3x - y = -6 #?

1 Answer

Answer:

The solutions to this liner system for "x" and "y" are #-1# and #3#, respectively.

Explanation:

Ultimately, you can go about solving this problem two different ways. The easiest and quickest way would be to use what's called the elimination method. As it's name implies, you eliminate terms from each equation. How do you do this?

You begin by stacking one equation over the other, like so.

#x + y = 2#
#3x - y = -6#

The goal of elimination is to eliminate all variables except for the one you are trying to solve for. In this instance, I'm picking #x# to solve for first, which means I'll eliminate the #y#'s.

So once you've stacked your equations, you need to check to see if anything can cancel out. Without doing any math, I immediately see that both #y#'s cancel if I add the two equations together. This leaves me with:

#4x = -4#

Solving for #x# leaves me with #x = -1#.

Now that we've found #x#, we can go back and do what's called substitution. Simply put, we substitute (hence the name "substitution") the #x#-value into one of our original equations. Pick whichever you like, but for simplicity purposes, I will choose

#x + y = 2#

I substitute.

#-1 + y = 2#

And solve, leaving me with #y = 3#. Tada. Congratulations. You have now solved the linear system.

For future reference, if you're using the elimination method, and nothing cancels out, you must multiply each equation by numbers that will allow one term from each equation to cancel out. For example:

#3x + 6y = 12#
#x + y = 6#

On first inspection, nothing seems to cancel out. And you would be right. So you must multiply each equation by numbers that will allow either #x#'s or #y#'s to cancel.

#1(3x + 6y = 12)#

#-6(x + y = 6)#

#3x + 6y = 12#

#-6x -6y = -36#

Now, you can see that both #y#'s will cancel.

#3x = 12#

#-6x = -36#

Solving for #x# yields #8# as the answer. Substitute back into one of the original equations and #y# will equal #-2#.

I hope that was helpful.