How do solve the following linear system?: # x + y = 2, 3x - y = -6 #?
1 Answer
The solutions to this liner system for "x" and "y" are
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
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
#4x = -4#
Solving for
Now that we've found
#x + y = 2#
I substitute.
#-1 + y = 2#
And solve, leaving me with
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
#1(3x + 6y = 12)#
#-6(x + y = 6)#
#3x + 6y = 12#
#-6x -6y = -36#
Now, you can see that both
#3x = 12#
#-6x = -36#
Solving for
I hope that was helpful.