How do you solve the system of equations #2x+y=9# and #8x-2y=6#?

1 Answer
Oct 28, 2017

#x = 2#
#y = 5#

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Explanation:

Having two unknowns in an equation is a problem. Hence, your plan of action should be to somehow "remove" one of those unknowns, which would bring it down to something you can easily solve.

This is where we can use the additional equation to our advantage. Now, there's two popular ways you could go about solving these kinds of problems: through elimination, or substitution. I'll walk you through each of these in turn:

Substitution
This is the one I feel makes more intuitive sense, but reality is that you will not use it as much as elimination, come future math classes. Nonetheless, the idea is that you solve for one variable in terms of the other in one of your equations, and plug it into the other equation.

Consider your first equation:

#2x + y = 9#

I'm going to solve for #y# in terms of #x#. Here's what it looks like:

#y = 9 - 2x#

Now, I'll plug this value in for #y# in my other equation, and this is what I'll get:

#8x - 2y = 6#
#=> 8x - 2(9-2x) = 6#

#=> 8x - 18 +4x = 6#
#=> 12x - 18 = 6#

Now, we've got one equation with just one variable! This is something we can easily solve:

#=> 12x =24#
#=> x = 2#

Now you can plug this into either starting equation to get #y#. I'll be using equation 1:
#=> y = 9 - 2(2)#
#=> y = 5#

That's substitution!

Elimination
This is what you will see more and more of, especially as you hit linear algebra and have to work with many equations with many variables. So, I'd recommend you try and master this.

The goal is to try and add or subtract our equations in such a way that one variable dissapears. So let's look at ours:

#2x + y = 9#
#8x - 2y = 6#

Now you could try adding or subtracting, but nothing's going to cancel out. Are we done? Well, there's one last trick we can pull: we are allowed to multiply our equations by any real numbers to make the variables line up.

I see a #+y# and a #-2y# there, so I'm going to multiply my first equation by 2, and add the equations:

#=> 4x + 2y = 18#
# + 8x - 2y = 6#

=> #12x = 24#
=> #x = 2#

Plug this into any of the original equations for y:

#y + 2(1)= 9#
#y = 5#

Same answer both ways!

Check your answer? Graph the equations and see their intersection (see short summary).

Note that you don't have to follow the exact same steps I did for these problems. You can eliminate #x# first if you want, for example. The only thing is you need to stay consistent throughout. Also, neither method is "better", but one or the other may be faster for different problems.

Sorry this was so long, but I promise, the knowledge is worth it.

Hope that helped :)