Given the following system, solve for x. x + 2y = 12; x - 4y = -18
- Thread starter jwk811
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pka
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If you subtract those two, you get \(\displaystyle 6y=30\).
There are a couple ways of doing this.
The easiest to understand is to take your first equation say and solve for y in terms of x.
In this case this turns out to be y = (12-x)/2. Do you understand why?
Then what you do is take that y you just solved for and plug it into your 2nd equation.
In this case you end up with
x - 4 ((12-x)/2) = -18
do all that arithmetic and you end up with
3x - 24 = -18, 3x = 6, x = 2
The second way to do this is to recognize that if I have two equations I can add each side to one another without changing the equality.
if I have A = B, and C = D, then A+C = B+D
if x + 2y = 12, then 2x + 4y = 24, all I've done is multiply both sides by 2.
Now add this to the second equation you get
2x + 4y = 24 +
x - 4y = -18
--------------
3x + 0 = 6
which is easily solved as x=2
You should come up with a few practice examples on your own to make sure you understand how to do this.
I'm learning all of Algebra 1 and 2 on my own so I can pass the placement test and not have to take them and move on to College Math
Thanks, I get it . I did something like that but don't know where I went wrong..I'll try to show it next time