elimination method?

sarahilc

New member
Joined
Feb 3, 2011
Messages
15
I find really confusing.
how would I work out this problem using the elimination method?

(1) 5x+2y=5
(2) 3x-4y=-23
 
I'm a student teaching myself through correspondence bookelts and I get stuck on some lessons so I thought I'd ask for an example that might help me understand it better.
 
(1) 5x+2y=5
(2) 3x-4y=-23

Ok, so elimination is very easy :).
First) Multiply any equation to be able to have either x or y eliminated. In this case the easiest is the top one. Multiply the top by 2
You now get:
5(2)x + 2(2)y = 5(2)
that is (1) 10x+4y=10

Second) Now it is time to eliminate :)!The problem now has turned into:
(1) 10x+4y=10
(2) 3x-4y=-23

So now you add them! so you get 10x + 3x .... +4y + (-4y) .... 10 + (-23)
So you should have 13x=-13 since you cancel out the +4y and -4y.Then just divide 13 to both sides as 13x turns to x by itself and also -13(you should have -13/13) which is equal to -1. Now you know that x=-1. TIME TO FIND THE Y :)

Ok so the y is easy to find :). Just plug in the x=-1 into the original equation you said
(1) 5x+2y=5
(2) 3x-4y=-23

i will use the top one (1)
so you get 5(-1)+2y=5
5 x (-1) = -5
So you now have:
-5+2y=5
Add 5 from both sides and you should get:
-5+5=0
5+5=10
2y=10/color]
Now divide 2 to both sides and you should get:
2y
---- = y
2

10
---- = 5
2


FINALLY :) YOU GOT Y=5 and X=-1

That means that the point of intersection is (-1,5) on a graph :)

i hope i helped.
 
sarahilc said:
Oh my gosh, yes that helped so much! :D Thank you!

Lol no problem mate :) I am here to help and ask for help as everyone does i believe :D Have a pleasant night.
 
CardinalHayes said:
i hope i helped.

No, CardinalHayes, you did *not* help! You did *all* of the work!

If you are to help, then refer the student to a site as was done by Subhotosh Khan, or give hints.

But do not give the solution to the user as that user has not shown any work, and was not
given a chance to show work after visiting a link first.
 
lookagain said:
CardinalHayes said:
i hope i helped.

No, CardinalHayes, you did *not* help! You did *all* of the work!

If you are to help, then refer the student to a site as was done by Subhotosh Khan, or give hints.

But do not give the solution to the user as that user has not shown any work, and was not
given a chance to show work after visiting a link first.

SORRY! I DID NOT KNOW! IT WONT HAPPEN AGAIN!
 
CardinalHayes said:
SORRY! I DID NOT KNOW! IT WONT HAPPEN AGAIN!

Don't be too harsh on yourself - we all did it at least once (may be not all - at least I did it). Anyway - and welcome .....
 
I understand where you're coming from but I did just basically ask for an example to be solved so that I could see the steps used and it was very helpful in solving the rest of my problems. I'd come across that website before and didn't understand because often websites like that only show examples of really easy convenient problems. Before I post something here I always try it for a long time by my self first.
 
It's ok, Sara and Cardinal...don't commit suicide.

Plus no need to "listen" to anyone here, except the moderators.
 
Denis said:
It's ok, Sara and Cardinal...don't commit suicide.

Plus no need to "listen" to anyone here, except the moderators.
oh, so we can give answers and explain how we got it ?
 
CardinalHayes said:
oh, so we can give answers and explain how we got it ?
I didn't say that!
Read the rules; "Read before posting".
 
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