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6.03] Part 1: Decide whether you would use the graphing, substitution, or elimination method to solve the following system of equations. Explain, in complete sentences, why you chose that method. Part 2: Solve the following system of equations and show all of your work. 4x + 3y = -1 3x + y = 3

Mathematics
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I would help you, BUT you hated my answer to your last question like this ;)
Oh, that is okay, I did not hete your answer anyhow. :) This is a new question, so it doesn't matter, If you post it, I can work it out. :)
OK. The first thing to understand is that you have two linear equations. They are lines. They expand across all of (X,Y) space. When you are asked to solve a system of equations, all they want you to do is figure out whether the two equations have any solutions. In other words, do the two lines intersect somewhere? Are you with me so far?

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Yes, but the thing is I don't knoiw, because I don't like inequality questions, they can be anything!
Miapoopoopoo, They aren't inequality questions :) Now I will help you proceed. NOTE: The form of the equations you have are USELESS. The information is in there, but it is difficult to see it. Solution: Get your two equations into slope intercept form, namely y=mx + b where m is your slope and b is the point where the line intersects the Y axis. Can you get the equations into y= (everything else)?
BEFORE you reply. Where I am taking you is toward the GRAPHING method. I can help you with any method you choose, though. You tell me what you would like to do...
I don't know, maybe elimination or substitution?; :)
I prefer ELIMINATION. Let's do that.
Okay, thx. :)
Quick question: Did they happen to give you the answer to the problem ahead of time? I recall you were not able to check the last time I helped ;)
No, actually, you are supposed to solve it, and then they tell you if you r right. :)
That is no problem :)
:D
4x+3y = -1 3x+ y = 3 I really really want you to realize that you can do ANYTHING to these equations, as long as you do the same thing to both sides. That maintains the equality. The key is to change the top and the bottom in such a way so as to make a cancellation (ELIMINATION) possible. Still following?
Yes, :)
COOL! Now can you think of something we can do to one equation (or BOTH) to make a cancellation possible? The key is to make a cancellation possible in X or in Y. 4x +3y = -1 3x + y = 3 For example, if we multiply the top equation by (-1) and multiply the bottom equation by (3) we can get the Y value to cancel. We will eliminate it.
(-1)(4x+3y=-1) (3) (3x+y = 3) Notice I am multiplying BOTH sides of each equation by the number we picked. Can you see that?
Yes. :)
Now let's multiply by distributing the numbers we picked. -4x-3y=1 9x+3y=9 Is that what you got when you multiplied (distributed)?
Yes., that's what you get. :) Had to look at the problem for a sec. :)
For Elimination method, you have to "add" the two equations. I prefer to say "combine" the equations. -4x-3y=1 9x+3y=9 ------------ 5x+0 =10 OR 5x=10
Now that we did the appropriate Elimination, can you figure out what X is?
2?
CORRECT!
So we figured out the value for X. Great! Now what? Remember, we solved a SYSTEM of equations. Our value of X is a "solution" for each equation. To figure out Y, simply plug your X into either equation. Now make it happen and tell me what you get...
Ok, thx so much! :D
We aren't done :)
I know, let's see, hold on a sec...
Is it -3?
YES!
So you figured out that X=2 and Y=-3 We know better so we also know that the point (2,-3) is the point at which both lines intersect.
Ok, is that it?
Now verify using the link below. Start liking Wolfram :) http://www.wolframalpha.com/input/?i=4x%2B3y%3D-1%2C+3x%2By%3D3 Note the two lines and one point of intersection. Also note the solutions they give you for X and for Y. THE SAME!
Ok, thx! LOL! I will check it out!
Enter your answers in the way they ask. IF they ask for an ordered pair (X,Y) give them that. If they ask for X= and Y= give them that.
Ok. :)
Do it and report back here with the results. MK.
Hey, they say it is right! :D
So I was probably right last time, too, huh? LOL
LOL, :) Oh well, :)
You got a medal this time, 2 from me. :)
I hope you learned a little about how to THINK about this kind of problem. Sure it took a long time, but it might help during exams. You are welcome and good luck :)
thx, lol! :D
are you in calculus?
Yes.
It is the bomb :)
So, you like it, I'm dreading it!
Nooooooo! It was the first time I saw some real application of math. All the math you have done leading until now is all in preparation for Calculus. I used to hate math. Now I love it. I'm in Cal 2. Cal 3 next semester.
Really, that is so cool! :D I hope I do good in it, lol!
Very practical stuff. It sounds difficult, but when you really get down to it, it is all about ADDING ( an infinite amount). That is the hard part. lol.
oh, wow, whattopics do you learn, lol?
Well, I am signing off now. If you ever get stuck, shoot me an email. Byeeeeeeeeee
Ok, byeee!

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