#!!
1 messages · Page 1 of 1 (latest)
- Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
- Wait patiently for a helper to come along.
- Once someone helps you, say thank you and close the thread with:
+close - Feel free to nominate the person for helper of the week in #helper-nominations
- Do not ping the mods, unless someone is breaking the rules.
- If you're happy with the help you got here, and the server overall, you can contribute financially as well:
@tacit orchid
okay well what is your first thought
X + Y + Z = 25
4x - 2y - z = 55
4 if he got it right, -2 if he got it wrong, and - 1 if he didn't answer any questions
X = 25 - Y - Z?
I'm not sure on what to do next
Solve the system generally. I assume the rest can be checked with brute force.
I dont think i quite understand how it works
Solve the system using Gauss's method first.
1 + 1 + 1 = 25
4 + -2 - 1 = 55 (?)
I just searched google
I dont think so
We have:
x + y + z = 25
4x - 2y - z = 55
Start by writing the matrix of this system.
Actually, do you mind explaining the Gauss method, if it's too long then it's fine
Oh.
I assume you haven't studied linear algebra.
Then just solve the system using an approach you know. Substitution, for example.
4(25 - y - z) - 2y - z = 55
I did but only the simple ones
I dont understand how to find y tho through this
Simplify it, then express y in terms of z.
Wait, where did x come from?
You only have y and z here.
Right. So, now you can express y in terms of z.
can you help me with this one
You have a linear equation in y and z. Express y in terms of z.

-6y = 5z + 45 (?)
Well, -6y = 5z - 45, rather. So, what is y?
There's more to count?
Well, you have expressed -6y. We need to express just y.
No.
are you free at the moment?
is it possible if we could get in a call, and you draw me some explanation?
Sorry, can't join a call right now.
I suggest you review the topic of solving linear systems.
Ah i see, no worries
Then, what's the right answer?
Well, I won't just tell the answer.
As I said, review the topic first, then you can solve this problem.
but it's 23:28
Am i almost finished with the questions, if i learned linear systems and got that part right, am i close to finishing the question?
Yes. You express x and y in terms of z, then just try various values of z, noting that they must be integers between 0 and 25.
Well, ok...
This makes no sense
Y = 7.5 - 5z
YES?
Is it correct this way?
We have the system:
x + y + z = 25
4x - 2y - z = 55
We express x from the first equation.
x = 25 - y - z
Then we substitute it into the first and simplify.
4(25 - y - z) - 2y - z = 55
100 - 4y - 4z - 2y - z = 55
6y + 5z = 45
Now we express y.
6y = 45 - 5z
y = 15/2 - (5/6)z
No offense, but this is a very easy question.
okay
So, we have y in terms of z. Now, substitute that into one of the initial equations (first one would probably be easier). Then express x in terms of z.
I dont understand how you get 15/2 - (5/6)z
I divided both parts of 6y = 45 - 5z by 6.
Like 45 ÷ 6?
Well, that and also -5/6.
Okay i get it
but how 15/2
in what way
like i know it both has a total of 7.5, but like how you think of putting 15/2
Because 45/6 = 15/2.
You should review the topic and solve this problem later.
right
there's no way it's wrong
Nice! So, we have:
x = 35/2 - (1/6)z
y = 15/2 - (5/6)z
Now, recall that x, y, z are all integers between 0 and 25. So, what values of z do we need to check?
What?
No. And we need to check several values of z.
Again, remember that x, y and z must be integers.
was it a close guess
Look at the first equation, for example. What form must z have so that x is an integer?
Better to start by thinking about what form (1/6)z must have.
5?
Yes?
Um...
Yeah, it isn't. It is, however, a half-integer.
So, if we want to add (or subtract) another thing to it, it must also be a half-integer.
So, (1/6)z has the form n + 1/2, where n is an integer. What form does z have, then?
I dont know
Well, if (1/6)z = n + 1/2, what is z?
I dont know simple math
That's why I recommend reviewing this topic first.
We may be able to finish this problem eventually, but it doesn't matter if you don't understand how to solve it.
I'll learn the topic more after this question
See it as if I'm trying to solve the exercise question
Z = 6( n + 1/2) ?
No. Those are the numbers of the form 6n. You need 6n + 3.
Right. So, now check for which value of z we get integer values of both x and y.
3
Well, 3 works. Anything else? Make sure to write the values of all the three variables: x, y, z.
So, for z = 3 we get (x, y, z) = (17, 5, 3).
z = 3 (x, y, z) = (17, 5, 3)
z = 9 (x , y, z) = (16, 0, 9)
z = 15 (x , y, z) = (15, -5, 15)
z = 21 (x , y, z) = (14, -10, 21)
Well, as x, y, z must be between 0 and 25, the latter two don't count.
So, just (17, 5, 3) and (16, 0, 9).
Now, read the statement of the problem again.
okay
James got 17 questions right ?
Thank you so much for teaching me, even though i was a little hard headed for not learning the topics beforehand but i really appreciate your thoughts and efforts.
Well, notice that James didn't think he attempted a lot of questions.
(17, 5, 3) corresponds to him attempting 22 questions, while (16, 0, 9) corresponds to 16 attempted questions. So, I'd say the latter works better.
Ah, wait.
