Is the first 30 marbles included in the second 50 marbles?
If Juan gives Isaih 30 marbles, than Isaih will have twice as many marbles as Juan. If Isaih gives Juan 50 marbles, Juan will have 3 times more than Isaih. How many marbles does Juan have?
(What is the equation?)
Rep. Power: 126
Minion: 11/28/12-3/15/14
Is the first 30 marbles included in the second 50 marbles?
is the equation set equal to each other? and there's a homework section if you didn't know.
Crossfire Minion: 28 October 2010Official Middleman: 20 September 2012Trusted Member: 15 February 2013
Official Middleman: 11 April 2013
Official Middleman: 10 January 2015---------------------------------------------------
Global Moderator: 25 December 2012
Head Administrator: 21 March 2013
x is Juan, y is Isiah
System of equations.
2(x- 30) = y + 30; 3(y-50) = x +50
~
I'll show you how to do the work in a follow-up post.
---------- Post added at 11:09 PM ---------- Previous post was at 10:53 PM ----------
If Juan (x) gives 30 of his marbles, he is subtracting from his own amount, and they are being added to the person he's giving them to, Isiah (y). Isiah will then have twice as many marbles after Juan loses 30 marbles (x-30) and those 30 marbles are given to Isiah (y+30). Now that Juan has lost 30 marbles (x-30) and Isiah has gained 30 marbles (y+30), Isiah has twice as many marbles, that is 2 times Juan's current amount which is (x-30), therefore 2(x-30) = y
If Isiah (y) loses 50 marbles (y-50) and Juan gains 50 marbles (x + 50), Juan will have 3 times more than Isiah now has 3(y-50).
Last edited by Arhk; 11-06-2012 at 08:57 PM.
"If the world hates you, keep in mind that it hated me first." John 15:18
2(x- 30) = y + 30; 3(y-50) = x +50
Your teacher may have some sort of short version of doing this, but as a future engineer I don't believe in not working step-by-step. And being honest all of those short-cut methods are what fuck you up in advanced mathematics.
Solving for 'y' in the first equation; subtracting 30 from each side to attain 'y' -> 2(x- 30) - 30 = y; substitute 'y' in the next equation with it's equivalent "2(x- 30) - 30"
3 { [2(x- 30) - 30] - 50 } = x + 50: to make things more simple we're going to perform the multiplication operations on the left-side ->
3[ (2x-60-30) - 50] = x + 50: Let's go ahead and perform the simple subtraction operation now in the left-hand side of the problem before going on with multiplying ->
3[(2x-90)-50] = x + 50; Now let's perform the left-side operation of multiplying by 3 -> (6x-270) - 150 = x + 50:keep going -> 6x - 420 = x + 50: subtracting 50 from both sides -> 6x - 470 = x: subtracting x-> 5x - 470 = 0:5x must be equal to 0, aka -> 5x = 470; x = 94
"If the world hates you, keep in mind that it hated me first." John 15:18
Bernard (11-07-2012)
Wow I completely forgot about the HW section. Never really used it since... well you know google.
2(x- 30) = y + 30; 3(y-50) = x +50: we've established x = 94; -> 3(y-50) = 144: performing the left-side multiplication operation -> 3y - 150 = 144:
adding 150 to both sides 3y = 294: -> dividing by 3, both sides -> y = 98
~
And for future reference you can go to https://www.wolframalph*****m
and simply enter the system of equations you found, 2(x- 30) = y + 30; 3(y-50) = x +50; and it'll give you the variable values.
Last edited by Arhk; 11-07-2012 at 05:43 AM.
"If the world hates you, keep in mind that it hated me first." John 15:18
Lol that @Arhk trying his best haha
Actually for me this is very simple, I already had the actual answer in a minute, it's just that I want him to understand but I was too tired to work-out the problem quickly last night.
~
Engineering mathematics is much more complex algebraically.
---------- Post added at 07:53 AM ---------- Previous post was at 07:47 AM ----------
It made me laugh at first, even in middle-school I could have solved this easily.
~
Last edited by Arhk; 11-07-2012 at 05:48 AM.
"If the world hates you, keep in mind that it hated me first." John 15:18
some sort of make-shift terrorist bomb.