1. Math problem

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?)

2. Is the first 30 marbles included in the second 50 marbles?

3. Originally Posted by iFresh
Is the first 30 marbles included in the second 50 marbles?
No, its separate situations, but in both, they have the same amount.

4. is the equation set equal to each other? and there's a homework section if you didn't know.

5. Originally Posted by Vice-Versa
If Juan gives Isiah* 30 marbles, then* Isiah* will have twice as many marbles as Juan. If Isiah gives Juan 50 marbles, Juan will have 3 times more than Isiah. How many marbles does Juan have?

(What is the equation?)
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 ----------

Originally Posted by Vice-Versa
If Juan gives Isiah* 30 marbles, then* Isiah* will have twice as many marbles as Juan
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

Originally Posted by Vice-Versa

If Isiah gives Juan 50 marbles, Juan will have 3 times more than Isiah. How many marbles does Juan have?

(What is the equation?)
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).

6. If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate at which the diameter decreases when the diameter is 8 cm. (Give your answer correct to 4 decimal places.)

Since we're on the topic of hw, can someone help me solve this?

7. 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

8. The Following User Says Thank You to Arhk For This Useful Post:

[MPGH]Bernard. (11-07-2012)

9. Wow I completely forgot about the HW section. Never really used it since... well you know google.

10. Originally Posted by Alday
Wow I completely forgot about the HW section. Never really used it since... well you know google.
It's kinda hard to find it.

11. 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 http://www.wolframalpha.com
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.

12. Lol that @Arhk trying his best haha

13. Originally Posted by Alday
Wow I completely forgot about the HW section. Never really used it since... well you know google.
i personally like the bing search engine better.

14. Originally Posted by Slendyy
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.
~

15. Originally Posted by Lemongrab
i personally like the bing search engine better.
I think the name Bing is retarded. Loool. But I barely use it so I wouldn't know xD

16. some sort of make-shift terrorist bomb.

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