Hmm... Can you solve this?

[MASTERBAKER]

Hmm... Can you solve this?

 

[Snyckerbar76]

I don't remember a bit of calculus or precalculus, but this looks familiar:

This is called "Systems of Equations" solved with substitution (http://www.sosmath.com/soe/SE/SE.html)

Trade fruit for variables:
Apples: x
Bananas: y
Coconuts: z

Here's how I did it:

x+x+x = 30
Simplify: 3x = 30

Bananas (y) in terms of x
x + 2y = 18
x = 18 - 2y

Solve for y using 3x = 30 (plug x value from bananas into 3x = 30 equation)
3(18-2y) = 30

(Use distributive property):
54-6y = 30
-6y = 30 - 54
-6y = -24
y = 4

Solve for z using y value (plug y value into y-z = 2 equation)
y - z = 2
4-z = 2
z = 2

Solve for x using x + 2y = 18
x + 2(4) = 18
x + 8 = 18
x = 10

Solve final portion of equation:
10 + 2 + 4 =16

Check:
3(10) = 30 (True!)
10 + 2(4) = 18 (True!)
4 - 2 = 2 (True!)

 

[Toro Negro]

22

 

[soze]

Apples=10
Bananas=4
Coconuts=2

 

[trappstarr82]

14. There is only 1 coconut in the last equation and 3 bananas in the bunch.

 

[Toro Negro]

Sorry gave the wrong answer previously.
The correct answer is 16 !!!!

 

[Commish]

Hmm... Can you solve this?

I simplified the equation

Apple=10
Banana=4
Coconut=2

Apple+Apple+Apple=30 or 10+10+10=30
Apple+Banana+Banana=18 or 10+4+4=18
Banana-Coconut=2 or 4-2=2
Coconut(2)+Apple(10)+Banana(4) or 2+10+4=16

??=16

 

[D24OHA]

14. There is only 1 coconut in the last equation and 3 bananas in the bunch.

This!!

it's 14

 

[Black James Bond]

I got 16 at first, but if you take in account the actual number of bananas in the bunch then each banana is equal to one and the last equation there are only three bananas then the last equation is 14 as the same logic applies to the coconuts.

 

[Commish]

Hmm... Can you solve this?

Correction: I didn't notice that there is one less coconut and banana in the last equation, so I guess my answer changes to 14. My bad. Lol