What is the most efficient way to solve:
P + G = 9
2*G + A = 19
A + P = 7

I can do it but it takes too many steps.

>>16714061 (OP)
(2) - (1) - (3)
G = 3
EZPZ; took me 5 seconds

Based on the image you can guess that there are whole number solutions for Pineapple, Grape, and Apple.
Based on two Grapes and an Apple = 19, you can guess that grape is pretty big. So, do guess and check starting with Grape = 8 and work down.

>>16714061 (OP)
> Original equations
P + G = 9 <--- Equation 1
2G + A = 19 <--- Equation 2
A + P = 7 <--- Equation 3

> Isolate A:
A = 7 - P

> Substitute A on second Equation
2G + (7 - P) = 19
2G - P = 19 - 7
2G - P = 12 <--- Equation 4

> Add Equation 1 and Equation 4:
(P + G) + (2G - P) = 9 + 12
3G + P - P = 21
3G = 21
G = 21/3
G = 7

> Now the rest is easy, since we know G:
P + G = 9 <--- Equation 1
Then P = 2

A + P = 7 <--- Equation 3
then A = 5

>>16714121
I forgot to mention, you can also isolate G or P on step 1 (where we isolated A).
All 3 methods will give the same result.
This is the most efficient way to resolve it, and it's called "elimination method for solving systems of equations"

What you're all missing is that the image isn't asking for the individual values, but just the sum, A+G+P.
Then it's just:
A+P+G
=(1/3)(3A+3G+3P)
=(1/3)((P+G)+(2G+A)+2(A+P))
=(1/3)(9+19+2*7)
=(1/3)(42)
=14

>>16714061 (OP)
A + B = 9
2B + C = 19
A + C = 7
A + B + C = X

>A + B + A + C = 2A + B + C = 16
2A + B + C = 16
>2B + C - ( A + C ) - ( A + B ) = -2A + B = 3
-2A + B = 3
>2A + B + C - 2A + B = 2B + C = 13
2B + C = 13
but
2B + C is 19

>>16714061 (OP)
G-A=2
3G=21
G=7, P=2, A=5, sum=14

>>16714061 (OP)
Not a math fag but here goes:

>P+G=9
Assuming nothing is zero, both values can't be higher than 8.
A can be anything
P can be 1,2,3,4,5,6,7,8
G can be 1,2,3,4,5,6,7,8

>2G+A = 19
If G max is 8, then 2G max is 16, thus A cannot be 1 or 2.
A can be 3,4,5,6,7,8,9, 10?
P can be 1,2,3,4,5,6,7,8
G can be 1,2,3,4,5,6,7,8

2G must always be even, therefore A must be odd
A: 3,5,7,9, 11?
P: 1,2,3,4,5,6,7,8
G: 1,2,3,4,5,6,7,8

>A+P=7
Again, assuming no 0, neither the apple or pair can be above 6
A: 3,5
P: 1,2,3,4,5,6
G: 1,2,3,4,5,6,7,8

SECOND PASS
>2G+A=19
Apples have been wittled down to just 2 values, therefore we can test them. We either have:
>2*7 Grapes + 5 Apples = 19
>2*8 Grapes + 3 Apples = 19
Grapes therefore must = 7 or 8:

A: 3,5
P: 1,2,3,4,5,6
G: 7,8

>P+G = 9
If grapes are either 7 or 8, then pineapples must be either 1 or 2:
A: 3,5
P: 1,2
G: 7,8

>A+P = 7
Given apples are 3 or 5, and pineapples are 1 or 2, the only logical conclusion is:
APPLE = 5
PINEAPPLE = 2
Therefore, GRAPE = 7

5+2+7 = 14

>>16714061 (OP)
Whoever keeps making these fucking things needs to be kicked in the nuts really hard.

>>16714061 (OP)
14

>>16714061 (OP)
Assuming 2 grapes equals 2 x grape not grape squared,

The best way to solve it is

3 apples + 3 grapes + 3 pineapples equals 42, divide that by 3 you get 14, so the last equation is 14

You get 3 of each fruit by adding the top equation, adding the second, doubling the third and then adding it.

so 9 + 19 + (2x7) = 42 and that represents the total cost of all 3 of each fruit (and then you divide it by 3)

>>16714061 (OP)
>Count the grapes. There are seven of them. Grapes = 7
>7 + pineapple =9
> pineapple = 9-7= 2
>14 + apple= 19
>apple = 19-14 = 5
>5 + 7 + 2= 14

x+y+z=4.5+/-sqrt(21.25)

>>16714191
Based

>>16714191
How the fuck wouls u realize this rearrange after multiplying by 3 trick quickly?

>>16714542
>Try adding the 3 sums together
>(P+G)+(2G+A)+(A+P)
>3G+2A+2P

>Huh, if I add another (A+P) I'll have exactly 3x the target
Pretty simple, really.

>>16714545
Ah I see. My first thought was to substitute into 4 but that only gave me what P was.

I'll remember your trick. Just sum all the equations that dont add in ?

>>16714061 (OP)
[math] \displaystyle
\left \{ \begin{matrix}
x+y &= 9 & \Leftrightarrow x=9-y \\
2y+z &= 19 & \Leftrightarrow z=19-2y \\
z+x &=7 & \Leftrightarrow 19-2y+9-y=7 \\
\end{matrix} \right.
\\
3y=19+9-7=21 \Leftrightarrow y=7 \Rightarrow x=2,z=5
[/math]

>>16714582
You failed

>>16714584
ok einstein

>>16714061 (OP)
Well, linear algebra. You could try using the determinant instead, but good ol' Gaussian seems to be efficient to me.
[math]
\left[
\begin{array}{ccc|c}
1 & 1 & 0 & 9 \\
0 & 2 & 1 & 19 \\
1 & 0 & 1 & 7
\end{array}
\right]\approx
\left[
\begin{array}{ccc|c}
1 & 0 & 0 & 2 \\
0 & 1 & 0 & 7 \\
0 & 0 & 1 & 5
\end{array}
\right]\implies 2 + 7 + 5 = 14[/math]

>>16715124
what approximations did you make to get the exact answer

>>16715132
Kek I used approx instead of sim whoops

>picrel
Keep it simple

Imqgine writing down so much stuff to count this, lmao. Smart people just move stuff in their heads and get an answer.

>>16715264
This is what I did. Anons wrote down responses to show how they did it in their minds.

>>16714491
I like this troll answer.

I did same >>16714221:
P is constant in both (1) and (2).
Therefore G = A + 2.
Replace A in (2) with G. 3G = 21=> G=7
=> A = 5
(3) 7-5=2=P

Sum = 14


Using total sums is a lot harder in versions of these puzzles without numbers.

1st line + 2nd line - 3rd line and you'll get it from there.

>>16714061 (OP)
I didn't notice the 2 grapes in the second equation, either way it is:
Pineapple = 2
Grape = 7
Apple = 5

2+7=9
2×7+5=19
5+2=7

And
5+7+2=14

>use algebra
>get wrong answer
>brute force it
>get right answer
explain this

>>16714061 (OP)

>>16714061 (OP)
(1) P + G = 9
(2) 2G + A = 19
(3) A + P = 7
(4) A + G + P = x

(5) A = x - G - P (rearrange (4) for A)

(6) x + G - P = 19 (substitute (5) into (2) to eliminate A)
(7) x - G = 7 (substitute (5) into (3) to eliminate A)

(8) x + 2G = 28 (add (1) and (6) to eliminate P)
(9) 3x = 42 (add (8) and 2*(7) to eliminate G)

(10) x = 14

>>16714061 (OP)

( (2) + (1) - (3) )/3 = G = 7
(3) + 7 = 14

4 steps is the best I can do.

>>16714061 (OP)
Once times the first equation +
Twice times the third equation +
Once twice the second equation.
Divide result by 3. Done.

>>16716469
*Once times the second equation.

>>16714061 (OP)
[eqn]\pmatrix{1,1,0\\ 0,2,1\\ 1,0,1}\pmatrix{a\\b\\c} = \pmatrix{9\\19\\7}[/eqn]
[eqn]\pmatrix{a\\b\\c} = \pmatrix{1,1,0\\ 0,2,1\\ 1,0,1}^{-1}\pmatrix{9\\19\\7} = \pmatrix{\frac{2}{3}, -\frac{1}{3}, \frac{1}{3}\\ \frac{1}{3},\frac{1}{3},-\frac{1}{3}\\ -\frac{2}{3},\frac{1}{3},\frac{2}{3}}\pmatrix{9\\19\\7} = \pmatrix{2\\7\\5}[/eqn]

apple + grape + pinapple = 5 + 2 + 7 = 14

After 35 replies I think guess and check wins.

>>16714061 (OP)
What do you mean by "too many"? How many steps do you think it should take, and why?

fruit?

If grape, apple, pineapple = the total number of fruit in the universe, than Pi = G grape, + A apple, + P pineapple.

Pi = G.A.P.

Therefore the ratio of 1 grape to an apple+pineapple = 1/3 Pi G.A.P.

Which as you know, when we need to use the number Pi we just use 1, so 1 = Pi G.A.P. which is symbolized by the double Dryphus symbol ^^ in point collision Matrix math.

>>16714061 (OP)
2+7=9
2x7+5=19
5+2=7
5+7+2=14

>>16714061 (OP)
It all adds up to 35. If you remove all the apples & pineapples, you subtract 14, because an apple plus a pineapple equals 7, and there are two of each. Then you're left with three grapes equals 21, so each grape must be 7.