Math problem;
"Jane made one out of two free throws in the first half and one out of three in the second. So, she made two out of five in the game."
so, 1/2+1/3=2/5?
Hello, turkmath!
Jane made $\displaystyle \tfrac{1}{2}$ of her free throws in the first half
and $\displaystyle \tfrac{1}{3}$ in the second half.
So, she made two out of five in the game.
so: $\displaystyle \frac{1}{2} + \frac{1}{3} \:=\:\frac{2}{5}$ ? . . . . . You're kidding, right?
Hello, turkmath!
No, it is not kidding!
Too bad . . . It's gruesomely incorrect.
By your reasoning, she will always average $\displaystyle \tfrac{2}{5} \,=\,40\%$
First of all: .$\displaystyle \frac{1}{2} + \frac{1}{3} \;=\;\frac{5}{6}$ . . . we can't add fraction like you did.
Secondly, it depends on how many attempts were made in each half.
Suppose in the first half, she made 5 out of 10 . . . That's $\displaystyle \tfrac{1}{2}$
And in the second half, she made 10 out of 30 . . . That's $\displaystyle \tfrac{1}{3}$
Then she made 15 out of 40 during the whole game . . . That's $\displaystyle \frac{15}{40} \,=\,\frac{3}{8} \:=\:37\tfrac{1}{2}\%$
Suppose in the first half, she made 1 out of 2 . . . That's $\displaystyle \tfrac{1}{2}$
And in the second half, she made 16 out of 48 . . . That's $\displaystyle \tfrac{1}{3}$
Then she made 17 out of 50 during the whole game . . . That's $\displaystyle \frac{17}{50}\:=\:34\%$
Is -1/3+4/6=3/9 correct?
Yest, it is.
However, it is correct due to the common rules of the addition - and for no other reason.
As Soroban noted, -1/3+4/6 = -1/3 + 6/9 which, acording to your rule would be 5/12 quite different from 3/9.
Sometimes things come out correct for wrong reasons. This does not make wrong reasons right.
For example,
0.5 + 0.2*0.3 = (0.5 + 0.2)(0.5 + 0.3), or
3^(2/3) * 9^(7/6) = (3*9)^(9/9) = 27.
(The examples are from E. J. Barbeau, Mathematical Fallacies, Flaws, and Flimflam.)