Question 12229
Let the no. of apples be x, the number of bananas be y and the number of oranges be z.
then x+y+z = 80 (given total of all these fruits)

Z = (1/3)x  + 5   given in question that oranges are 5 more than one-third of apples

y = (1/4)x  - 1   given in question

Now we have three equations and consider the ques solved cause in simple equations we can solve as many varibles as are the number of equations e.g.

here let us substitute the vlue of x & y in original equation.
x + (1/4)x + 5 + (1/3)x - 1  = 80
(19/12)x + 4  =  80
(19/12)x =80-4
x  = 76 x (12/19)
x = 48  thus no. of apples is 48

z = (1/3)x + 5  from earliar eqn. thus z =21 hence oranges are 21

and balance 80 - (48 + 21) = 11 would be bananas
alternatively we can also find bananas by substituting value of x in the eqn
y = (1/4)x - 1

So apples = 48, bananas = 11, oranges =  21