Apples are worth a=1 and bananas b=2
I end with x2=4 apples and y2=5 bananas
Thus i have a total of p2=9
Worth a total of t2=14
p1 is the total number of fruits (x1 + y1)
t1 is the total value of fruits (equal to t2=14)
we know that initially:
x1 = k(p1) (where k=0.6)
x1 = k(x1 + y1)
(1/k)x1 = x1 + y1
((1/k)-1)x1 - y1 = 0
some other things we know:
x1(a) + y1(b) = t1
x1(1) + y1(2) = 14
so now we just have some sim equations:
x1((1/k)-1) - y1(1) = 0
x1(a) + y1(b) = t1
let's flip the bottom row:
x1((1/k)-1) - y1(1) = 0
-x1(a) - y1(b) = -t1
and devide by b=2
x1((1/k)-1) - y1 = 0
x1(-a/b) - y1 = -t1/b
and join the two:
-x1(a) - y1(b) + 14 = 0
...
x1((1/k)-1) - y1 = x1(-a/b) - y1 + (t1/b)
then we can drop y1:
x1((1/k)-1) = x1(-a/b) + (t1/b)
and reduce:
x1((1/k)-1+(a/b)) = (t1/b)
and again:
x1 = (t1/b) / ((1/k)-1+(a/b))
so:
x1 = (14/2) / ((1/0.6)-1+(1/2))
x1 = 6
and then we know that:
x1(a) + y1(b) = t1
6(1) + y1(2) = 14
y1(2) = 14 - 6(1)
y1 = (14 - 6(1)) / 2
y1 = 4
so now we know:
x1 = 6
y1 = 4