一家糕点店每天烘焙 20 个香蕉蛋糕、50 个巧克力蛋糕和 30 个榴莲蛋糕。1 个香蕉蛋糕、1 个巧克力蛋糕和 1 个榴莲蛋糕的总价为 $29.50。在某一天下午 7 点，该糕点店通过销售蛋糕已收入 $730，此时还剩下一半的香蕉蛋糕、十分之一的巧克力蛋糕和三分之一的榴莲蛋糕。为了尽可能多地售出蛋糕，从下午 7 点开始，所有蛋糕在各自售价的基础上降价 40% 销售。到歇业时，所有蛋糕全部售出，整天的总营业额为 $880。求每种蛋糕折前的售价。

1. Define variables and formulate an equation for the total price of one of each cake.

Let $x$, $y$, and $z$ be the selling price of a banana cake, chocolate cake, and durian cake respectively.
$x + y + z = 29.50 \quad \text{--- (1)}$

2. Calculate the number of cakes sold by 7 pm and formulate an equation for the revenue collected.

Number of cakes left at 7 pm:
Banana: $\frac{1}{2} \times 20 = 10$
Chocolate: $\frac{1}{10} \times 50 = 5$
Durian: $\frac{1}{3} \times 30 = 10$

Number of cakes sold by 7 pm:
Banana: $20 - 10 = 10$
Chocolate: $50 - 5 = 45$
Durian: $30 - 10 = 20$

$10x + 45y + 20z = 730 \quad \text{--- (2)}$

3. Calculate the number of cakes sold after 7 pm, apply the 40% discount, and formulate an equation for the remaining revenue.

Revenue collected after 7 pm $= \char36{}880 - \char36{}730 = \char36{}150$.
The remaining cakes are sold at a $40\%$ discount, which means they are sold at $60\%$ ($0.6$) of their original price.

$10(0.6x) + 5(0.6y) + 10(0.6z) = 150$
$6x + 3y + 6z = 150$
Dividing by 3:
$2x + y + 2z = 50 \quad \text{--- (3)}$

4. Solve the system of three linear equations algebraically to find the original prices.

From (3), we can express $y$ in terms of $x$ and $z$:
$2(x + z) + y = 50$
From (1), we know $x + z = 29.5 - y$.
Substitute this into the expression from (3):
$2(29.5 - y) + y = 50$
$59 - 2y + y = 50$
$y = 9$

Substitute $y = 9$ into (1) and (2):
$x + z = 29.5 - 9 \implies x + z = 20.5 \quad \text{--- (4)}$
$10x + 45(9) + 20z = 730$
$10x + 405 + 20z = 730$
$10x + 20z = 325$
Divide by 10:
$x + 2z = 32.5 \quad \text{--- (5)}$

Subtract (4) from (5):
$(x + 2z) - (x + z) = 32.5 - 20.5$
$z = 12$

Substitute $z = 12$ into (4):
$x + 12 = 20.5$
$x = 8.5$