有兩題Algebra，請幫忙回答

1) Suppose he sold x apples and y pears.

The original price of each apple : $ 2 / 5 = $ 0.4

The original price of each pear : $ 12 / 16 = $ 0.75

Consider the case when both are sold at $ 0.5 @,

0.5 x + 0.5 y = 0.4 x + 0.75 y - 1

0.1 x - 0.25 y = -1 --- ( 1 )

Consider the case when both are sold at 3 for $ 2 ( i.e. $ 2 / 3 @ ),

2x / 3 + 2y / 3 = 0.4 x + 0.75 y + 9

2x + 2y = 1.2 x + 2.25y + 27

0.8 x - 0.25y = 27 --- ( 2 )

( 1 ) - ( 2 ) : 0.1 x - 0.8 x = -1 - 27

0.7 x = 28

x = 40

0.8 ( 40 ) - 0.25 y = 27 { put into ( 2 )}

y = 20

So he sold 40 apples and 20 pears.

2) Let the time needed for Q to overtake P be t hours.

The distance travelled by P in 2 hours:

25 x 2 = 50km

Then,

50 + 25 t = 30 t

5t = 50

t = 10

So 10 hours are needed for Q to overtake P.

In 10 hours, Q has travelled:

30 x 10 = 300 km

The time required for R to travel 300 km:

300 / 40 = 7.5 h

Then,

10h - 7.5h = 2.5h

So, R must start 2.5 h after Q so that R can take over P at the same moment.