De: 10x+ 15y=300。

8x+ 18y=300

The solution is x= 15.

y= 10

A: A model car is 6.5438+0.5 million yuan each, and B model car is 6.5438+0.5 million yuan each.

(2) Buy X A-class cars and 30 B-class cars. ..

de: 15x+ 10(30-x)≤400。

0.8x+0.5(30-x)≥20.4

Solution: 18≤x≤20

∴x= 18、 19、20

∴ When x= 18, (30-x)=30- 18= 12 (vehicle)

At this time, the profit is18× 0.8+12× 0.5 = 20.4 (ten thousand yuan).

When x= 19, (30-x) = 30-19 =11(vehicle)

At this time, the profit is19× 0.8+1× 0.5 = 20.7 (ten thousand yuan).

When x=20, (30-x)=30-20= 10 (vehicle)

At this time, the profit is 20×0.8+ 10×0.5=2 1 (ten thousand yuan).

A: * * * There are three car purchase schemes, and the profits are 204,000 yuan, 207,000 yuan, 265,438 yuan +0.000 yuan respectively.

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