读懂题。不是一共花了9100，是有9100可以花，就是可以实际的比这个少。。。汗。

1）通过电脑和打印机的倍数关系确定二者大小关系，电脑C最多是打印机P的3倍，也就是C ≤ 3P
The number of computers can be at most 3 times the number of printers
Let C = the number of computers purchased
Let P = the number of printers purchased
So, we can write: C ≤ 3P
Subtract 3P from both sides of the inequality to get: C - 3P ≤ 0

2）买电脑和打印机一起的钱z最多不超过是$9100
Computers cost $1,500 each, and printers cost $300 each.
So, the total cost of purchasing C computers and P printers = 1500C + 300P

What is the greatest number of computers can the company buy if it has a total of $9,100 to spend on computers and printers?
This tells us that the total cost cannot exceed $9100
So, we can write, 1500C + 300P ≤ 9100

3）联立不等式，求得C的范围
We now have the following system of inequalities:
C - 3P ≤ 0
1500C + 300P ≤ 9100

Take the top inequality and multiply both sides by 100 to get:
100C - 300P ≤ 0
1500C + 300P ≤ 9100

Since the inequality symbols are facing the same direction, we can add the two inequalities to get: 1600C ≤ 9100
Divide both sides of the inequality by 1600 to get: C ≤ 9100/1600
Simplify to get: C ≤ 5.something

C最大是5

不是完全按照3：1买的。。剩下那么多钱还能继续买

9100-4800=4300，剩下的4300够买一个打印机和2个电脑。。。

电脑最多买到尹书记的3倍，假设尹书记买b个，电脑最多买3b个，电脑1500一个，尹书记300一个，现在一共有9100元，问电脑最买几个？（不是一共华妃9100元的定值，别理解错。。。）
1500*3b+300*b=9100

300x+1500y=9100,y

1c [1500] + 1p [300] = 1800
2c [3000] + 1p [300] = 3300
3c [4500] + 1p [300] = 4800
4c [6000] + 2p [600] = 6600
5c [7500] + 2p [600] = 8300
6c [9000] + 2p [600] = 9600 > 9100