Question

解一元一次不等式组:当a是什么整数是 方程组{x+y=a 5x+3y=31}的解为正数

Answer 1

x + y = a ~ (1)
5x + 3y = 31 ~ (2)
联立两式,
(2) - (1) * 3 : 2x = 31 - 3a
x = (31 - 3a) / 2
得 y = (5a - 31)/2

要令方程组的解为正数,
即x > 0, y > 0
得(31 - 3a) / 2 > 0, (5a - 31) / 2 > 0
化简为 a < 31/3 且 a > 31/5
∴ 31/5 < a < 31/3
且a为整数, ∴a = 7, 8, 9.

Answer 2

解：{x+y=a
{5x+3y=31
x=(31-3a)/2 y=(5a-31)/2
(31-3a)/2>0 a<31/3
(5a-31)/2>0 a>31/5
所以31/5<a<31/3