The non negative numbers a, b, and c satisfy a b c 30, 3a b c 50, and find the maximum and minimum v

3a+3b+3c=90, 3a+b-c=50, so n=140-a

It is solved by a+b+c=30, 3a+b-c=50.

b=40-2a

c=a-10

Because the non-negative numbers a, b, c

So 40-2a>=0, a<=20, a-10>=0, a>=10, so 120<=n<=130

The maximum and minimum values are 130 and 120, respectively

5x+4y+2z=3(x+y+z)+(3x+y-z)-x=3*30+50-x=140-x.

Take x as a known number to get the equation:

y+z=30-x，--1

y-z=50-3x。--2

Solution: y=40-2x, z=x-10.

Because: x,y,z>=0,40-2x>=0,x-10>=0,x>=0.

10<=x<=20.

5x+4y+2z=140-x, the maximum is 140-10=130. The minimum is 140-20=120.

a+b+c=30, then 2a+2b+2c=60 and because 3a+b-c=50, the two equations are added together, that is: 5a+3b+c=110=n+a+b

A+B=30-C,a,B,C are non-negative, so 30>=C>=0 then 30>=A+B>=0

n=110-(a+b)

So n is a maximum of 110 and a minimum of 80

3a+3b+3c=90, 3a+b-c=50, the two formulas are added to give 6a+4b+2c=140

So s=140-a

b=40-2a is solved by a+b+c=30, 3a+b-c=50

c=a-10

Because the non-negative numbers a, b, c

So 40-2a 0, a 20, a-10 0, a 10, so 10 a 20

20≤-a≤-10

So 120 n 130

The maximum and minimum values of n are 130 and 120, respectively

This is an inequality. First, find the unary expression about m=3a+b-7c and solve the system of equations 3a+2b+c=5（1） 2a+b-3c=1.

2) A-7c=-3(3) b+11c=7.(4) From (1)-(4):

3a+b-10c=-2, i.e. 3a+b-7c=3c-2 So: m=3a+b-7c=3c-2(5) Step 2:

Find C's.

Because a+b+c=30 so b+c=30-a, (1) because 3a+b-c=50 so b-c=50-3a, (2) from (1) and (2) b=40-2a, c=-10+a, and then substituted p 5a+4b+2c to obtain: p manuscript base ridge-a+140 because a, b, c are frontal non-negative, and b=40-2a, c=-10+a, so the minimum value of a is 10, the maximum value is 20, so when a 1....

Solution: Think of c as a constant first.

3a+2b=5-c

2a+b=1+3c

Solve systems of equations. a=7c-3

b=7-11c

And there is, c=(a+3) 7;c = (7-b) 11 Because abc is greater than or equal to 0, the minimum value of c is 3 7; The maximum value of c is 7 11, and AB denoted by c is substituted for m=3a+b-7c

m=3c-2

So the minimum value of m is 9 7-2 = -5 7

The maximum value is 21 11-2 = -1 11

First, find the unary expression about m=3a+b-7c.

Solve systems of equations. 3a+2b+c=5...1）

2a+b-3c=1...2）

Get a-7c=-3....3)

b+11c=7...4)

From (1)-(4):

3a+b-10c=-2, that is, 3a+b-7c=3c-2

So: m=3a+b-7c=3c-2....5) Step 2: Find the range of values of c.

Since a, b, and c are all non-negative numbers, we get from (3): a=7c-3 0

c 3 7 is obtained from (4): b = 7-11c 0

c≤7/11

So 3 7 c 7 11 7 11

Step 3: Discuss.

When c=7 11, the value of substitution (5)m is the largest, which is -1 11 When c=3 7, the value of substitution (5)m is the smallest, which is -5 7

Approximately: 3a+3b+3c=90, 3a+b-c=50, so n=140-a

It is solved by a+b+c=30, 3a+b-c=50.

b=40-2a

c=a-10

Because the non-negative numbers a, b, c

So 40-2a>=0, a<=20, a-10>=0, a>=10, so 120<=n<=130

The maximum and minimum values are 130 and 120, respectively

a+b—c=2

a—b+2c=1 a+b=2+ca-b=1-2c, add 2a=3-ca=(3-c), 2 0c, subtract 2b=1+3cb=(1+3c), 2 c-1 3, so 0 c, 3 s=a+b+c=2+c=2+c=2+2c, when c=0, minimum=2c=3, maximum=8

The synaptic formula yields c = 3-2a and b = 5-3a

s=a+b+c=2+c+2c=2+3c>=2s=a+5-3a

3-2a=8-4a<=8

So a is a maximum of 8 and a minimum of 2

a+b=2+c

a-b=1-2c

The two equations are added together.

2a=3-c

i.e. c = 3-2a

Substituting c=3-2a into a+b-c=2 gives a+b-(3-2a)=2, so b=5-3a

So s=a+b+c=a+5-3a+3-2a=8-4a, because a>=0, so 4a>=0, so when a=0, 8-4a is the maximum.

The value is 8 and the minimum value is - (negative infinity).

a+b=c+2

a-b=1-2c

Solve a=(3-c) 2, b=(3c+1) 2 from a,b,c>=0 to get 0<=c<=3

s=a+b+c=2c+2, so 2<=s<=8 so the maximum value of s is 8 and the minimum value is 2