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Prove that 3<a+b+c<5, for 5a+4b+3c=60?

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majaarmour | Student, Grade 11 | eNotes Newbie

Posted November 23, 2010 at 5:57 PM via web

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Prove that 3<a+b+c<5, for 5a+4b+3c=60?

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giorgiana1976 | College Teacher | Valedictorian

Posted November 23, 2010 at 5:59 PM (Answer #1)

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If a,b,c are natural numbers, we have:

3a<5a (1)

3b<4y (2)

3c=3c (3)

We'll add (1) + (2) + (3):

3a+3b+3c<5a+4b+3c=60

We'll take the inequality 3a+3b+3c<60 and we'll divide it by 3:

a+b+c < 20

We'll do the same way:

5a=5a (4)

4b<5b (5)

3c<5c (6)

We'll add (4) + (5) + (6):

5a+4b+3c=60<5a+5b+5c

Well divide by 5:

12<a+b+c

But a+b+c<20

12<a+b+c<20

We'll divide by 4 the double inequality:

3<a+b+c<5 q.e.d.

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krishna-agrawala | College Teacher | Valedictorian

Posted November 23, 2010 at 7:50 PM (Answer #2)

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It is not true that the double inequality:

3<(a + b + c)<5

necessarily holds true, when:

5a + 4b + 3c = 60.

We can find many different set of values for a, b and c for which the conditions given in the problem are not applicable. For example:

When a = 9, b = 3 and c = 1

5a + 4b + 3c = 5*9 + 4*3 + 3*1 = 45 + 12 + 3 = 60

But

a + b + c = 9 + 3 + 1 = 13

Thus the given double inequality does not apply.

As a matter of fact the correct applicable limits for (a + b + c) that satisfy the condition:

5a + 4b + 3c = 60

has been correctly identified in the answer posted above as double inequality:

12<(a + b + c)<20

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rbeaker91 | Student, Undergraduate | eNoter

Posted November 23, 2010 at 7:39 PM (Answer #3)

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12<a+b+c<20

We'll divide by 4 the double inequality:

3<a+b+c<5 q.e.d.

Doesn't this actually have to be 3 < (a+b+c)/4 < 5.

How can we say that if a+b+c < 20, it is also less than 5.

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neela | High School Teacher | Valedictorian

Posted November 23, 2010 at 10:02 PM (Answer #4)

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To prove that  3 <a+b+c< 5, for 5a+4b+3c = 60.

First let us verify whether the inequality holds  a = 15, b= 0 and c = -5:

5a+4b+3c = 5*15+4*b*0+ 3(-5) = 75 - 15 = 60.

But a+b+c = 15-5 = 10 ,  which does not satisfy the inequality 3 <a+b+c< 5.

So the given inequality 3 < a+b+c < 5  is disproved dispite 5a+4b+3c = 60 for some a = 15 , b= 0 and c = -5.

Now the condtion must be something like  a>=0, b >= 0 and c > =0 , whether the given inequality holds ?

5a+4b+3c = 60

We can write this as :

5a+(5b-b)+(5c-2c) = 60. Or

5(a+b+c)- (b+2c)  = 60.

We add 2b+c to both sides:

5(a+b+c) = 60+(b+2c) = 60 + a positive quantity, as b>0 and c > 0.

Therefore 5(a+b+c) > 60.

So a+b+c > 60/5 = 12....(1).

Similarly , we can write 5a+4b+3c = 60 as below:

(3a+2a) +(3b+b)+3c = 60 = 60.

 3(a+b+c) +(2a+b) = 60.

We subtract (2a+b) from both sides:

3(a+b+c) = 60 - (2a+b)

3(a+b+c) = 60 - positive quantity, as a >0 and b > 0.

3(a+b+c) < 60.

(a+b+c) < = 60/3 = 20.

Therefore a+b+c <= 20.......(2).

Combining the cases at (1) and(2), we get:

12 < = a+b+c  < =20 , if 5a+4b+3c = 60 for non negative a , b , and c.

Hope this helps.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted November 23, 2010 at 8:12 PM (Answer #5)

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Actually, just because, 5a + 4b + 3c = 60, we cannot say that a + b + c < 5.

We could only say that a + b +c >3 as for no value of a, b and c where 5a + 4b + 3c = 60 is a + b + c not greater than 3.

But to say that a + b + c < 5 for all values of a, b and c that satisfy 5a + 4b + 3c = 60 is incorrect. As pointed out, one example is a= 10, b= 1 and c = 2, where 5a + 4b +3c = 60, but a + b + c = 13 which is not less than 5.

Another example is a= 8, b= 2 and c= 4, where 5a + 4b + 3c = 60 but a + b + c = 14 which is greater than 5.

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