Question

if a and b are whole numbers then the numbers of ordered pairs (a,b) which satisfy the equation 3a+7b =87 is​

Answer 1

Let's find the constraint for the variable "b":

87 = 3a + 7b;

87 - 3 > 7b;

84 > 7b;

b < 14.

Let's find the constraint for the variable "a":

87 = 3a + 7b;

80 = 3a;

a < 26.

We should consider the following:

3a = 87 - 7b;

87 - 7b | 3;

-b | 3;

b | 3.

Since 7b must be a multiple of 3, therefore 87 // 21 = 4 ⇾ 4 + 1 and that means ...here are five ordered pairs.

3a = 87;

(29, 0).

3a + 21 = 87;

(22, 3).

3a + 42 = 87;

(15, 6).

3a + 63 = 87;

(8, 9).

3a + 84 = 87;

(1, 12).

{(29, 0), (22, 3), (15, 6), (8, 9), (1, 12)}.

Thanks to @amitrw for the correction, my bad, I always be a little incorrect for some reason.

I, from now on, will always consider zero as the possibility (since I forgot that 87 | 3).

Answer 2

Given :- if a and b are whole numbers then the numbers of ordered pairs (a,b) which satisfy the equation 3a+7b =87 is ?

Answer :-

we have,

→ 3a + 7b = 87

→ 3a = (87 - 7b)

→ a = (87 - 7b)/3

Putting possible values of b such that, a will be a whole number are ,

• At b = 0 => 87/3 = 29 = a => A whole number.
• At b = 3 => (87 - 21)/3 = 66/3 = 22 = a => A whole number.
• At b = 6 => (87 - 42)/3 = 45/3 = 15 = a => A whole number.
• At b = 9 => (87 - 63)/3 = 24/3 = 8 = a => A whole number.
• At b = 12 => (87 - 84)/3 = 3/3 = 1 = a => A whole number.

then the numbers of ordered pairs (a,b) which satisfy the equation are :-

• (29, 0)
• (22, 3)
• (15, 6)
• (8, 9)
• (1, 12) .

Learn more :-

let a and b positive integers such that 90 less than a+b less than 99 and 0.9 less than a/b less than 0.91. Find ab/46

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