Question

The no of ordered pairs (x,y) satisfying 3x+2y=27 is a, then find the value of a,if x and y are non negetive integers

Answer 1

Given that ordered pairs $(x,y)$ satisfy the equation:

$3x+2y=27$

Also, x and y are non negative integers i.e. values of x and y can not be negative or in fractions.

Let us put values of x and then let us find the value of y:

1. Putting x = 0:

$\Rightarrow 3\times 0+2y=27\\\Rightarrow y = \dfrac{27}{2}$

Hence, this pair not possible because y is in fractions.

2. Putting x = 1:

$\Rightarrow 3\times 1+2y=27\\\Rightarrow 2y = 27-3\\\Rightarrow y = 12$

Solution set: {(1,12)}

3.  Putting x = 2:

$\Rightarrow 3\times 2+2y=27\\\Rightarrow 2y = 27-6\\\Rightarrow y = \dfrac{21}{2}$

y can not be in fractions.

Solution set: {(1,12)}

4. Putting x = 3:

$\Rightarrow 3\times 3+2y=27\\\Rightarrow 2y = 27-9\\\Rightarrow y = 9$

Solution set: {(1,12), (3,9)}

5. Putting x = 4:

$\Rightarrow 3\times 4+2y=27\\\Rightarrow 2y = 27-12\\\Rightarrow y = \dfrac{15}{2}$

y in fractions is not possible.

Solution set: {(1,12), (3,9)}

6. Putting x = 5:

$\Rightarrow 3\times 5+2y=27\\\Rightarrow 2y = 27-15\\\Rightarrow y = 6$

Solution set: {(1,12), (3,9), (5,6)}

7. Putting x = 6:

$\Rightarrow 3\times 6+2y=27\\\Rightarrow 2y = 27-18\\\Rightarrow y = \dfrac{9}{2}$

y in fractions is not possible.

Solution set: {(1,12), (3,9), (5,6)}

8. Putting x = 7:

$\Rightarrow 3\times 7+2y=27\\\Rightarrow 2y = 27-21\\\Rightarrow y = 3$

Solution set: {(1,12), (3,9), (5,6),(7,3)}

9. Putting x = 8:

$\Rightarrow 3\times 8+2y=27\\\Rightarrow 2y = 27-24\\\Rightarrow y = \dfrac{3}{2}$

y in fractions is not possible.

Solution set: {(1,12), (3,9), (5,6),(7,3)}

10. Putting x = 9:

$\Rightarrow 3\times 9+2y=27\\\Rightarrow 2y = 27-27\\\Rightarrow y = 0$

Solution set: {(1,12), (3,9), (5,6),(7,3),(9,0)}

So, number of ordered pairs are 5.

Hence, a = 5