The value of A is 3 and the value of B is 2.
They are the numbers represented by positive real numbers where the fractions and decimal numbers are not included.
The question gives:
A and B are whole numbers;Thus, you need to find A and B for these conditions.
When we calculated the Least Common Multiple ( LCM ) between 3 and 11, we can rewrite the equation as:
[tex]\frac{A}{11}+\frac{B}{3} =\frac{31}{33}\\ \\ \frac{3A+11B}{33}=\frac{31}{33}[/tex]
Thus, we have:
[tex]3A+11B=31\\ \\ 3A=31-11B\\ \\ A=\frac{31-11B}{3}[/tex] (1)
From this condition, for that A is a whole number, then 31-11B >0.
Then,
[tex]31-11B > 0\\ \\ -11B > -31\\ \\ 11B < 31\\ \\ B < 2.8[/tex]
In this case, for that B should be a whole number, B can be 0,1 and 2.
Now, you should replace probable numbers for B in equation (1). The value of B will be the number from that you will find the whole number for A. See below.
For B=0, you have [tex]A=\frac{31-11*0}{3}=\frac{31}{3}=10.33[/tex] . In this case, A is not a whole number. Then, B can't be 0 (zero).
For B=1, you have [tex]A=\frac{31-11*1}{3}=\frac{31-11}{3}=\frac{20}{3}=6.67[/tex] . In this case, A is not a whole number. Then, B can't be 1 (one).
For B=2, you have [tex]A=\frac{31-11*2}{3}=\frac{31-22}{3}=\frac{9}{3}=3[/tex] . In this case, A is a whole number (3). Then, B can be 2 (two).
Let's replace this information in the given equation in your question.
[tex]\frac{A}{11}+\frac{B}{3} =\frac{31}{33}\\ \\ \frac{3}{11}+\frac{2}{3} =\frac{31}{33}\\ \\ \frac{9+22}{33}= \frac{31}{33}\\ \\ \frac{31}{33}=\frac{31}{33}[/tex]
Hence, A=3 and B=2.
Read more about the whole number here:
https://brainly.com/question/19896422
