Question

Find two consecutive integers such that one fifth of the first less one-ninth of the second is 3. Pls answer.​

Answer 1

Answer:

35 and 36

Step-by-step explanation:

To solve this problem, we must interpret it line by line;

• We are dealing with consecutive integers; these are integers with a difference of 1 between them.

   let the first number = x

               second number  = x + 1

• one - fifth of first;

             $\frac{1}{5}$x

•   one - ninth of the second;

            $\frac{1}{9}$(x + 1)

for the "less" ;

         $\frac{1}{5}$x -  $\frac{1}{9}$(x + 1) = 3

Multiply through by 45

       45($\frac{1}{5}$x) - (45) $\frac{1}{9}$(x + 1) = 3 x 45

           9x - 5(x + 1) = 135

           9x - 5x -5 = 135

             4x = 135 + 5

              4x = 140

                x = 35

Now, x+1  = 35 + 1 = 36

The consecutive integers are 35 and 36