Question 1201890
Find three consecutive odd integers...<pre>1st odd integer = x
2nd odd integer = x+2
3rd odd integer = x+4</pre>such that four times the sum of the first and second<pre>4[x + (x+2)]

is 

   =</pre>11 more than 3 times the third.<pre>3(x+4) + 11

4[x + (x+2)] = 3(x+4) + 11
4[x + x + 2] = 3x + 12 + 11
   4[2x + 2] = 3x + 23
      8x + 8 = 3x + 23
          5x = 15
           x = 3

1st odd integer = x = 3
2nd odd integer = x+2 = 3+2 = 5
3rd odd integer = x+4 = 3+4 = 7

Checking:</pre>four times the sum of the first and second<pre>4(3 + 5) = 4(8) = 32</pre>is 11 more than 3 times the third.<pre>3(7) + 11 = 21 + 11 = 32

So it checks.

Edwin</pre>