Find two consecutive odd numbers such that the sum of the larger number and twice the smaller number is 27 less then foir times the smaller number

Step-by-step explanation:

x      - some integer

2x+1      - an odd integer (the smaller one)

2x+1+2=2x+3       - the next odd integer consecutive to 2x+1  (the larger)

2×(2x+1)       - twice the smaller number

2x+3+2•(2x+1)     -the sum of the larger number and twice the smaller number

4•(2x+1)      - four times the smaller number

4•(2x+1)-27     - 27 less then four times the smaller number

2x + 3 + 2•(2x + 1) = 4•(2x + 1) - 27

2x + 3 + 4x + 2 = 8x + 4 - 27

            6x + 5  =  8x - 23

                  -5            -5

              6x  =  8x - 28

             -8x       -8x

            -2x  =  - 28

          ÷(-2)      ÷(-2)

              x  = 14

2x+1 = 2•14 + 1 = 29

2x+3 = 2•14 + 3 = 31

Check:  31+2•29=31+58=89;  4•29-27=116-27=89

41*67/2=1373.5

step-by-step explanation:

i just used a calculator