Question 838995
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Hi,
three consecutive integers: {{{highlight_green(x)}}}, {{{highlight_green(x+1)}}}, {{{highlight_green(x+2)}}}
Question states***
4 times the third is 51 less than 3 times the sum of the first and second integers.
 ***4(x+2) = 3[x + (x+1)] -51   | Solving for x
    4x + 8 = 3(2x+1) - 51
    4x + 8 = 6x + 3 - 51
      56 = 2x
      28 = x , the first Integer.  The third integer is {{{highlight( 30)}}}
CHECKING our answer***
{{{4*30 = 3(28+29) - 51}}}
    120 = 120
Wish You the Best in your Studies.