Question

Ten times the smallest of three consecutive integers is twenty two more than three times the sum of integers. Find the integers

Answer 1

Answer:

i. let x is the smallest integer

ii. let x+1 is the middle integer

iii. let x+2 is the largest integer

10x= 22+3(x+x+1+x+2)=22+3(3x+3)

10x=22+9x+9

10x=9x+31

x=31

x+1=32

x+3=33

Step-by-step explanation:

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Answer 2

Let the smallest integer be x

Let the next consecutive integers be x+1 and x+2

         10 (x)  =  3 [x+x+1+x+2] +22

(It is written this way because if you say 20 is 15 more than 5, it's written as   20 = 5 + 15 or 20 - 15 = 5)

           

            10x    =  3x +3x+3+3x+6 +22

               10x = 9x + 31

         10x - 9x = 31

                   x  = 31

Therefore the first integer is 31, the second is (31+1) = 32, and the third is (31+2)  = 33

Proof:

If x = 31,

10 (31) = 3 (31+31+1+31+2) +22

310    = 3 (96) + 22

 310   = 288 +22

  310 = 310