Let the integers be n and n + 1.n + 3(n+ 1) = 47
4n + 3 = 47
4n = 44
n = 11
n+ 1 = 12.
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6x + 5 = 53 6x = 48 x = 8 8 + 45 = 53 8 and 9
this is a algebraic word problem: we know we have two consecutive integers so x + 1 = y and we know that the smaller one (x) added to three times the larger, the results is 23 so x + 3y = 23 substitute (x+1) for y: x + 3(x+1) = 23 4x + 3 = 23 4x = 20 x = 5 y = 6
Let the integers by 'm' & 'm+1' m +3(m+1) = 23 m + 3m + 3 = 23 4m = 20 m = 5 Hence m+1 = 6
Let x equal the smaller integer. Three times the larger is 3(x + 1) or 3x + 3. Added all together, 4x + 3 = 31. Solve for x. 4x = 28 x = 7 Check it. 7 + 24 = 31 It checks.
The sum of any three consecutive odd integers is going to give an odd result. It is impossible for the sum of an odd number of odd integers to equal an even number.