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Q: Find three consecutive even integers such that if the largest integer is subtracted from four times the smallest the result is 8 more than twice the middle integer?

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The integers are 47, 49, 51, and 53. 47 * 53 = 2491

-1

18

One possible answer is -4 and -3.

First Integer = x2nd Integer = x+13rd Integer = x+2(x) + (x+1) + (x+2) = 963x + 3 = 963x = 93x = 31Correct integers are 31, 32, and 33 (total 96)

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The smallest is 121.

consecutive integers

59,61,63,65 It is 59

"Consecutive" integers are integers that have no other integer between them.

For x, which is the largest integer of nconsecutive positive integers of which the smallest is m:x = m + n - 1

The sum of three consecutive integers is -72

If the largest integer is subtracted from four times the smallest, the result is 4 more than twice the middle integer. Let the smallest integer be x, then the others are x + 2 and x+ 4. Therefore 4x - (x + 4) = 2 (x + 2 ) + 4 Expanding, we get 4x -x -4 = 2x + 4 + 4 Gathering terms: x = 12 Thus the three integers are 12, 14 and 16.

1 is the smallest positive integer. But if you include negative integers, there is no smallest.

The smallest six consecutive composite integers are:90, 91, 92, 93, 94, 95.(And 96 is also composite, for a run of seven consecutive.)Is that what you were asking ?

The integers are 47, 49, 51, and 53. 47 * 53 = 2491

The first integer is 17.

the sum of two consecutive integers is -241, what is the larger integer?

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