Q: Find 3 consecutive integers such that twice the smallest is 12 more than the largest?

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5872345098234783904672083946728390752430689723409687298290843 theres your answer

The answer would be 10 12 and 14... 14 x 3 = 42 and 2(10 + 12) = 44. So the product of the largest integer and three is two less than twice the sum of the lower integers.

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.

27, 29

integers are x and x + 2, so x + 2 = 2x - 10 ie x = 12. Integers are 12 and 14

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5872345098234783904672083946728390752430689723409687298290843 theres your answer

-- Call the three consecutive integers (x-1), x, and (x+1).-- Twice the smallest is 2(x-1) or (2x-2).-- 12 more than the largest is (x+1)+12 or (x+13).-- These are equal, so2x - 2 = x + 13Subtract 'x' from each side of the equation:x - 2 = 13Add 2 to each side:x = 15 .-- The smallest of the three numbers is (x-1) = 14.-- The three consecutive integers are 14, 15, and 16.-- Twice the smallest is 28.-- 28 is 12 more than 16.QED, man, QED !

three consecutives numbers: a = smallest a+2 a+4 14 less = -14 than twice the smallest = 2a so... a+a+2+a+4=-14+2a 3a+6=-14+2a 3a-2a=-14-6 a=-20 answer: smallest = -20 greatest = -16

98

This equation works out to 2x = x + 11. This is because the three consecutive odd integers can be represented by x, x + 2, and x + 4. From the equation 2x - x = 11, so x = 11. Checking, 2 times 11 = 22 which is 7 more than 15. The three integers are 11, 13 and 15.

7, 8, 9 Let x be the smallest of the three integers; thus, the integers are x, x+1, x+2. From the problem, we get: 2x=(x+2)+5=x+7 2x-x=7 x=7

Suppose the middle integer is 2a. Then the smallest is 2a-2 and the biggest is 2a+2. 4 times the smallest is 8a-8 So largest subtracted from the smallest is (8a-8) - (2a+2) = 6a-10 So, 6a-10 = 2*2a = 4a so that 2a = 10 So the integers are 8, 10 and 12.

The answer would be 10 12 and 14... 14 x 3 = 42 and 2(10 + 12) = 44. So the product of the largest integer and three is two less than twice the sum of the lower integers.

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.

Suppose the smallest integer is A. The next two even numbers are A+2 and A+4. Using the information supplied we can form an equation: 2A - 14 = A + A+2 + A+4 Rearranging: 2A - 14 = 3A + 6 -20 = A So the three integers are -20, -18 and -16.

Let the smallest integer be x. Since the consecutive even integers differ by 2, we havex + 2(x + 2) = (x + 4) + 203x + 4 = x + 24 (subtract x and 4 from both sides)3x - x - 4 + 4 = x - x - 4 + 242x = 20 (divide by 2 to both sides)x = 10Thus, the three consecutive even integers are 10, 12, and 14.

Since we know that the integers are even and consecutive, we can call them x, x+2, x+4, and x+6, with x being the smallest of the four. twice the sum of the second and third can be written as 2(x+2+x+4)=4x+12 the sum of the first and fourth increased by 14 is x+x+6+14=2x+20 Then we can solve 4x+12=2x+20-->2x=8-->x=4 4