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Q: Twice the smallest of three consecutive odd integers is seven more than the largest find the integers?

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98

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

The numbers are 14, 16 and 18.

2 [ x + (x+1) ] = 246x + (x+1) = 1232x + 1 = 1232x = 122x = 61x+1 = 62ho hum

The three integers, since they are consecutive, can be listed as a, a+1, and a+2. Twice the first is 2a. Three times the third is 3(a+2) = 3a+6. First make a formula of the information given: 2a+(3a+6)= -24 Next, solve the formula: 5a + 6 = -24 Subtract 6 from each side. 5a = -30 Divide each side by 5. a = -6 The three consecutive numbers are -6, -5, and -4.

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5872345098234783904672083946728390752430689723409687298290843 theres your answer

They are (14, 15, 16).

-- 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

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

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.

(x)(x+2)(x+4)

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.

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.

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.

-10 and -11.

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