Answers

2014-11-20T19:11:54+08:00
1.) Three consecutive odd integers
Let 'x' be the first number
      'x+2' be the second number
     'x+4' be the third number
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-Three times the largest is seven times the smallest
7x = 3(x+4)
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Distribute 3 to the quantity (x+4)
7x = 3x + 12
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Transpose 3x to the left side of the equation
7x - 3x = 12
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Combine like terms
4x = 12
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Divide the whole equation with 4
x = 3
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x = 3
x + 2 = 3 + 2
x + 2 = 5
x + 4 = 3 + 4
x + 4 = 7
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Therefore the three consecutive odd integers are 3, 5, 7.
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2.) Two numbers are consecutive even
Let 'x' be the first number
      'x+2' be the second number
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-If twice the smaller is 18 more than the larger
2x = (x+2) + 18
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Remove parenthesis
2x = x + 2 + 18
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Transpose x to the left side of the equation 
2x - x = 2 + 18
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Combine like terms
x = 20
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x = 20
x + 2 = 20 + 2
x + 2 = 22
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Therefore the two consecutive even numbers are 20, 22.
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2014-11-22T15:59:18+08:00
1. The integers be x(smallest), x + 2 and x + 4(largest) since they are consecutive odd integers

3 times the largest is 7 times the smallest.

3(x + 4) = 7x

Multiply

3x + 12 = 7x

Transpose 6 and 7x

3x - 7x = -12

Subtract

-4x = -12

Divide by -4 both sides

-4x/-4 = -12/-4

Therefore

x = 3 (first odd integer)

Let's now substitute the value of x to determine the other two odd integers

second odd integer

3 + 2 = 5

third odd integer

3 + 4 = 7

Therefore the odd integers are 3, 5 and 7

2. Let the consecutive even be x and x + 2(largest)

Twice the smallest is 18 more than the larger

2x = (x + 2) + 18

Add 2 to 18

2x = x + 20

Transpose x

2x - x = 20

Therefore

x = 20

Let's now substitute the value of x to determine the value of the largest even

x + 2

20 + 2

Therefore

22 is the largest even and 20 is the smallest

I hope it helps

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