consecutive integers help

**BeBeMala** · December 2nd 2009, 04:28 PM

find three consecutive integers such that the sum of the first and the third is 80?

the difference of two numbers is 46.the larger of the two numbers is 3 times the smaller. find the value of the two numbers.

**skeeter** · December 2nd 2009, 04:53 PM

Originally Posted by BeBeMala

find three consecutive integers such that the sum of the first and the third is 80?

let x = the first integer ...

x + (x+2) = 80

the difference of two numbers is 46.the larger of the two numbers is 3 times the smaller. find the value of the two numbers.

x - y = 46

x = 3y

...

**Bacterius** · December 2nd 2009, 04:57 PM

Say your two consecutive numbers are :

$x$ , $x + 1$ and $x + 2$

The sum of the first and the third is $80$ , so $x + x + 2 = 80$ , which simplifies to $2x + 2 = 80$ , or $2x = 78$ , and finally $x = 39$ . Substitute back this value into your consecutive integers. Your consecutive integers are :

$39$ , $39 + 1$ and $39 + 2$ , that is, your integers are $39$ , $40$ and $41$ .

--

Say the two numbers are $x$ and $y$ , such as $x > y$ .

- Their difference is $46$ , so $x - y = 46$ .

- The larger of the two numbers ( $x$ ) is $3$ times the smaller ( $y$ ), so $x = 3y$ .

You have $x - y = 46$ , and $y = 3x$ . Substitute the second into the first :

$(3y) - y = 46$

Which simplifies to :

$2y = 46$

Thus $y = 23$ .

Plug this back into the first equation to find $x$ :

$x - (23) = 46$ , so $x = 46 + 23 = 69$ .

Your two numbers are $23$ and $69$ .

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