How do you find three consecutive even integers whose sum is 48?

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How do you find three consecutive even integers whose sum is 48?

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Answer 1 · 2018-03-23T02:44:18

$1st Integer = 15$ $"1st Integer"=15$
$2nd Integer = 16$ $"2nd Integer"=16$
$3rd Integer = 17$ $"3rd Integer"=17$

Explanation:

Let's use $n$ $n$ to represent an integer (whole number). Since we need three integers, let's define them like this:

$n =$ $color(blue)(n)=$ 1st integer
$n + 1 =$ $color(red)(n+1)=$ 2nd integer
$n + 2 =$ $color(green)(n+2)=$ 3rd integer

We know we can define the second and third integers as $n + 1$ $n+1$ and $n + 2$ $n+2$ due to the problem telling us that the integers are consecutive (in order)

Now we can make our equation since we know what it's going to equal:

$n + n + 1 + n + 2 = 48$ $color(blue)(n)+color(red)(n+1)+color(green)(n+2)=48$

Now that we've set up the equation, we can solve by combining like terms:

$3 n + 3 = 48$ $3n+3=48$

$3 n = 45$ $3n=45$ $Subtract 3 from both sides$ $color(blue)(" ""Subtract " 3 " from both sides")$

$n = 15$ $n=15$ $\frac{45}{3} = 15$ $color(blue)(" "45/3=15)$

Now that we know what $n$ $n$ is, we can plug it back into our original definitions:

$n = 15$ $color(blue)(n)=15$ $1st Integer$ $color(blue)(" 1st Integer")$
$15 + 1 = 16$ $color(red)(15+1)=16$ $2nd Integer$ $color(red)(" 2nd Integer")$
$15 + 2 = 17$ $color(green)(15+2)=17$ $3rd Integer$ $color(green)(" 3rd Integer")$

$15 + 16 + 17 = 48$ $color(blue)(15)+color(red)(16)+color(green)(17)=48$ $True$ $" True"$

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How do you find three consecutive even integers whose sum is 48?

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