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Basile [38]
3 years ago
8

The lesser of two consecutive even integers is 10 more than one-half the greater. Find the integers.

Mathematics
1 answer:
harkovskaia [24]3 years ago
6 0
Hello,

Here's my solution:

Let n = the lesser of the two consecutive even integers; so what will be a good way to represent the greater of the two consecutive even integers? I say n + 2.

Let's write the equation:
n = ((n+2)/2) + 10

I'll multiply both sides of the equation by 2 to eliminate the denominator, getting:
2n = 2[((n+2)/2) + 10]

This reduces to:
2n = n+2 + 20

Which can be further simplified to:
2n = n+22
 
Subtracting n from both sides we get:
n = 22

So n, the lesser consecutive even integer is 22. The greater consecutive even integer is 24.


Let's check this solution by substituting n = 22 into our original equation:

n = ((n+2)/2) + 10

22 = ((22+2)/2) + 10
22 = ((24)/2) + 10
22 = (12) + 10
22 = 22 - it checks!
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