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Kruka [31]
3 years ago
6

Find two consecutive even integers such that twice the smaller is 16 more than the larger

Mathematics
2 answers:
drek231 [11]3 years ago
6 0
<u>Define x:</u>
Let one of the numbers be x.
The other number is x + 2

<u>Construct equation:</u>
<span>twice the smaller is 16 more than the larger
</span>⇒2x = x + 2 - 16

<u>Solve x:</u>
2x = x + 2 - 16
x = -14

<u>Find the two numbers:</u>
Smaller number = x = -14
Larger number = x + 2 = -12

Answer: The two numbers are -14 and -12
jarptica [38.1K]3 years ago
5 0
Let's make an equation. X will be the smaller number while y is the bigger.
x+2=y; since they're consecutive even numbers. We will use that equation to represent y.
Twice the smaller; 2x
16 more than the larger; 16+x+2
2x=16+x+2; let's add like terms.
2x=18+x
Let's subtract x from both sides.
x=18
If the smaller number is 18, the bigger number is 20.

The two numbers are 18 and 20.
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