There are $5$ girls and $5$ boys in a chess club. The club holds a round-robin tournament in which every player plays against ev
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Answer:
13 and 14.
Step-by-step explanation:
So we have two consecutive integers.
Let's call the first integer a.
Since the integers are consecutive, the other integer must be (a+1) (one more than the last one).
We know that the sum of the greatest integer (or a+1) and twice the lesser integer (a) is 40. Therefore, we can write the following equation:
The first term represents the greatest integer. The second term represents 2 times the lesser integer. And together, they equal 40.
Solve for a. Combine like terms:
Subtract 1 from both sides. The 1s on the left cancel:
Divide both sides by 3:
Therefore, a or the first integer is 13.
And the second integer is 14.
And we can check:
14+2(13)=14+26=40