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.
Say your two consecutive numbers are :
$\displaystyle x$, $\displaystyle x + 1$ and $\displaystyle x + 2$
The sum of the first and the third is $\displaystyle 80$, so $\displaystyle x + x + 2 = 80$, which simplifies to $\displaystyle 2x + 2 = 80$, or $\displaystyle 2x = 78$, and finally $\displaystyle x = 39$. Substitute back this value into your consecutive integers. Your consecutive integers are :
$\displaystyle 39$, $\displaystyle 39 + 1$ and $\displaystyle 39 + 2$, that is, your integers are $\displaystyle 39$, $\displaystyle 40$ and $\displaystyle 41$.
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Say the two numbers are $\displaystyle x$ and $\displaystyle y$, such as $\displaystyle x > y$.
- Their difference is $\displaystyle 46$, so $\displaystyle x - y = 46$.
- The larger of the two numbers ($\displaystyle x$) is $\displaystyle 3$ times the smaller ($\displaystyle y$), so $\displaystyle x = 3y$.
You have $\displaystyle x - y = 46$, and $\displaystyle y = 3x$. Substitute the second into the first :
$\displaystyle (3y) - y = 46$
Which simplifies to :
$\displaystyle 2y = 46$
Thus $\displaystyle y = 23$.
Plug this back into the first equation to find $\displaystyle x$ :
$\displaystyle x - (23) = 46$, so $\displaystyle x = 46 + 23 = 69$.
Your two numbers are $\displaystyle 23$ and $\displaystyle 69$.