Consecutive Integers Help

[RVCA123]

Find two consecutive even integers such that half of the smaller integer is two more than the larger integer.

How would this be set up? Solved?

 

[soroban]

Hello, RVCA123!

Find two consecutive even integers such that half of the smaller integer is two more than the larger integer.

Let \(\displaystyle x\) = smaller integer.
Let \(\displaystyle x\!+\!1\) = larger integer.

We have: .\(\displaystyle \underbrace{\text{half of smaller}}\:\underbrace{\text{is}}\;\underbrace{ \text {two more than}}\;\underbrace{\text{larger}}\)
. . . . . . . . . . . . \(\displaystyle \dfrac{1}{2}x\) . . . . \(\displaystyle =\) . . . . . \(\displaystyle 2\;+ \) . . . .. \(\displaystyle x+1\)


You're right, lookagain!
I overlooked the word "even" . . . *blush*

 

[lookagain]

find two consecutive > > > > even integers < < < < such that half of the smaller integer is two more than the larger integer.

How would this be set up? Solved?

.

let \(\displaystyle x\) = smaller integer.
Let \(\displaystyle x\!+\!1\) = larger integer.

No, soroban, they are consecutive even integers, not consecutive integers.

rvca123,

let x = the smaller integer

let x + 2 = the larger integer

The phrase "half of the smaller integer" would then mean \(\displaystyle \dfrac{1}{2}x. \)

The word "is" corresponds to the equals sign.

The phrase "two more than the larger integer" can mean "(x + 2) + 2." \(\displaystyle \ \ \ \)*

The equation can be:



\(\displaystyle \dfrac{1}{2}x \ = \ (x + 2) \ + \ 2, \ \ \ or \ \ \ \dfrac{1}{2}x \ = \ 2 \ + \ (x + 2)\)







* \(\displaystyle \ \ Or, \ \ you \ \ could \ \ also \ \ use \ \ \ 2 \ + \ (x + 2) \)


-----------> Continue and solve one of the above equations for x, and then add 2 to that result to get the other number.