Translating a Word Problem into an Algebraic Equation

Question

Find two consecutive odd integers such that three times the smaller one exceeds two times the larger one by $7$.

Answer 1

Let $x$ be the least of two consecutive odd numbers. So the greater of the two consecutive odd numbers must be $x+2$.

Now, we want [three times the smaller odd number] to exceed [twice the larger odd number] by $7$. So the difference between $3x$ and $2(x+1)$ should equal seven.

$$3x -2(x+2) = 7 \quad\iff \quad x = 11$$

So we have the least of the odd numbers. What is the next consecutive odd number after $x$?

Answer 2

"Find two consecutive odd integers..."

OK, that's $2k+1, \space2k+3$, where $k\in \mathbb{Z}$.

"...such that three times the smaller one..."

$$3(2k+1)$$

"...exceeds two times the larger one..."

$$ 2(2k+3)$$

..."by 7."

$$3(2k+1)=2(2k+3)+7$$

Solve for $k=5$, and then plug that into $2k+1$ and $2k+3$ to get $11$ and $13$, respectively.

Answer 3

Three times the smaller one exceeds two times the larger one by 7.

$\text{Three times the smaller one} = \text{(two times the larger one)} + 7$.

$3\times\text{the smaller one} = 2\times\text{(the larger one)} + 7$.

$3\times\text{the smaller one} = 2\times(\text{the smaller one}+2) + 7$.

$3x = 2(x+2) + 7$

$\dots$