Translating a Word Problem into an Algebraic Equation Find two consecutive odd integers such that three times the smaller one exceeds two times the larger one by $7$.
 A: 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 $x$?
A: "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. 
A: Three times the smaller one exceeds two times the larger one by 7.
$\text{Three times the smaller one} = \text{(two times the larger one)} + 7$.
$3\times\text{the smaller one} = 2\times\text{(the larger one)} + 7$.
$3\times\text{the smaller one} = 2\times(\text{the smaller one}+2) + 7$.
$3x = 2(x+2) + 7$
$\dots$
