I'd like to know how you go about solving the problem below. Does a trick exist other than evaluating substituted values until a solution is found?
Furthermore, is there a generic name for such problems? I'll happily change the title of the question once I know if a name exists for such a class of problems.
A car dealership sells used cars for \$7,000 and new cars for \$16,000. If a total of 17 cars were sold for \$191,000, how many of the cars sold were used?
You're right, such problems do have a specific name: simultaneous equations, or more formally, these ones are a system of linear equations.
Let's think about the maths of it with a little bit of algebra. If the number of new cars the dealership sold can be labelled $n$, and the number of used cars $u$, then the statement of the problem can be written as a pair of equations:
$$ (1) $$ $$ n + u = 17 $$
and
$$ 16000n + 7000u = 191000 $$
or
$$ (2) $$ $$ 16n + 7u = 191 $$
by simply dividing both sides by $1000$.
The fact that these equations are both true and both involve the same variables ($n$ and $u$) is what gives the pair the name simultaneous.
Now, for how to go about solving them:
Solving Simultaneous Equations
The key is to realise that both equations are true. Therefore, anything that can be said using one can be assumed to be true in the other (after all, if we have some number of used cars sold, knowing an extra fact about it doesn't change that number!). We use this to our advantage to cast the problem in terms of a single variable.
Equation $(1)$ tells us:
$$ n + u = 17 $$
so, by re-arranging it:
$$ u = 17 - n $$
Now, we know that this is true in both equations, so we can go ahead and substitute $u$ for $17 - x$ wherever it appears in equation $(2)$! From there, it's easy to reduce things down:
$$ 16n + 7u = 191 $$
$$ 16n + 7(17 - n) = 191 $$
$$ 16n + 119 - 7n = 191 $$
$$ 9n = 72 $$
$$ n = 8 $$
Et voila! $n = 8$. We take this and put it back into equation $(1)$ (or our re-arranged form) to get a value for $u$:
$$ u = 17 - n = 9 $$
And there we have it; the dealership sold $8$ new cars and $9$ used cars.
Summary
The overall approach we've got here holds for any such problem - we've got two variables and two known equations in those variables. The key is to re-arrange one to get a representation of one of the variables in terms of the other, and then plug that into the other equation.
In general, you might like to know that any set of $N$ equations involving $N$ variables can be solved by doing the exact same thing over and over again. But even one less equation to give you information, and you'll never manage it!
Historical Note: We can use "algebra." But before algebra, there were techniques available to solve such problems, and more complicated ones. The technique had various names. For example, in China it was called by a name that roughly translates to "too much and not enough." It was a standard method in India, and in the Islamic world. In European languages, a name that translates to "Rule of False Position" was sometimes used. The technique was taught routinely in European schools until the $19$th century.
Let us "guess" that the dealership sold $17$ used cars. Now let's check whether we are right. If they sold $17$ used cars, the income was $(17)(7000)=119000$.
Our guess is not right. We missed the income by $72000$.
If we replace a used car by a new one, the income increases by $9000$. So there must have been $\frac{72000}{9000}$ new cars sold.
Note that we did not use "Guess and Check." From the analysis of one (deliberately wrong) guess, we reasoned our way directly to the answer.
This kind of problem is called a system of linear equations in two variables. In particular it's also a word problem; many similar problems exist, for example here.
The method for solving them is to make variables for the unknown quantities. Let $U$ be the number of used cars sold, and $N$ be the number of new cars sold. Then you get $$ U + N = 17 $$
$$ 7000U + 16000N = 191000 $$
and solving the problem amounts to solving the system of equations.