Algebraic Aproach For this word problem How can the followin question be solved algebraically?
A certain dealership has a total of 100 vehicles consisting of cars and trucks. 1/2 of the cars are used and 1/3 of the trucks are used. If there are 42 used vehicles used altogether, how many trucks are there?

$
\newcommand{\xu}{x^{\text{used}}}
\newcommand{\yu}{y^{\text{used}}}
\newcommand{\xn}{x^{\text{new}}}
\newcommand{\yn}{y^{\text{new}}}
$
The key in converting text problems into algebraic expressions is to  write an expression for every sentence or phrase which contains quantifiable information.

For example, consider your problem

A certain dealership has a total of $100$ vehicles consisting of cars and trucks. 
  $1/2$ of the cars are used and $1/3$ of the trucks are used. 
  If there are $42$ used vehicles used altogether, how many trucks are there?

Let us disassemble it into set of statements working with on at a time:


*

*
A certain dealership has a total of $100$ vehicles consisting of cars and trucks.

Let  $x, y$ be the total number of cars and trucks respectively. 
Then we write the first equation
$$
x + y = 100
$$

*
$1/2$ of the cars are used and $1/3$ of the trucks are used.    

Note that in this sentence we have two statemnts, so let us deal with them separately.
Denote $\xn$ and $\yn$ the number of new cars and trucks, $\xu$, $\yu$ – number of used cars and trucks, then we can write the second and the third equations


$1/2$ of the cars are used $(\dots)$


$$
  \xu = \frac{1}{2} x 
  $$


$(\dots)$ and $1/3$ of the trucks are used


$$
  \yu = \frac{1}{3} y
  $$

*
If there are $42$ used vehicles used altogether, how many trucks are there?

The last sentence contains one quantitate statement and states the question for problem


If there are $42$ used vehicles used altogether, $( \dots ) $


$$\xu + \yu = 42$$
and states the question for problem


how many trucks are there?


which can be written in our notation as "find $y$""
$$
    y \quad - \quad  ?
    $$

Finally, combining items $1$ to $3$, we write the system of equations, which is the algebraic formulation of the original text problem:

$
\begin{aligned}
\text{Find } y \text{ given } \qquad  \qquad  \qquad  \qquad 
\begin{cases}
x+ y = 100, \\
\xu = \dfrac{1}{2} x, \\
\yu = \dfrac{1}{3} y, \\
\xu + \yu = 42,  
\end{cases}
\end{aligned}
$

Furthermore, we can simplify the system and write 
$$
\begin{cases}
x + y = 100, \\
\dfrac{1}{2} x + \dfrac{1}{3} y = 42,
\end{cases} 
$$
which is easy to solve: 
$$
\begin{cases}
x + y = 100, \\
\dfrac{1}{2} x + \dfrac{1}{3} y = 42,
\end{cases} 
\implies
\begin{cases}
\dfrac{1}{2} x + \dfrac{1}{2}y = 50, \\
\dfrac{1}{2} x + \dfrac{1}{3} y = 42,
\end{cases} 
\implies
\begin{cases}
\left(\dfrac{1}{2} x - \dfrac{1}{2} x  \right) + \left(\dfrac{1}{2} y - \dfrac{1}{3} y  \right) = 50 - 42, \\
\dfrac{1}{2} x + \dfrac{1}{3} y = 42,
\end{cases} 
\implies
\dfrac{1}{6}y = 8 \implies y = 48
$$
Thus, the final answer is 

There are $48$ trucks at the dealership.

