Systems of linear equations word problems

A word problem often gives you two unknowns and two facts that connect them.
That is exactly a system of two equations.
The challenge is translating the words into math.

Start by naming the unknowns with letters.
Suppose a store sells pens for $\$2$ each and notebooks for $\$5$ each.
Let $p$ be the number of pens and $n$ the number of notebooks.

Turn each fact into an equation.
"Someone buys $8$ items in total" becomes $p + n = 8$.
"The total cost is $\$28$" becomes $2p + 5n = 28$.
Now you have a system.

Solve the system with substitution or elimination.
From $p + n = 8$, we get $p = 8 - n$.
Substituting into $2p + 5n = 28$ gives $2(8 - n) + 5n = 28$, so $16 + 3n = 28$, and $n = 4$.
Then $p = 4$.

Finally, answer what was actually asked, with units.
If the question wants the number of notebooks, the answer is $4$ notebooks, not just "$4$".
Always re-read the question so you report the right variable.

A movie theater sells adult tickets for $\$12$ and child tickets for $\$8$. One family buys $5$ tickets for $\$52$. How many adult tickets did they buy?
Let $a$ be adult tickets and $c$ be child tickets.
The two facts give $a + c = 5$ and $12a + 8c = 52$.
From the first, $c = 5 - a$; substituting gives $12a + 8(5 - a) = 52$, so $4a + 40 = 52$ and $a = 3$.
They bought $3$ adult tickets.

A cafeteria prepared $30$ lunches, some vegetarian and some not, and there were $6$ more non-vegetarian lunches than vegetarian ones. How many vegetarian lunches were there?
Let $v$ be vegetarian and $n$ be non-vegetarian lunches.
The facts give $v + n = 30$ and $n = v + 6$.
Substituting gives $v + (v + 6) = 30$, so $2v = 24$ and $v = 12$.
There were $12$ vegetarian lunches.

Two numbers add up to $40$, and one number is $4$ times the other. What are the two numbers?
Let the numbers be $x$ and $y$, with $x + y = 40$ and $y = 4x$.
Substituting gives $x + 4x = 40$, so $5x = 40$ and $x = 8$.
Then $y = 4(8) = 32$, so the numbers are $8$ and $32$.