Each adult pass is $15 and each senior pass is $10. How can you write an equation involving $a$ and $s$ that equals the total money collected, 530?

Once you have the two equations, decide whether substitution or elimination will be faster. Can you multiply one equation so that adding or subtracting the equations cancels one variable?

After solving, plug your values for $a$ and $s$ back into both equations to make sure they satisfy both the total number of passes and the total money collected.

In Desmos, use $x$ for adult passes and $y$ for senior passes. Type the equation x + y = 40.

Type the second equation 15x + 10y = 530. You should now see two lines on the graph.

Tap or click on the point where the two lines intersect. The coordinates will appear as $(x, y)$. The $y$-value of this point is the number of senior passes sold.

Let $a$ be the number of adult passes and $s$ be the number of senior passes.

From the problem:

• Total passes: $a + s = 40$
• Total money: $15a + 10s = 530$

So we have a system of two equations in two variables.

We want to eliminate either $a$ or $s$. Eliminate $s$ by making the $s$ terms cancel.

Multiply the first equation by $-10$:

Now add this to the second equation:

which simplifies to

From $5a = 130$, divide both sides by $5$:

So there were $26$ adult passes sold.

Use $a + s = 40$ and $a = 26$:

Subtract $26$ from both sides:

So the number of senior passes sold was 14.