The second equation is just $2$ times the first equation:

So both equations represent the same line. When two equations in a system represent the same line, every point on that line satisfies both equations. That means the system has infinitely many solutions.

Let $s$ be the number of student tickets sold on Friday and $a$ be the number of adult tickets sold on Friday. Then

and

Substitute $s=180-a$ into the revenue equation:

So on Friday, $60$ adult tickets and $120$ student tickets were sold.

On Saturday, the number of adult tickets was $15$ more, so adult tickets sold was

Since the total number of tickets was still $180$, student tickets sold was

Saturday's revenue was

So the school collected \1{,}890$onSaturday.$

Let $s$ be the number of student tickets sold on Friday and $a$ be the number of adult tickets sold on Friday. Then

and

Substitute $s=180-a$ into the revenue equation:

So $s=180-60=120$. On Saturday, the school sold $120+20=140$ student tickets and $60-10=50$ adult tickets. The total money collected was

So the school collected \1{,}340$onSaturday.$

Since both equations equal $y$, set them equal to each other: $2x+1=7$. Then $2x=6$, so $x=3$.

The intersection point of a system is the point that makes both equations true. Substitute $(3,2)$ into each pair. For $y=x-1$, $2=3-1$, which is true. For $y=-x+5$, $2=-3+5$, which is also true. So this system could represent two lines that intersect at $(3,2)$.