CPhill

OK....let's look at the cost for one person...

We have the total cost = $63   .....this includes a $13 concert ticket plus the cost of two passes, each of which cost the same.

Call the cost of one of the passes, P.   So we have

63 =  13 + 2P  subtract 13 from both sides

50 = 2P          divide both sides by 2

25 = P 

So the cost of one pass = $25

Thanks, Alan, for that answer....

A little more on this strange graph.....

Apparently, if this can be factored into this form....(x + y) (x - y) = 0, we have a graph of intersecting lines...let's see...

2x^2 - 3y^2 = 5xy 

2x^2 - 5xy - 3y^2 = 0

(2x + y) (x - 3y)  = 0

Since this "reducible" to this form, this is a degenerate conic that will form two intersecting lines.

2x^2-3y^2=5xy and-3x+y=5     rearranging the second equation, we have y = 3x + 5

And putting this into the first equation, we have

2x^2 - 3(3x + 5)^2 = 5x(3x + 5)  simplify

2x^2 - 3(9x^2 + 30x +25) = 15x^2 + 25x

2x^2 -27x^2 - 90x - 75  = 15x^2 + 25x

-13x^2 - 27x^2 - 115x - 75  = 0   multiply through by -1

13x^2 + 27x^2  + 115x + 75 = 0

40x^2 + 115x + 75 = 0      divide through by 5

8x^2 + 23x + 15 = 0    factor

(x + 1 )(8x + 15) = 0

And setting each factor to 0, we have that x = -1 and x = -15/8

And using y = 3x + 5 when x = -1, y = 2  and when x = -15/8, y = -5/8

So....our solutions are (-1, 2) and (-15/8, -5/8)

x-y=3 and xy-5x+y=-13

Using the first equation we can rearrange it as y = x -3    and substituting this in the second, we have

x(x -3) - 5x + (x -3) = -13   simplify

x ^2  - 3x - 5x + x -3 = -13

x^2 - 7x + 10  = 0      factor

(x -5) ( x-2) = 0   and setting each factor to 0, we have that x = 5 and x =2

And using y = x -3, when x = 5, y = 2   and when x = 2 , y = -1

So...our solutions (intersection points) are (5,2) and (2, -1)

2x^2-px+q=0

x=1/2, -2

what is p-q?

We have that

2(1/2)^2 -(1/2)p + q = 0  →  (1/2) - (1/2)p + q = 0       (1) 

And .... 2(-2)^2 - (-2)p + q  = 0     →     8 + 2p  + q  = 0   (2)

And subtracting (1) from (2) we have  15/2 + (5/2)p = 0    →  15 + 5p =0  → p = -3

And using (2), we have 8 - 6 + q = 0   → 2 + q  = 0    →  = -2

So, our function is ..y =  2x^2 -3x - 2

So....p - q  = -3 - (-2)  = -1

y ≤ -2x^2

x^2+y^2=24 and y=2x+3  

Substituting  for y in the first equation, we have

x^2 + (2x + 3)^2  = 24    simplify

x^2 + 4x^2 + 12x + 9 = 24

5x^2 + 12x  - 15  = 0     this won't factor..so using the onsite solver, we have

$${\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{15}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{111}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}\right)}{{\mathtt{5}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{111}}}}{\mathtt{\,-\,}}{\mathtt{6}}\right)}{{\mathtt{5}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{3.307\: \!130\: \!750\: \!570\: \!547\: \!8}}\\
{\mathtt{x}} = {\mathtt{0.907\: \!130\: \!750\: \!570\: \!547\: \!8}}\\
\end{array} \right\}$$

Let's round these to -3.31 and .91

And using y = 2x+ 3,  when x = -3.31, y = -3.614 and when x =.91, y = 4.814

So the solutions are (-3.31, -3.614) and ( .91, 4.814)

Let's try this a slightly different way...

12 x 1/8   =  12/1 x 1/8 =  12/8   = [ 4 x 3 ] / [4 x 2 ]  =  3 / 2

The fractions you're adding just have the denominators of 4, 6 and 8

We only need worry about the 6 and the 8 because 4 is already a factor of 8

So 8 factors as 2^3 and 6 factors as 2 x 3

So...taking the highest power of each different factor, the  common denominator = 2^3 x 3 = 8 x 3 =  24

So we have

3/4 = 18/24

-5/6  = - 20/24   and

-3/8 = -9/24

Now...just add.subtract the fractions on the right side of the equal signs...