Multiply the bottom equation by $\displaystyle 5$, then add to the top equation:

$\displaystyle 6x-y = -9$

$\displaystyle 5 \left (6x-y \right ) =5\cdot (-9)$

$\displaystyle 30x-5y = -45$

$\displaystyle \underline{\textrm{\; } 3x + 5y = \; \; 23}$

$\displaystyle 33x \;\;\;\;\;\; \; =-22$

Divide both sides by $\displaystyle 33$

$\displaystyle \frac{33x}{33}=-\frac{22}{33}$

$\displaystyle \frac{22}{33}=-\frac{2}{3}$

When we add the two equations, the $\displaystyle y$ variables cancel leaving us with:

$\displaystyle 8x=-8$

Solving for $\displaystyle x$ we get:

$\displaystyle \frac{8x}{8}=\frac{-8}{8}$

$\displaystyle x=-1$

We can then substitute our value for $\displaystyle x$ into one of the original equations and solve for $\displaystyle y$:

$\displaystyle \begin{matrix} -4-2y=-3\\ -4+4-2y=-3+4\\ -2y=1\\ y=-\frac{1}{2}\\ \end{matrix}$

We are given

$\displaystyle y=-3x+1$

$\displaystyle y=2x-9$

We can solve this by using the substitution method.  Notice that you can plug $\displaystyle -3x+1$ from the first equation into the second equation and then get

$\displaystyle -3x+1=2x-9$

Add $\displaystyle 3x$ to both sides

$\displaystyle -3x+3x+1=2x+3x-9$

$\displaystyle 1=5x-9$

Add 9 to both sides

$\displaystyle 1+9=5x-9+9$

$\displaystyle 10=5x$

Divide both sides by 5

$\displaystyle \frac{10}{5}=\frac{5x}{5}$

$\displaystyle 2=x$

So $\displaystyle x=2$. We can use this value to find y by using either equation. In this case, I'll use $\displaystyle y=-3x+1$.

$\displaystyle y=-3(2)+1$

$\displaystyle y=-6+1$

$\displaystyle y=-5$

So the solution is 

$\displaystyle (x,y)=(2,-5)$

Let $\displaystyle A$ be the number of adult tickets sold. Then the number of child tickets sold is $\displaystyle 5\textup,450 - A$.

The amount of money raised from adult tickets is $\displaystyle 9A$; the amount of money raised from child tickets is $\displaystyle 5 \left (5\textup,450 - A \right )$. The sum of these money amounts is $\displaystyle $36,330$, so the amount of money raised can be defined by the following equation:

$\displaystyle 9A + 5 (5\textup,450-A) = 36\textup,330$

To find the number of adult tickets sold, solve for $\displaystyle A$:

$\displaystyle 9A + 27\textup,250-5A = 36\textup,330$

$\displaystyle 4A + 27\textup,250 = 36\textup,330$

$\displaystyle 4A + 27\textup,250 - 27\textup,250= 36\textup,330- 27\textup,250$

$\displaystyle 4A = 9,080\textup$

$\displaystyle 4A \div 4 = 9\textup,080 \div 4$

$\displaystyle A = 2\textup,270$

$\displaystyle 2\textup,270$ adult tickets were sold.

Let's call "$\displaystyle x$" the cost of one jersey (this is the value we want to find)

Let's call the amount of money Jack starts with "$\displaystyle J$"

Let's call the amount of money Aaron starts with "$\displaystyle A$"

We know Jack buys a glove for $\displaystyle \$15$ and $\displaystyle 3$ bats for $\displaystyle \$8$ each, and then has $\displaystyle \$11$ left over after. Thus:

$\displaystyle J=15+3(8)+11$

simplifying, $\displaystyle J=$50$ so Jack started with $\displaystyle \$50$

We know Aaron buys $\displaystyle 2$ hats for $\displaystyle \$10$ each and $\displaystyle 3$ jerseys (unknown cost "$\displaystyle x$") and spends all his money.

$\displaystyle A=2(10)+3x$

The last important piece of information from the problem is Aaron starts with $\displaystyle 30$ dollars more than Jack. So:

$\displaystyle A=30+J$

From before we know:

$\displaystyle J=50$

Plugging in:

$\displaystyle A=30+50$

$\displaystyle A=80$

so Aaron started with $\displaystyle \$80$

Finally we plug $\displaystyle \$80$ into our original equation for A and solve for x:

$\displaystyle A=2(10)+3x$

$\displaystyle 80=2(10)+3x$

$\displaystyle 80=20+3x$

$\displaystyle 60=3x$

$\displaystyle 20=x$

Thus one jersey costs $\displaystyle \$20.00$

Use the elimination process to determine the value of $\displaystyle y$.

Subtract the second equation with the first equation.  This will eliminate the x-variable in order to find the y-variable.

The equation becomes:  $\displaystyle y=-2$

Use this value and substitute into either the first or second equation and solve for the x-variable.

Let's select the first equation and solve for $\displaystyle x$.

$\displaystyle x+3(-2)=6$

$\displaystyle x-6=6$

Add six on both sides.

$\displaystyle x=12$

The result will be the same if we substitute $\displaystyle y=-2$ into the second equation.

The answer is:  $\displaystyle 12$

In order to solve for $\displaystyle y$, multiply the first equation by four in order to eliminate the fractions in front of the coefficients.

$\displaystyle 4(\frac{1}{2}x+\frac{1}{4}y=2)$

The first equation becomes:

$\displaystyle 2x+y=8$

The second equation remains the same:  

$\displaystyle 2x+3y=4$

Subtract the second equation with the first equation:

The result is:  $\displaystyle 2y=-4$

Divide by two on both sides.

$\displaystyle \frac{2y}{2}=\frac{-4}{2}$

Simplify both sides.

The answer is:  $\displaystyle y=-2$

In order to solve for the $\displaystyle x$ variable, we will need to use the elimination method.

Subtract the first equation with the second equation in order to eliminate the $\displaystyle x$ variable to solve for $\displaystyle y$.

After subtracting equation one with equation two:

$\displaystyle y=-6$

Use this value and either equation to solve for $\displaystyle x$. Let's choose the first equation.

$\displaystyle x-2(-6) =-4$

$\displaystyle x+12=-4$

Subtract twelve from both sides.

$\displaystyle x+12-12=-4-12$

Simplify both sides.

The answer is: $\displaystyle x=-16$

6,035 total tickets were sold, and the total number of tickets is the sum of the adult and child tickets, $\displaystyle A + C$.

Therefore, we can say $\displaystyle A + C = 6,035$.

The amount of money raised from adult tickets is $11 per ticket mutiplied by $\displaystyle A$ tickets, or $\displaystyle 11A$ dollars; similarly, $\displaystyle 7C$ dollars are raised from child tickets. Add these together to get the total amount of money raised:

$\displaystyle 11A + 7C = 50,713$

These two equations form our system of equations.

This system can be solved a variety of ways, including graphing. To solve algebraically, write an equation for each of the different trains. We will use y to represent the distance from San Francisco, and x to represent the time since 8AM.

The blue train travels 80 miles per hour, so it adds 80 to the distance from San Francisco every hour. Algebraically, this can be written as $\displaystyle y=80x$.

The green train starts 380 miles away from San Francisco and subtracts distance every hour. This equation should be $\displaystyle y=380-60x$.

To figure out where these trains' paths will intersect, we can set both right sides equal to each other, since the left side of each is $\displaystyle y$.

$\displaystyle 80x=380-60x$ add $\displaystyle 60x$ to both sides

$\displaystyle 140x = 380$ divide both sides by 140

$\displaystyle x=2.714$

Since we wrote the equation meaning time for $\displaystyle x$, this means that the trains will cross paths after 2.714 hours have gone by. To figure out what time it will be then, figure out how many minutes are in 0.714 hours by multiplying $\displaystyle 60*0.714 = 42.857$. So the trains intersect after 2 hours and about 43 minutes, so at 10:43AM.

To figure out how far from San Francisco they are, figure out how many miles the blue train could have gone in 2.714 hours. In other words, plug 2.714 back into the equation $\displaystyle y=80x$, giving you an answer of $\displaystyle y=217.12$.