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To find whether the dollar belongs inside the cash box, a system of equations will be written and solved. Let $x$ and $y$ be the number of tickets sold to individuals and to couples, respectively. Considering that $47$ tickets were sold, an equation can be written. Next, since each individual's ticket is $\$5,$ the total amount of money made by selling tickets to individuals is $5x.$ Similarly, since each ticket sold to couples is $\$4,$ the total amount of money made by selling tickets to couples is $4y.$ Considering that the cash box contains $\$200$ total, a second equation can be written. These two equations can be written together to form a system of equations. Finally, the above system will be solved. For simplicity, the Substitution Method will be used. It has been found that $x=12.$ Now, this value will be substituted into the first equation to find the value of $y.$ The solution to the system is $x=12$ and $y=35.$ Considering the context, this means that without the extra dollar, $35$ tickets were sold to couples and $12$ tickets were sold to individuals. However, this is not possible, since couples tickets were only sold in pairs. Therefore, the dollar found on the floor should be placed inside the cash box.

To find out how many of the older, heavier pennies and how many of the newer, lighter pennies this roll contains, a system of equations will be written and solved. Let $x$ be the number of older pennies and $y$ the number of newer pennies in the roll. Since the roll contains $50$ coins, an equation can be written. Next, since older pennies weigh $3.11$ grams, the total weight of the older pennies in the roll is $3.11x.$ Similarly, since a newer penny weighs $2.5$ grams, the total weight of the newer pennies is $2.5y.$ With this information, and knowing that the $50\text{-}$coin roll weighs $138.42$ grams, a second equation can be written. These two linear equations form a system of linear equations. For simplicity the system will be solved by the Elimination Method. To do so, the first equation will be multiplied by $2.5$ in order for the $y\text{-}$variable to have the same coefficient in both equations. Now, Equation (I) will be subtracted from Equation (II). It has been found that $x=22.$ Finally, this value will be substituted in the first equation to find the value $y.$ The solution to the system is $x=22$ and $y=28.$ In the context of the situation, this means that in the $50\text{-}$coin roll there are $22$ older, heavier pennies and $28$ newer, lighter pennies.

To find out how many oranges and peaches Tiffaniqua used for a jug, a system of equations will be written and solved. Let $x$ and $y$ be the number of oranges and peaches, respectively. A first equation can be written knowing that Tiffaniqua used $13$ fruits. Next, consider the fact that the jug contains $128$ grams of carbohydrates. It is also known that an average orange has about $12$ grams of carbohydrates and that an average peach has about $8$ grams of carbohydrates. With all this information, a second equation can be written. These two equations form a system of linear equations. For simplicity, the system will be solved by graphing. To do so, both linear equations must first be written in slope-intercept form. Now that both equations are written in slope-intercept form, their slopes and $y\text{-}$intercepts can be used to draw the lines on the same coordinate plane.

Finally, the point of intersection can be determined.

The lines intersect at the point with coordinates $(6,7).$ Therefore, the solution to the system is $x=6$ and $y=7.$ In the context of the situation, this means that Tiffaniqua used $6$ oranges and $7$ peaches to prepare a jug of juice.

To determine which of the groups can safely take a safari, a system of linear inequalities will be written and solved. Let $x$ be the number of adults and $y$ the number of children. Knowing that each boat can hold at most $8$ people, the first inequality can be written. Next, a second inequality will be written. The company assumes that an adult weighs $150$ and a child weighs $75$ pounds. Also, for each person there are considered $10$ extra pounds of gear. Furthermore, each group requires $200$ pounds of gear, no matter the number of people. Knowing that each boat can only carry $1200,$ the second inequality can be written. The two inequalities obtained form a system of inequalities. The above system will be solved by graphing. To do so, the equations of the boundary lines will be written in slope-intercept form. To obtain the equations of the boundaries lines, the inequality symbols will be replaced with equals signs. Now both lines can be drawn on the same coordinate plane. Since the number of adults and children cannot be negative, only the first quadrant will be considered. Also, because neither inequality is strict, the lines will be solid. Next, a point not on the boundary lines will be tested to determine the region to be shaded. For simplicity, $(1,2)$ will be used. To start, this point will be tested in the first inequality. If a true statement is obtained, the region containing the point will be shaded. Otherwise, the opposite region will be shaded. Since a true statement was obtained, the region that includes this point will be shaded.

By following the same procedure, the region that corresponds to the second inequality can be determined.

Test Point: $(1,2)$
Inequality	Substitute	Simplify
$x+y\le 8$	$1+2\stackrel{?}{\le }8$	$3\le 8✓$
$160x+85y+200\le 1200$	$160(1)+85(2)+200\stackrel{?}{\le }1200$	$230\le 1200✓$

The second inequality is also satisfied by $(1,2).$ Therefore, the region that contains this point will be shaded.

To fully see the region that satisfies both inequalities, the unwanted regiones will be removed.

Finally, a point that represents each of the three groups will be plotted to see if they belong to the shaded area.

Group	Adults $\&$ Children	Point
A	$4$ adults and $2$ children	$(4,2)$
B	$3$ adults and $5$ children	$(3,5)$
C	$8$ adults	$(8,0)$

These points will be plotted on the coordinate plane.

The points that represent groups A and B are in the shaded area, and the point that represents group C is not in the shaded area. Therefore, A and B are the only groups that can safely take a river safari.

To find the number of a cars and motorcycles in the agency, a system of linear equations will be written and solved. Let $x$ be the number of cars and $y$ the number of motorcycles. Because there are $18$ vehicles, the sum of the two variables equals $18.$ Next, since spare tires are not considered, each car has $4$ tires and each motorcycle has $2$ tires. Knowing that Mark counted $50$ tires, a second equation can be written. These two linear equations form a system. Finally, the system will be solved. For simplicity, the Substitution Method will be used. The solution to the system is $x=7$ and $y=11.$ In this context, it means that there are $7$ cars and $11$ motorcycles in the agency.