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In this lesson, systems of equations will be used to model and solve real-life situations.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Challenge

Mark has just started working with his father at a car dealership. They sell cars and motorcycles.

Mark counted $18$ vehicles and $50$ tires on the lot. Without considering spare tires, use this information to find the number of cars and motorcycles.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Number of cars:","formTextAfter":null,"answer":{"text":["7"]}}

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Example

Zain volunteered to work at the reception desk at the school concert.

At the end of the evening, they picked up the cash box and noticed a dollar lying on the floor next to it. Zain wonders whether the dollar belongs inside the cash box or not.

The price of tickets for the concert was one ticket for $$5$ for individuals, or two tickets for $$8$ for couples. Zain looked inside the cash box and found $$200$ and ticket stubs for the $47$ people in attendance. Does the dollar belong inside the cash box?{"type":"choice","form":{"alts":["Yes","No"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}

Without considering the extra dollar found on the floor, write and solve a system of equations. Does the solution make sense in this context?

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.
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.

$x+y=47 $

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.
$5x+4y=200 $

These two equations can be written together to form a system of equations.
${x+y=475x+4y=200 (I)(II) $

Finally, the above system will be solved. For simplicity, the Substitution Method will be used.
${x+y=475x+4y=200 $

SubEqn

$(I):$ $LHS−x=RHS−x$

${y=47−x5x+4y=200 $

Substitute

$(II):$ $y=47−x$

${y=47−x5x+4(47−x)=200 $

${y=47−xx=12 $

The answer can also be found by solving a system considering the extra dollar.
This means that, considering the extra dollar, $34$ tickets were sold to couples and $13$ tickets were sold to individuals. In the context of the situation, this is a perfectly valid answer.

${x+y=475x+4y=201 (I)(II) $

The Substitution Method will be used.
${x+y=475x+4y=201 $

▼

Solve by substitution

SubEqn

$(I):$ $LHS−x=RHS−x$

${y=47−x5x+4y=201 $

Substitute

$(II):$ $y=47−x$

${y=47−x5x+4(47−x)=201 $

Distr

$(II):$ Distribute $4$

${y=47−x5x+188−4x=201 $

SubTerm

$(II):$ Subtract term

${y=47−xx+188=201 $

SubEqn

$(II):$ $LHS−188=RHS−188$

${y=47−xx=13 $

Substitute

$(I):$ $x=13$

${y=47−13x=13 $

SubTerm

$(I):$ Subtract term

${y=34x=13 $

Example

Pennies used to be made of $95%$ copper. In $1983,$ the cost of this metal significantly increased, so the US government changed the composition of the coin. Since $1983,$ pennies have been made of $97.5%$ zinc and plated with a thin copper coating.
### Hint

### Solution

Pennies made in $1983$ and after weigh $2.5$ grams, while pennies made before $1983$ weigh $3.11$ grams. Knowing that a roll of pennies contains $50$ coins and weighs $138.42$ grams, how many of the older, heavier pennies and newer, lighter pennies does this roll contain?

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Write and solve a system of equations.

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.
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 $50-$coin roll there are $22$ older, heavier pennies and $28$ newer, lighter pennies.

$x+y=50 $

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 $50-$coin roll weighs $138.42$ grams, a second equation can be written.
$3.11x+2.5y=138.42 $

These two linear equations form a system of linear equations.
${x+y=503.11x+2.5y=138.42 (I)(II) $

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 $y-$variable to have the same coefficient in both equations. ${x+y=503.11x+2.5y=138.42 $

MultEqn

$(I):$ $LHS⋅2.5=RHS⋅2.5$

${(x+y)(2.5)=50(2.5)3.11x+2.5y=138.42 $

${2.5x+2.5y=1253.11x+2.5y=138.42 $

${2.5x+2.5y=1253.11x+2.5y=138.42 $

SysEqnSub

$(II):$ Subtract $(I)$

${2.5x+2.5y=1253.11x+2.5y−(2.5x+2.5y)=138.42−125 $

${2.5x+2.5y=125x=22 $

Example

Tiffaniqua is selling juice to make some money for a trip to the beach. To prepare a big jug, she used oranges and peaches.

Tiffaniqua's math teacher stopped by to buy some juice and told her the amount of carbohydrates that oranges and peaches have.

Then, the teacher asked Tiffaniqua how many oranges and peaches were used in the jug of juice. Tiffaniqua decided to quiz her teacher by telling her that she used a total of $13$ fruits and that the jug contains $128$ grams of carbohydrates. How many oranges and peaches did she use?{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"Oranges:","formTextAfter":null,"answer":{"text":["6"]}}

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Write and solve a system of equations.

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.
Now that both equations are written in slope-intercept form, their slopes and $y-$intercepts can be used to draw the lines on the same coordinate plane.

$x+y=13 $

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.
$12x+8y=128 $

These two equations form a system of linear equations.
${x+y=1312x+8y=128 (I)(II) $

For simplicity, the system will be solved by graphing. To do so, both linear equations must first be written in slope-intercept form.
${x+y=1312x+8y=128 $

▼

$(I), (II):$ Write in slope-intercept form

SubEqn

$(I):$ $LHS−x=RHS−x$

${y=-x+1312x+8y=128 $

SubEqn

$(II):$ $LHS−12x=RHS−12x$

${y=-x+138y=-12x+128 $

DivEqn

$(II):$ $LHS/8=RHS/8$

${y=-x+13y=8-12x+128 $

WriteSumFrac

$(II):$ Write as a sum of fractions

${y=-x+13y=8-12x +8128 $

MovePartNumRight

$(II):$ $ca⋅b =ca ⋅b$

${y=-x+13y=8-12 x+8128 $

MoveNegNumToFrac

$(II):$ Put minus sign in front of fraction

${y=-x+13y=-812 x+8128 $

ReduceFrac

$(II):$ $ba =b/4a/4 $

${y=-x+13y=-23 x+8128 $

CalcQuot

$(II):$ Calculate quotient

${y=-x+13y=-23 x+16 $

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.

Example

Adventures on the Water is a company that organizes river safaris in Sri Lanka. They take tourists on boats along a river in the middle of the jungle for a full day.

Each boat can hold at most $8$ people and can only carry $1200$ pounds of weight, including passengers and gear, for safety reasons. The company assumes that, on average, an adult weighs $150$ pounds and a child weighs $75$ pounds. It is also assumed that each group will require $200$ pounds of gear plus $10$ pounds of gear per person. There are three groups who wish to take a river safari.

- Group A has $4$ adults and $2$ children.
- Group B has $3$ adults and $5$ children.
- Group C has $8$ adults.

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Write and solve a system of inequalities.

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.
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.

$x+y≤8(I) $

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.
$Adult:Child: 150x+10x=160x175y+10y=85y $

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.
$160x+85y+200≤1200(II) $

The two inequalities obtained form a system of inequalities.
${x+y≤8160x+85y+200≤1200 (I)(II) $

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.
${x+y=8160x+85y+200=1200 $

▼

$(I), (II):$ Write in slope-intercept form

SubEqn

$(I):$ $LHS−x=RHS−x$

${y=-x+8160x+85y+200=1200 $

SubEqn

$(II):$ $LHS−200=RHS−200$

${y=-x+8160x+85y=1000 $

SubEqn

$(II):$ $LHS−160x=RHS−160x$

${y=-x+885y=-160x+1000 $

DivEqn

$(II):$ $LHS/85=RHS/85$

${y=-x+8y=85-160x+1000 $

WriteSumFrac

$(II):$ Write as a sum of fractions

${y=-x+8y=85-160x +851000 $

MoveNegNumToFrac

$(II):$ Put minus sign in front of fraction

${y=-x+8y=-85160x +851000 $

MovePartNumRight

$(II):$ $ca⋅b =ca ⋅b$

${y=-x+8y=-85160 x+851000 $

ReduceFrac

$(II):$ $ba =b/5a/5 $

${y=-x+8y=-1732 x+17200 $

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

Test Point: $(1,2)$ | ||
---|---|---|

Inequality | Substitute | Simplify |

$x+y≤8$ | $1+2≤? 8$ | $3≤8✓$ |

$160x+85y+200≤1200$ | $160(1)+85(2)+200≤? 1200$ | $230≤1200✓$ |

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.

Closure

The challenge presented at the beginning of the lesson can also be modeled by a system of equations. It is known that Mark and his father sell cars and motorcycles.

In the agency, Mark counted $18$ vehicles and $50$ tires. Without considering spare tires, use this information to find the number of cars and motorcycles.Write and solve a system of equations.

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 $18.$
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.

$x+y=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.
$4x+2y=50 $

These two linear equations form a system.
${x+y=184x+2y=40 (I)(II) $

Finally, the system will be solved. For simplicity, the Substitution Method will be used.
${x+y=184x+2y=50 $

▼

$(I), (II):$ Solve by substitution

SubEqn

$(I):$ $LHS−x=RHS−x$

${y=-x+184x+2y=50 $

Substitute

$(II):$ $y=-x+18$

${y=-x+184x+2(-x+18)=50 $

Distr

$(II):$ Distribute $2$

${y=-x+184x−2x+36=50 $

SubTerm

$(II):$ Subtract term

${y=-x+182x+36=50 $

SubEqn

$(II):$ $LHS−36=RHS−36$

${y=-x+182x=14 $

DivEqn

$(II):$ $LHS/2=RHS/2$

${y=-x+18x=7 $

Substitute

$(I):$ $x=7$

${y=-7+18x=7 $

AddTerms

$(I):$ Add terms

${y=11x=7 $

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