Would you rather have a coupon for \( \$ 20 \) off or \( 20 \% \) off?
From work by Dan Meyer
http://www.101qs.com/1936 |
In Writing Linear Functions Target C we learned to model a contextual situation in two variables with a linear equation. We can also model a contextual situation that involves two variable with a system of linear equations.
Example 1: You own a Great Adventure Mini-Golf course. It costs \( \$ 7 \) per child and \( \$ 10 \) for an adult. You know there were \( 42 \) games of mini-golf played today and you collected \( \$ 324 \). How many children played and how many adults played today?
As we begin to look at the problem we can set up our variable first based on the question being asked, let's let \( c = \text{the number of children} \) and \( a = \text{the number of adults} \). We know these are our variables because it is what we are trying how many of each played that day. Also, notice their are numbers representing amounts of money and also numbers representing rounds of golf. Let's look at each of these numbers:
There are two distinct types of equations being formed. One that is dealing with money and one that is dealing with how many people played.
The linear equation dealing with money will be the amount collected based on the number of games children played plus the amount collected based on the number of games that adults played and totaling the amount of money collected that day. Note that it is often easier to write equations without labels such as \( \$ \), as long as we keep in mind what each coefficient represents.
\( 7c + 10a = 324 \)
The linear equation dealing with total number of games will be the number of games played by children added to the number of games played by adults and totaling the number of people who played that day.
\( c + a = 42 \)
Now we will solve the system of equations using one of the methods we learned in Linear Systems Target A- substitution, linear combinations or graphing. We will show the use linear combination to solve the equation.
Example 1: You own a Great Adventure Mini-Golf course. It costs \( \$ 7 \) per child and \( \$ 10 \) for an adult. You know there were \( 42 \) games of mini-golf played today and you collected \( \$ 324 \). How many children played and how many adults played today?
As we begin to look at the problem we can set up our variable first based on the question being asked, let's let \( c = \text{the number of children} \) and \( a = \text{the number of adults} \). We know these are our variables because it is what we are trying how many of each played that day. Also, notice their are numbers representing amounts of money and also numbers representing rounds of golf. Let's look at each of these numbers:
- \( \$ 7 \) per child is a rate. \( 7c \) will give you the amount collected from children playing.
- \( \$ 10 \) per adult is a rate. \( 10a \) will give you the amount collected from adults playing.
- \( \$ 324 \) is the total amount of money collected.
- \( 42 \) is the total number of games of mini-golf played.
There are two distinct types of equations being formed. One that is dealing with money and one that is dealing with how many people played.
The linear equation dealing with money will be the amount collected based on the number of games children played plus the amount collected based on the number of games that adults played and totaling the amount of money collected that day. Note that it is often easier to write equations without labels such as \( \$ \), as long as we keep in mind what each coefficient represents.
\( 7c + 10a = 324 \)
The linear equation dealing with total number of games will be the number of games played by children added to the number of games played by adults and totaling the number of people who played that day.
\( c + a = 42 \)
Now we will solve the system of equations using one of the methods we learned in Linear Systems Target A- substitution, linear combinations or graphing. We will show the use linear combination to solve the equation.
\(7c+10a=324\)
\(c+a=42\)
\begin{align}&-7(c+a=42)\ & \ \ \ & \text{Multiply the second equation by -7 to solve for a.}\\\\
&\ 7c+10a=324\ & \ \ \ & \text{Rewrite the first equation.}\\
&\underline{-7c-7a=-294}\ & \ \ \ & \text{Distribute -7 to the second equation.}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \underline{3a}=\underline{30}\ & \ \ \ & \text{Add the two equations together.}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ \ \ \ \ \ 3\ & \ \ \ &\text{Divide by 3 on both sides.}\\
&\ \ \ \ \ \ \ \ \ \ \ \ a=10\\\\
&c+10=42\ & \ \ \ &\text{Sove for c by substituting 10 in for a.}\\
&c=32\end{align}
\(c+a=42\)
\begin{align}&-7(c+a=42)\ & \ \ \ & \text{Multiply the second equation by -7 to solve for a.}\\\\
&\ 7c+10a=324\ & \ \ \ & \text{Rewrite the first equation.}\\
&\underline{-7c-7a=-294}\ & \ \ \ & \text{Distribute -7 to the second equation.}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \underline{3a}=\underline{30}\ & \ \ \ & \text{Add the two equations together.}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ \ \ \ \ \ 3\ & \ \ \ &\text{Divide by 3 on both sides.}\\
&\ \ \ \ \ \ \ \ \ \ \ \ a=10\\\\
&c+10=42\ & \ \ \ &\text{Sove for c by substituting 10 in for a.}\\
&c=32\end{align}
There were \( 32 \) children games of mini-golf and \( 10 \) adult games of mini-golf.
Example 2: You have been saving your change and decided to take it to the bank to cash it in. It is only made of dimes and nickels. The teller said you had \( 8 \) more nickels then dimes. You had a total of \( \$ 1.75 \). How many nickels did you have and how many dimes did you have?
So we can determine our variables, nickels and dimes, let's let \( n = \text{nickels} \) and \( d = dimes \). We should have two equations, one that represents the amount of money and a second one giving the relationship between dimes and nickels.
The first equation:
The second equation:
We know we have "\( 8 \) more nickels than dimes", which means you have more nickels. If you are trying to write an equation we need to have two sides equal to one another, right now nickels than dimes. So in order to write a balanced equation you will need to add \( 8 \) more to the dimes side. See below:
Example 2: You have been saving your change and decided to take it to the bank to cash it in. It is only made of dimes and nickels. The teller said you had \( 8 \) more nickels then dimes. You had a total of \( \$ 1.75 \). How many nickels did you have and how many dimes did you have?
So we can determine our variables, nickels and dimes, let's let \( n = \text{nickels} \) and \( d = dimes \). We should have two equations, one that represents the amount of money and a second one giving the relationship between dimes and nickels.
The first equation:
- A dime is worth \( \$ 0.10 \). If we multiply it by the \( d = \text{number of dimes} \), we will get an expression for the amount of money we have for dimes only, \( 0.10d \).
- A nickel is worth \( \$ 0.05 \). If we multiply by the \( n = \text{number of dimes} \), we will get an expression for the amount of money we have for nickels only, \( 0.05n \).
- We have a total of \( \$ 1.75 \), so add the previous two values together it should give us our total: \( 0.10d + 0.05d = 1.75. \)
The second equation:
We know we have "\( 8 \) more nickels than dimes", which means you have more nickels. If you are trying to write an equation we need to have two sides equal to one another, right now nickels than dimes. So in order to write a balanced equation you will need to add \( 8 \) more to the dimes side. See below:
To solve this system we will use substitution since the second equation has the variable "\( n \)" already isolated. Before we do this, to make the equation easier to handle, we will multiply the first equation by \( 100 \) so to "clear the decimals". This is not necessary, and the system could be solved without doing this, though we would have to work with decimals rather than integers. Follow below:
\begin{align}&100(0.10d+0.05n=1.75)\ & \ \ & \text{Multiply the first equation by 100.}\\
&10d+5n=175\\\\
&10d+5(d+8)=175\ & \ \ & \text{Substitute the second equation into the first for n.}\\
&10d+5d+40=175\ & \ \ & \text{Distribute 5.}\\
&15d+40=175\ & \ \ & \text{Simplify.}\\
&15d=135\ & \ \ & \text{Subtract.}\\
&d=9\ & \ \ & \text{Divide.}\\\\
&n+9+8\ & \ \ & \text{Solve for n by substituting 9 in for d into the second equation.}\\
&n=17\end{align}
&10d+5n=175\\\\
&10d+5(d+8)=175\ & \ \ & \text{Substitute the second equation into the first for n.}\\
&10d+5d+40=175\ & \ \ & \text{Distribute 5.}\\
&15d+40=175\ & \ \ & \text{Simplify.}\\
&15d=135\ & \ \ & \text{Subtract.}\\
&d=9\ & \ \ & \text{Divide.}\\\\
&n+9+8\ & \ \ & \text{Solve for n by substituting 9 in for d into the second equation.}\\
&n=17\end{align}
We have \( 9 \) dimes and \( 17 \) nickels. You can check the equation easily, \( 9 \) dimes is \( \$ 0.90 \) and \( 17 \) nickels is \( \$ 0.85 \) which adds to \( \$ 1.75 \)!
You could have solved this with a spreadsheet of all possible combinations. Click on the diagram to go to the spreadsheet.
You could have solved this with a spreadsheet of all possible combinations. Click on the diagram to go to the spreadsheet.
You could have solved this problem by graphing. Click on the diagram to go to DESMOS.
Quick Check
The Jet Ski Super Store rents jets skis and life jackets. Jet skis rent for \( \$ 60 \) and life jackets rent for \( \$ 5 \). The store knows they rented a total of \( 63 \) items and collected \( \$ 2405 \). How many Jet Skis did the store rent and how many life jackets?
The Jet Ski Super Store rents jets skis and life jackets. Jet skis rent for \( \$ 60 \) and life jackets rent for \( \$ 5 \). The store knows they rented a total of \( 63 \) items and collected \( \$ 2405 \). How many Jet Skis did the store rent and how many life jackets?