Applications of linear systems

Blue candle

You have to find the parameters of the equation $h_{blue}(t)=m_{blue}\cdot t+c_{blue}$.

In one hour it burns down by $0.5cm$ . Thus the slope is $m_{blue}=-0.5$.

The y-axis intercept is the initial height of the candle: $c_{blue}=6$.

That way, you get the equation of the function: $h_{blue}(t)=-0.5\cdot t+6$

Blaue Kerze

$h_{red}(t)=m_{red}\cdot t+c_{red}$

The red candle burns down by $0.9cm$ per hour. The slope is therefore $m_{red}=-0.9$.

The red candle initially has a height of $c_{red}=13$.

Hence, you get the function equation: $h_{red}(t)=m_{red}\cdot t+c_{red}$

First, you create a system of linear equations with two variables from the functions:

$\begin{array}{lrll}\phantom{\mathrm{I}}\mathrm{I}& h& =& -0.5t+6\\ \mathrm{II}& h& =& -0.9t+13\end{array}$

To find the time $t$ at which both candles have the same height, you look for the solution of the system of equations. The elimination by equating coefficients is suitable, here, since both equations are solved for $h$.

1. Equate

$\Rightarrow -0.5t+6=-0.9t+13$

2. Solve for t

$\begin{array}{rlll}-0.5t+6& =& -0.9t+13& |+0.9t\\ 0.4t+6& =& 13& |-6\\ 0.4t& =& 7& |:0.4\\ t& =& 17.5& \end{array}$

So the candles are the same height after $17.5$ hours, that is $17$ hours and $30$ minutes.

Answer: After $17$ hours and $30$ minutes, the candles have the same height $h$.

Height of the candles

According to $b)$, the two candles have the same height after $17.5$ hours. But how tall are they?

To find this out, plug $t=17.5$ into one of the equations from b) and calculate $h$.

$\begin{array}{lll}h& =& -0.5\cdot (17.5)+6\\ & =& -8.75+6\\ & =& -2.75\end{array}$

So, after $17.5$ hours, both candles should be $-2.75cm$ tall.

What does $-2.75$ mean?

Mathematically, the two straight lines from a) have an intersection point, there.

But in reality, this solution makes no sense. From the straight line equation of the red candle $h_{red}=-0.5t+6$ we can calculate that it has already burned down after 12 hours. (see calculation of zeros).

Alternatively, you can draw the graphs of the candle heights, to interpret your solution.

Answer:

Although a solution arises mathematically, the two candles are never the same size: they will have burnt down, befor this may happen.

First, define variables for the quantities in the text:

gallons of of premium gasoline $\widehat{=}$ $x$

gallons of of regular gasoline $\widehat{=}$ $y$

Next, you need to organize all the information. On the one hand you have gallons loaded and on the other, money spent.

Important: $x$ and $y$ are included in both data.

So:

Gallons: $x$ gallons, $y$ gallons, $5$ gallons

Expenses: $\$1.35\cdot x$ , $\$1.20\cdot y$ , $\$6.50$

In the last step you set up one equation from each set of respective data.

You get the first equation by the information that $x$ gallons and $y$ gallons add up to $5$ gallons.

You get the second equation by the information that $\$1.35\cdot x$ and $\$1.20\cdot y$ add up to $\$6.50$ .

Both equations $\mathrm{I}$ and equation $\mathrm{II}$ must be satisfied according to the problem setting. So, you can represent the problem from the setting by the following system of equations.

$\begin{array}{rrlll}\mathrm{I}& x& +& y& =5\\ \mathrm{II}& 1.35\cdot x& +& 1.2\cdot y& =6.5\end{array}$

A system of linear equations is given, which can be solved via elimination by equating coefficients

$\phantom{\mathrm{I}}\mathrm{I}x+y=5$

$\mathrm{II}1.35\cdot x+1.2\cdot y=6.5$

1. Solve both equations for y

Solve both equations for one variable. In this case, it does not matter whether you take $x$ or $y$.

2. Equate

Set the two equations $\mathrm{I}$ and $\mathrm{I}{\mathrm{I}}^{\prime }$ equal.

$\Rightarrow 5-x=\frac{65}{12}-1.125\cdot x$

3. Solve equation for x

4. Plug in x to find out y

Plug $x$ into ${\mathrm{I}}^{\prime }$ or $\mathrm{I}{\mathrm{I}}^{\prime }$.

$\phantom{\mathrm{I}}{\mathrm{I}}^{\prime }y=5-\frac{10}{3}=\frac{5}{3}$

Write down the solution set

$${\displaystyle L=\left\{(\frac{10}{3};\frac{5}{3})\right\}}$$

Dave has loaded $x=\frac{10}{3}\approx 3.3$ gallons of regular gasoline and $y=\frac{5}{3}\approx 1.7$ gallons of premium gasoline.

First, label the quantities with appropriate variables:

number of hamburgers $\widehat{=}$ $x$

number of cheeseburgers $\widehat{=}$ $y$

Next, you need to organize all the information. On the one hand you have burger quantities and on the other hand you have costs.

Important: $x$ and $y$ are included in both pieces of information.

So:

quantities: $x$ hamburgers, $y$ cheeseburgers, total number: $12$ burgers.

money: $\$0.99\cdot x$ , $\$1.19\cdot y$ , $\$12.68$

In the last step, you set up an equation from the respective data.

You get the first equation by the information that $x$ hamburgers and $y$ cheeseburgers add up to $12$ burgers.

You get the second equation by the information that $\$0.99\cdot x$ and $\$1.19\cdot y$ add up to $\$12.68$ .

If you want to show that $x$ and $y$ solve a system of linear equations, you only need to substitute the variables into the system. In the end, each equation must yield a true statement.

So plug $x=8$ and $y=4$ into the equations and check:

$\begin{array}{rrll}\mathrm{I}& x+y& =& 12\\ \mathrm{II}& 0.99\cdot x+1.19\cdot y& =& 12.68\end{array}$

Plug in the values.

$\begin{array}{rrll}\mathrm{I}& 8+4& =& 12\\ \mathrm{II}& 0.99\cdot 8+1.19\cdot 4& =& 12.68\end{array}$

Calculate. If the same number is on the left and on the right, the statement is true.

$\begin{array}{rrll}\mathrm{I}& 12& =& 12\\ \mathrm{II}& 12.68& =& 12.68\end{array}$

You see immediately that $x=8$ hamburgers and $y=4$ cheeseburgers provide a true statement.

A linear system of equations can have either none, exactly one, or infinitely many solutions.

The equations are linear and not identical. In addition, there are only $2$ equations. Thus there is at most one intersection point. A maximum of one intersection point means that there is either no intersection point or one intersection point. From b) you already know that the former is not true, because $x=8$ and $y=4$ is a solution. Therefore, the system of equations has exactly one solution.