Problem Solving with Linear Systems

Section 1.6 Problem Solving with Linear Systems

Most of the problems we’ve seen so far have started with a given linear system, and we learned how to find the solution(s) to the system by performing Gauss-Jordan elimination and interpreting the resulting reduced row echelon form.

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In this section, we will consider problems where the linear system is “hidden”, that is, not stated in the problem. We will create a linear system to solve the problem, and we will also need to interpret the possible solution(s) in terms of the context of the question, where it’s possible that not all values of the variables make sense.

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Subsection Prepare

Let’s start with an example.

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Example 1.6.1. Seats in a Baseball Stadium.

A baseball stadium has seating arranged in three sections. As part of a special promotion, guests will receive two of three prizes. Guests seated in Sections A and B will receive a hat, guests seated in Sections B and C will receive a shirt, and guests seated in Sections A and C will receive a hot dog. The promoters told the stadium managers of their plans, and asked how many seats were in each section. The managers, thinking they were being helpful, told the promoters they would need $800$ hats, $608$ shirts, and $612$ hot dogs, and they have since been unavailable for further help. How many seats are in each section?

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In order to answer this question, we need to make we know what is being asked. The final sentence of the question asks: “How many seats are in each section?”

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Since we are trying to solve for the number of seats in each section, those should be our variables. Let $x_1$ be the number of seats in Section A, $x_2$ be the number of seats in Section B, and $x_3$ be the number of seats in Section C.

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Now we use the information given to create linear equations. We were told that hats go to guests in Sections A and B and that we’ll need $800$ hats, which means that

\begin{equation*} x_1+x_2 = 800. \end{equation*}

Guests in Sections B and C get shirts, which means that

\begin{equation*} x_2+x_3 = 608\text{.} \end{equation*}

Lastly, guests in Sections A and C get hot dogs, which means that

\begin{equation*} x_1+x_3 = 612\text{.} \end{equation*}

Putting these three equations into a linear system results in

\begin{align*} \amp x_1\amp \amp + \amp \amp x_2\amp \amp \amp \amp \amp \amp = 800 \\ \amp \amp \amp \amp \amp x_2\amp \amp +\amp \amp x_3\amp \amp = 608 \\ \amp x_1 \amp \amp +\amp \amp \amp \amp \amp \amp x_3\amp \amp =612 \text{,} \end{align*}

and the corresponding augmented matrix is

\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 0\amp 800\\0\amp 1\amp 1\amp 608\\1\amp 0\amp 1\amp 612\\ \end{array} \right]\text{.} \end{equation*}

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To solve our system, we put the augmented matrix into reduced row echelon form and get

\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 0\amp 800\\0\amp 1\amp 1\amp 608\\1\amp 0\amp 1\amp 612\\ \end{array} \right] \rrefarrow \left[\begin{array}{cccc} 1\amp 0\amp 0\amp 402\\0\amp 1\amp 0\amp 398\\0\amp 0\amp 1\amp 210\\ \end{array} \right]\text{.} \end{equation*}

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Since $x_1=402\text{,}$ this means that Section A has $402$ seats. Similarly, we see that Section B has $398$ seats, and Section C has $210$ seats. This is a reasonable solution because the number of seats in each section is nonnegative and a whole number.

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Let’s recap and generalize our steps.

We made sure we knew what the problem was asking. (“How many seats in each section?”)
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We gave variable names to the unknown number of seats in each section. (Section A had $x_1$ number of seats, Section B had $x_2\text{,}$ and Section C had $x_3$ many seats.)
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We used the information given in the problem to write down linear equations involving those variables.
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We constructed an augmented matrix from the system of linear equations and put the matrix into reduced row echelon form.
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We made sure the solution obtained from the reduced row echelon form made sense in the context of the problem. (The number of seats in a section couldn’t have been a negative or fractional number like $-4.78\text{,}$ for example.)
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This set of problem-solving steps is important enough to say again.

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Steps for Problem Solving with Linear Systems.

Identify what is being asked.
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Identify and give variable names to the unknown quantities whose values we need to find.
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Use the information in the problem to write down linear equations involving the variables.
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Form an augmented matrix from the system of linear equations and put the matrix into reduced row echelon form.
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Check that any solution(s) found from the reduced row echelon form make sense in the context of the problem.
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An example with points scoring in football.

The most common ways to score points in an American football game are by touchdowns, extra points, 2-point conversions, and field goals. Touchdowns are worth $6$ points and are immediately followed by either an extra point worth $1$ point or by a 2-point conversion worth $2$ points. Field goals are worth $3$ points.

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Suppose that in one game a team had $7$ scoring plays and scored a total of $24$ points. Each touchdown was followed by a successful extra point or 2-point conversion. How did the team score their points?

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Let’s go through our problem-solving steps.

Identify what is being asked 🔗: We are asked to find how the team scored their points.
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Identify unknowns and name variables 🔗: We need to focus on the unknowns that answer the question: how many touchdowns, extra points, 2-point conversions, and field goals the team had.
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Activity 1.6.2. Which quantities should be variables?

An American football team can score points via touchdowns, extra points, 2-point conversions, and field goals. We were given the total points scored and some other information and were asked to find how the team scored its points. What should the variables in this problem be?

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The variables should be the things we are solving for, the ways to score points. Other information in the problem will get used as part of the linear equations which involve the variables.

The number of touchdowns, $t$
The number of field goals, $f$
The number of extra points, $x$
The number of 2-point conversions, $c$
Variables in this problem
The number of points, $p$
Points earned through touchdowns, $6t$
Extra points and 2-point conversions only come after touchdowns, $x+c\leq t$
Not variables in this problem

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Form linear equations involving the variables 🔗: We are told that there were $7$ scoring plays, so

\begin{equation*} t+x+c+f=7\text{.} \end{equation*}

We were also told that the total number of points is $24\text{,}$ and that each touchdown is worth $6$ points, each field goal is worth $3\text{,}$ each extra point is worth $1$ and each 2-point conversion is worth $2$ points. This means that

\begin{equation*} 6t+x+2c+3f=24\text{.} \end{equation*}

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There is one more given piece of information: that every touchdown is followed by either an extra point or a 2-point conversion. This means that $t=x+c\text{,}$ or

\begin{equation*} t-x-c=0 \end{equation*}

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Form augmented matrix and row reduce 🔗: Putting the three equations above into an augmented matrix and row reducing gives us

\begin{equation*} \left[\begin{array}{ccccc} 1\amp 1\amp 1\amp 1\amp 7\\ 6\amp 1\amp 2\amp 3\amp 24\\1\amp -1\amp -1\amp 0\amp 0\\ \end{array} \right] \rrefarrow \left[\begin{array}{ccccc} 1\amp 0\amp 0\amp 0.5\amp 3.5\\0\amp 1\amp 0\amp 1\amp 4\\ 0\amp 0\amp 1\amp -0.5\amp -0.5\\ \end{array} \right]\text{.} \end{equation*}

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Interpret solutions which make sense 🔗: The reduced row echelon form has a free variable in the fourth column. Turning each row of the matrix back into an equation results in

\begin{align*} t \amp =3.5-0.5f\\ x\amp =4-f\\ c\amp =-0.5+0.5f\text{.} \end{align*}

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Mathematically, the value of $f$ can be any real number, which means there are infinitely many solutions. However, in this problem $f$ stood for the number of field goals, and there can’t be a negative or fractional number of field goals in a football game. The solution only makes sense if $f$ is $0,1,2, 3,\ldots\text{.}$
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Looking at the second equation $x=4-f\text{,}$ we see a further restriction on the value of $f\text{.}$ If $f\gt 4$ then the value of $x$ is negative, and it doesn’t make sense to have a negative number of extra points. Thus $f$ must be one of $0,1,2,3,4\text{.}$ Can we eliminate still more solutions?
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Looking at the first equation $t=3.5-0.5f\text{,}$ if $f=0$ then $t=3.5\text{,}$ but the number of touchdowns must be a whole number. We also get a fractional number of touchdowns if $f=2$ or if $f=4\text{.}$ The only remaining possibilities are $f=1$ or $f=3\text{.}$
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If $f=1\text{,}$ then

\begin{align*} t \amp =3.5-0.5\cdot 1=3\\ x\amp =4-1=3\\ c\amp =-0.5+0.5\cdot 1=0\\ f \amp = 1 \end{align*}

and $3$ touchdowns each followed by an extra point plus a field goal adds up to $24$ points in $7$ scoring plays.

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If $f=3\text{,}$ then

\begin{align*} t \amp =3.5-0.5\cdot 3=2\\ x\amp =4-3=1\\ c\amp =-0.5+0.5\cdot 3=1\\ f \amp = 3 \end{align*}

and $2$ touchdowns, one followed by an extra point and one by a 2-point conversion, plus $3$ field goals also adds up to $24$ points in $7$ scoring plays.

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We see that sometimes there’s a lot of work involved in interpreting the solutions which make sense. In the case of points in a football game, there were infinitely many solutions which made mathematical sense, but only two solutions which made sense as actual ways to score points meeting the necessary conditions.

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Activity 1.6.3. Which solutions make sense?

(a)

Consider a linear system whose augmented matrix and reduced row echelon form are shown below.

\begin{equation*} \left[\begin{array}{cccc} 1\amp 1\amp 1\amp 32\\ 1\amp 5\amp 10\amp 100\\ \end{array} \right] \rrefarrow \left[\begin{array}{cccc} 1\amp 0\amp -\frac54\amp 15\\0\amp 1\amp \frac94\amp 17\\ \end{array} \right] \end{equation*}

Note that this translates back to equations as

\begin{align*} x \amp =15+\frac54z\\ y\amp =17 - \frac94z\\ z \amp \text{ is free.} \end{align*}

Which of the following values is a possible solution for $x\text{,}$ $y\text{,}$ and $z\text{?}$

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$x=16.25\text{,}$ $y=14.75\text{,}$ $z=1$
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Correct!
$x=10\text{,}$ $y=26\text{,}$ $z=-4$
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Correct!
$x=15\text{,}$ $y=17\text{,}$ $z=0$
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Correct!
$x=20\text{,}$ $y=8\text{,}$ $z=4$
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Correct!

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(b)

A woman has $32$ total bills in her purse, consisting of $\$ 1\text{,}$ $\$ 5$ and $\$ 10$ bills, giving her a total of $\$ 100\text{.}$ How many bills of each denomination does she have?

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Let $x\text{,}$ $y$ and $z$ be the number of ones, fives and tens, respectively, and suppose that the problem has been solved as far as

\begin{align*} x \amp =15+\frac54z\\ y\amp =17 - \frac94z\\ z \amp \text{ is free.} \end{align*}

Which of the following values is a possible solution for $x\text{,}$ $y\text{,}$ and $z\text{?}$

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$x=16.25\text{,}$ $y=14.75\text{,}$ $z=1$
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She can’t have $16.25$ one-dollar bills, or $14.75$ five-dollar bills.
$x=10\text{,}$ $y=26\text{,}$ $z=-4$
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She can’t have a negative amount of ten-dollar bills.
$x=15\text{,}$ $y=17\text{,}$ $z=0$
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Correct! This is a total of $32$ bills, and $15$ ones plus $17$ fives add up to $\$ 100\text{.}$

If you had to pick one solution only, this one is likely not the best choice because the wording “consisting of” might imply that she has at least one of each type of bill.
$x=20\text{,}$ $y=8\text{,}$ $z=4$
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Correct! This is a total of $32$ bills, and $20$ ones, $8$ fives, and $4$ tens add up to $\$ 100\text{.}$

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Let’s do another example, one which involves balancing a chemical equation.

All chemical reaction problems found in this text are thanks to Gregory Bard and Gergely Sirokman.

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Example 1.6.4. Balancing a chemical equation.

When chemicals react, no elements are created or destroyed; instead they recombine into different substances. For example, a blow torch works by burning acetylene $C_2H_2$ in the presence of oxygen $O_2\text{.}$ The result is carbon dioxide $CO_2$ and water $H_2 O\text{.}$

\begin{equation*} \text{acetylene } + \text{ oxygen } \longrightarrow \text{ carbon dioxide } + \text{ water} \end{equation*}

If there were $4$ atoms of carbon (as part of the acetylene) before the reaction, then there have to be $4$ atoms of carbon (in the form of carbon dioxide) after the reaction also.

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In chemistry, balancing a chemical equation means finding the smallest numbers of the different molecules involved in the reaction. Most students of chemistry balance equations by guessing and checking, but there’s actually a system of linear equations involved, which means there’s an algorithm which always leads to a solution.

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Let’s balance the blow torch reaction, using our problem solving steps.

\begin{equation*} \fillinmath{X}C_2H_2 + \fillinmath{X}O_2 \longrightarrow \fillinmath{X} CO_2 + \fillinmath{X}H_2O \end{equation*}

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We need to find the coefficients of each of the molecules, so those are the unknowns we should give names to. Let’s say there are $x_1$ many molecules of $C_2H_2\text{,}$ and $x_2$ many molecules of $O_2\text{,}$ and $x_3$ many molecules of $CO_2\text{,}$ and $x_4$ many molecules of $H_2O\text{.}$

\begin{equation*} x_1 \, C_2H_2 + x_2 \, O_2 \longrightarrow x_3 \, CO_2 + x_4 \, H_2O \end{equation*}

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Now we consider each element and the number of atoms there are before and after the reaction.

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For carbon, which is denoted by $C\text{,}$ there are $x_1$ molecules of $C_2H_2\text{,}$ and there are $2$ carbon atoms in each of the $x_1$ molecules, which means there are $2x_1$ carbon atoms before the reaction takes place. After the reaction (or after the arrow), there are $x_3$ molecules of $CO_2$ and $1$ carbon atom in each molecule, which means there are $x_3$ many atoms of carbon after the reaction occurs. Because the number of carbon atoms must be the same before and after the reaction, we get the equation $2x_1=x_3\text{,}$ or restated in the form which has the constant after the equal sign as

\begin{equation*} 2x_1-x_3=0\text{.} \end{equation*}

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Next, we examine hydrogen, which is denoted by $H\text{.}$ Before the reaction there were $2$ hydrogen atoms in each of the $x_1$ molecules of $C_2H_2\text{,}$ for a total of $2x_1$ hydrogen atoms. After the reaction, there are $2$ hydrogen atoms in each of the $x_4$ molecules of $H_2O\text{,}$ for a total of $2x_4$ hydrogen atoms. This gives the equation $2x_1=2x_4\text{,}$ or moving the variables to the left side of the equal sign,

\begin{equation*} 2x_1-2x_4=0\text{.} \end{equation*}

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Finally, we look at oxygen, denoted by $O\text{.}$ Note that there is oxygen in both of the molecules that appear after the reaction. There were $2x_2$ oxygen atoms before the reaction took place, and after the reaction there are $2$ oxygen atoms in each of the $x_3$ molecules of $CO_2$ and $1$ oxygen atom in each of the $x_4$ molecules of $H_2O\text{,}$ which means there is a total of $2x_3+x_4$ oxygen atoms after the reaction. This gives us the equation $2x_2=2x_3+x_4\text{,}$ or

\begin{equation*} 2x_2-2x_3-x_4=0\text{.} \end{equation*}

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We put these equations into an augmented matrix and then find the reduced row echelon form

\begin{equation*} \begin{bmatrix} 1\amp 0 \amp -1\amp 0\amp 0\\2\amp 0 \amp 0\amp -2\amp 0\\0\amp 2\amp -2\amp -1\amp 0\\ \end{bmatrix} \rrefarrow \begin{bmatrix} 1\amp 0\amp 0\amp -1 \amp 0\\0\amp 1\amp 0\amp \frac{-5}{2}\amp 0\\0\amp 0\amp 1\amp -2 \amp 0\\ \end{bmatrix}\text{.} \end{equation*}

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Note that the variable $x_4$ is free, which means that mathematically, $x_4$ can take on any value. What happens if we let $x_4=1\text{?}$ The third row says that $x_3-2x_4=0\text{,}$ which means that $x_3=2\text{.}$ The second row says that $x_2-\frac{5}{2}x_4=0\text{,}$ which means that $x_2=\frac{5}{2}\text{,}$ and the first row says that $x_1-x_4=0\text{,}$ which means that $x_1=1\text{.}$

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However, it is impossible to have a fractional number of molecules. To clear the denominator, let’s multiply by $2$ and let $x_4=2$ instead of $1\text{.}$ Then the particular solution becomes

\begin{align*} x_1 \amp =2 \\ x_2 \amp =5 \\ x_3 \amp =4 \\ x_4 \amp =2 \end{align*}

and the balanced equation is

\begin{equation*} 2 C_2H_2 + 5 O_2 \longrightarrow 4 CO_2 + 2 H_2 O\text{.} \end{equation*}

Note that it makes sense to have infinitely many solutions to the linear system, since multiplying the whole thing by $2$ or $5$ or any whole positive number will keep the reaction balanced with equal numbers of each element’s atoms before and after the arrow.

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Reading Questions Reading Questions

1. Parsons Problem, Equation of a Quadratic Through 3 Points.

Select the correct steps to find the equation of the quadratic function that goes through the points $(-1,6)\text{,}$ $(1,2)$ and $(2,3)\text{.}$

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A quadratic function has equation \(y=ax^2+bx+c\text{.}\)🔗

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\(x\text{,}\) \(y\text{,}\) \(a\text{,}\) \(b\text{,}\) and \(c\) are all unknowns which need to be solved for.🔗
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\(x\) and \(y\) refer to \((x,y)\) points in the plane, so the unknowns to solve for are the coefficients \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\)🔗

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Passing through the point \((-1,6)\) means the equation holds when \(x=-1\) and \(y=6\text{,}\) so we get the equation \(6=a(-1)^2+b(-1)+c \text{.}\)🔗

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The two remaining points gives us the equations \(2=a(1)^2+b(1)+c\) and \(3=a(2)^2+b(2)+c \text{.}\)🔗

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Rewriting the set of linear equations gives us

\begin{align*}
a-b+c \amp = 6\\
a+b+c \amp = 2\\
4a+2b+c \amp = 3 \text{.}
\end{align*}

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Putting the equations into an augmented matrix and then row reducing gives

\begin{equation*}
\left[\begin{array}{cccc} 1\amp -1\amp 1\amp 6\\1\amp 1\amp 1\amp 2\\4\amp 2\amp 1\amp 3\\ \end{array} \right] \rrefarrow \left[\begin{array}{cccc} 1\amp 0\amp 0\amp 1\\0\amp 1\amp 0\amp -2\\0\amp 0\amp 1\amp 3\\ \end{array} \right]\text{.}
\end{equation*}

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The solution is \(a=1\text{,}\) \(b=-2\) and \(c=3\text{,}\) which is the quadratic function \(y=x^2-2x+3\text{.}\)🔗

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2. Parsons Problem, Balancing a Chemical Equation.

Select the steps to balance the chemical equation below.

\begin{equation*} \fillinmath{X}Ca(OH)_2 + \fillinmath{X}H_3PO_4 \longrightarrow \fillinmath{X} Ca_3(PO_4)_2 + \fillinmath{X}H_2O \end{equation*}

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We are looking for the values of \(x_1,x_2,x_3,x_4\) such that

\begin{equation*}
{\small x_1 \, Ca(OH)_2 + x_2 \, H_3PO_4 \longrightarrow x_3 \,  Ca_3(PO_4)_2 + x_4 \, H_2O }
\end{equation*}

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Looking at the element calcium \(Ca\text{,}\) we have \(x_1\) atoms before the reaction and \(3x_3\) atoms of calcium after the reaction takes place, so

\begin{equation*}
x_1-3x_3=0\text{.}
\end{equation*}

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Similarly, looking at the number of atoms of oxygen \(O\) before and after the reaction gives us the equation

\begin{equation*}
2x_1+4x_2-8x_3-x_4=0\text{,}
\end{equation*}

while examining hydrogen \(H\) gives the equation

\begin{equation*}
2x_1+3x_2-2x_4=0\text{,}
\end{equation*}

and considering phosphorus \(P\) results in

\begin{equation*}
x_2-2x_3=0\text{.}
\end{equation*}

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Putting these equations into an augmented matrix and then row reducing gives

\begin{equation*}
\left[\begin{array}{ccccc} 1\amp 0\amp -3\amp 0 \amp 0\\2\amp 4\amp -8\amp -1\amp 0\\2\amp 3\amp 0\amp -2\amp 0 \\ 0\amp 1\amp -2\amp 0\amp 0\end{array} \right] \rrefarrow \left[\begin{array}{ccccc} 1\amp 0\amp 0\amp \frac{-1}{2} \amp 0\\0\amp 1\amp 0\amp \frac{-1}{3} \amp 0\\0\amp 0\amp 1\amp \frac{-1}{6}\amp 0\\ \end{array} \right]
\end{equation*}

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Turning the reduced row echelon form back into equations results in

\begin{align*}
x_1 \amp =\frac{1}{2}x_4 \\
x_2 \amp =\frac{1}{3}x_4 \\
x_3 \amp =\frac{1}{6}x_4 
\end{align*}

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Letting \(x_4=36\) yields a final answer of

\begin{equation*}
{\small 18 \, Ca(OH)_2 + 12 \, H_3PO_4 \longrightarrow 6 \,  Ca_3(PO_4)_2 + 36 \, H_2O }\text{.}
\end{equation*}

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 #distractor
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Letting \(x_4=6\) yields a final answer of

\begin{equation*}
3 \, Ca(OH)_2 + 2 \, H_3PO_4 \longrightarrow   Ca_3(PO_4)_2 +  6\, H_2O\text{.}
\end{equation*}

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3. Reflection.

(a)

How confident do you feel with the material you just read about?

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1.
Not at all confident or didn’t do the reading.
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2.
Not very confident.
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3.
Somewhat confident.
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4.
Mostly confident.
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5.
Confident so far and ready to engage more deeply.
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(b)

Ask a question about the material. What additional information do you think someone would need to become more confident in their understanding?

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Worksheet Participate

Objectives

Be able to use linear systems and reduced row echelon form to solve problems in different contexts where the linear system is not immediately apparent.
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Be able to identify unknowns and constraints, or in other words, variables and equations involving those variables.
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Build on our past work interpreting solutions to a linear system from the reduced row echelon form by further restricting to solutions which make sense in the context of a given problem.
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1.

Suppose we know that in a football game there were $24$ points scored from $8$ scoring occasions and also that the number of successful extra point kicks was equal to the number of successful two point conversions. Find all ways in which the points may have been scored in this game.

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2.

Pure elemental phosphorous and hypochlorous acid $HClO$ react in water to make phosphoric acid $H_3PO_4$ and hydrochloric acid $HCl\text{.}$ Balance the chemical equation

Exercise Group.

In the following exercises, find the solution of the given problem by:

creating an appropriate system of linear equations
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forming the augmented matrix that corresponds to this system
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putting the augmented matrix into reduced row echelon form
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interpreting the reduced row echelon form of the matrix as a solution
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The general exponential function has the form $f(x)= ae^{bx}\text{,}$ where $a$ and $b$ are constants and $e$ is Euler’s constant ($\approx$ 2.718). We want to find the equation of the exponential function that goes through the points $(1,2)$ and $(2,4)\text{.}$

Show why we cannot simply substitute in values for $x$ and $y$ in the equation $y = ae^{bx}$ and solve using the techniques we used for polynomials.
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Show how the equality $y = ae^{bx}$ leads us to the equation $\ln(y) = \ln(a) + bx\text{,}$ which is linear in $x\text{.}$
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Use the techniques we developed to solve for the unknowns $\ln(a)$ and $b\text{.}$
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Knowing $\ln(a)\text{,}$ find $a\text{;}$ find the exponential function $f(x) = ae^{bx}$ that goes through the points $(1,2)$ and $(2,4)\text{.}$
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Answer.

$f(x) = e^{x\ln(2)}$

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16.

In a football game, $29$ points are scored from $8$ scoring occasions. There are $2$ more successful extra point kicks than successful two point conversions. Find all ways in which the points may have been scored in this game.

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Answer.

The augmented matrix from this system is $\left[\begin{array}{ccccc} 1\amp 1\amp 1\amp 1\amp 8\\6\amp 1\amp 2\amp 3\amp 29\\0\amp 1\amp -1\amp 0\amp 2\\ \end{array} \right]\text{.}$ From this we find the solution

\begin{align*} t\amp =4-\frac13f\\ x\amp =3-\frac13f\\ w\amp =1-\frac13f\text{.} \end{align*}

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The only time each of these variables are nonnegative integers is when $f=0$ or $f=3\text{.}$ If $f=0\text{,}$ then we have $4$ touchdowns, $3$ extra points and $1$ two point conversion (no field goals). If $f=3\text{,}$ then we have $3$ touchdowns, $2$ extra points and no two point conversions (and $3$ field goals).

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17.

In a basketball game, where points are scored either by a $3$ point shot, a $2$ point shot or a $1$ point free throw, there were $110$ points scored from $70$ successful shots. Find all ways in which the points may have been scored in this game.

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Answer.

Let $x_1\text{,}$ $x_2$ and $x_3$ represent the number of free throws, 2 point and 3 point shots taken. The augmented matrix from this system is $\left[\begin{array}{cccc} 1\amp 1\amp 1\amp 70\\1\amp 2\amp 3\amp 110 \end{array} \right]\text{.}$ From this we find the solution

\begin{align*} x_1\amp =30+x_3\\ x_2\amp =40-2x_3\text{.} \end{align*}

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In order for $x_2$ to be nonnegative, we need $x_3\leq 20\text{.}$ Thus there are $21$ different scenerios: the “first” is where $0$ three point shots are taken ($x_3=0\text{,}$ $30$ free throws and $40$ two point shots; the “last” is where $20$ three point shots are taken, $50$ free throws, and no two point shots.

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18.

Describe the equations of the linear functions that go through the point $(1,3)\text{.}$ Give $2$ examples.

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Answer.

Let $y = ax+b\text{;}$ all linear functions through $(1,3)$ come in the form $y = (3-b)x+b\text{.}$ Two possible examples could be: setting $b=0$ yields $y = 3x\text{;}$ and setting $b=2$ yields $y=x+2\text{.}$

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19.

Describe the equations of the linear functions that go through the point $(2,5)\text{.}$ Give $2$ examples.

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Answer.

Let $y = ax+b\text{;}$ all linear functions through $(2,5)$ come in the form $y = (2.5-\frac12b)x+b\text{.}$ Two possible examples could be: setting $b=1$ yields $y = 2x+1\text{;}$ and setting $b=-1$ yields $y=3x-1\text{.}$

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20.

Describe the equations of the quadratic functions that go through the points $(2,-1)$ and $(1,0)\text{.}$ Give $2$ examples.

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Answer.

Let $y = ax^2+bx+c\text{;}$ we find that $a = -\frac12+\frac12 c$ and $b = \frac12-\frac32c\text{.}$ Two possible examples could be: setting $c=1$ yields $y = -x+1\text{;}$ and setting $c=3$ yields $y=x^2-4x+3\text{.}$

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21.

Describe the equations of the quadratic functions that go through the points $(-1,3)$ and $(2,6)\text{.}$ Give $2$ examples.

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Answer.

Let $y = ax^2+bx+c\text{;}$ we find that $a = 2-\frac12 c$ and $b = -1+\frac12c\text{.}$ Two possible examples could be: setting $c=0$ yields $y = 2x^2-x\text{;}$ and setting $c=-2$ yields $y=3x^2-2x-2\text{.}$

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Section 1.6 Problem Solving with Linear Systems

Subsection Prepare

Example 1.6.1. Seats in a Baseball Stadium.

Steps for Problem Solving with Linear Systems.

An example with points scoring in football.

Activity 1.6.2. Which quantities should be variables?

Activity 1.6.3. Which solutions make sense?

(a)

(b)

Example 1.6.4. Balancing a chemical equation.

Reading Questions Reading Questions

1. Parsons Problem, Equation of a Quadratic Through 3 Points.

2. Parsons Problem, Balancing a Chemical Equation.

3. Reflection.

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(b)

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Objectives

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Summary.

Exercises Additional Practice

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Worksheet Participate