Word problems involving linear systems

If you are asked to solve the linear system

(1) $\begin{equation*} x+y=7 \end{equation*}$

(2) $\begin{equation*} 2x-y=2, \end{equation*}$

you will very easily accomplish this using any of the methods we considered before. What if the question was presented differently, in words, for example? It could be something like “Find two numbers whose sum is $7$ and when the second number is subtracted from twice the first number gives $2$ ”. Technically, the question hasn’t changed; it’s just the presentation or formulation — and the solver’s subsequent perception and interpretation — that may change.

As $x$ and $y$ begin to have physical “meanings”, the class of problems that a single pair of equations can model can be many.

Solution strategy

ensure that you understand the question;
identify the unknowns in the question (there will usually be two unknowns, since we’re dealing with linear systems in two variables);
use letters to represent the two unknowns;
formulate two linear equations based on the information provided in the question (this is a crucial step, and is one of the reasons you should endeavour to understand the question);
solve the linear equations formulated in the previous step and interpret the solution.

In the examples below, our focus will be on step 4 (since we’ve already covered the three standard methods of solving linear systems: elimination, substitution, and graphing).

Word problems involving integers

Example 1

Mr. Monday is thinking of two numbers. If he adds them together he obtains $16$ . If he multiplies the smaller number by $3$ and then subtracts the bigger number, he gets $12$ . What are the two numbers?

Let the two numbers be $m$ and $n$ . Assume that $n$ is the smaller number. Since their sum is $16$ , we have that $m+n=16$ . If we multiply $n$ by $3$ , we’ll have $3n$ ; then if we subtract $m$ we’ll have $3n-m$ . According to the question, $3n-m=12$ , and so we have the linear system

(3) $\begin{equation*} n+m=16 \end{equation*}$

(4) $\begin{equation*} 3n-m=12 \end{equation*}$

whose solution is $n=7,~m=9$ . Check that $7+9=16$ and that $3\times 7-9=21-9=12$ . So, Mr. Monday was thinking of $7$ and $9$ .

Example 2

The sum of the digits of a 3-digit number is $9$ . If the first and third digits are reversed, the number increases by $198$ . Find the 3-digit number, given that it’s middle digit is $3$ .

Let $x$ and $z$ be the other two digits. We’re already provided with the middle digit, so the number can take the form $x3z$ , where $x$ is the hundreds digit and $z$ is the ones digit. We have that $x+3+z=9$ , or $x+z=6$ . So we have our first equation.
To get the second equation, first note that any 3-digit number abc is actually $100a+10b+c$ . With this in mind, our original number is $100x+10\times 3 + z=100x+30+z$ . When the first and third digits are reversed, the number becomes $z3x$ , which is effectively $100z+10\times 3+x=100z+30+x$ . Since reversing the first and third digits increases the number by $198$ , we have that

$100z+30+x-(100x+30+z)=198;$

that is, $99z-99x=198$ , or just $z-x=2$ . Thus we have the linear system

(5) $\begin{equation*} z+x=6 \end{equation*}$

(6) $\begin{equation*} z-x=2 \end{equation*}$

whose solution is $z=4,~x=2$ . The 3-digit number is $234$ . Check that $2+3+4=9$ , and $432-234=198$ .

Word problems involving costs

A certain store sells $2$ pencils and $5$ pens for 3 dollars, while $6$ pencils and $1$ pen are sold for 2 dollars. Find the cost of a pencil and the cost of a pen.

Let the cost of a pencil be $c$ dollars and let the cost of a pen be $n$ dollars. According to the question, the cost of $2$ pencils and $5$ pens is three dollars, so we have our first equation: $2c+5n=3$ . The cost of $6$ pencils and $1$ pen is two dollars, so we have our second equation: $6c+n=2$ . The resulting linear system is:

(7) $\begin{equation*} 2c+5n=3 \end{equation*}$

(8) $\begin{equation*} 6c+n=2 \end{equation*}$

whose solution is $c=0.25,n=0.5$ . So the store sells a pencil for $25$ cents and a pen for $50$ cents.

Example 4

On a weekly basis, a STEM tutor normally spends seven hours tutoring both math and science, and makes $200$ dollars overall. If the hourly charges for math and science are $30$ and $25$ dollars, respectively, determine the number of hours spent on each subject.

Let’s suppose that the STEM tutor spends $m$ hours tutoring math and spends $s$ hours tutoring science each week. Since the tutor spends $7$ hours overall, we have that $m+s=7$ . Since the hourly rate for math is $30$ dollars, the total amount from math is $30m$ dollars each week; similarly, since the hourly rate for science is $25$ , the tutor makes $25s$ dollars from science tutoring. These amounts should add up to $200$ dollars, so we have $30m+25s=200$ , and then the linear system:

(9) $\begin{equation*} m+s=7 \end{equation*}$

(10) $\begin{equation*} 30m+25s=200 \end{equation*}$

Notice that the second equation can be reduced to $6m+5s=40$ . Solving the above linear system we find that $m=5$ and $s=2$ . So the tutor spends $5$ hours tutoring math, and spends $2$ hours tutoring science, each week.

Word problems involving coins

Example 5

If you have $30$ coins made up of quarters and nickels, and the total value of your money is $5.5$ dollars, how many quarters and how many nickels do you have?

Let the number of quarters be $q$ , and let the number of nickels be $n$ . Since you have $30$ coins, it follows that $n+q=30$ . Moreover, since a quarter is worth $25$ cents and a nickel is worth $5$ cents, the total value of your money, in cents, is $5n+25q$ . Therefore, $5n+25q=550$ (we’ve converted $5.5$ dollars to cents by multiplying it by $100$ ). So we have the following linear system

(11) $\begin{equation*} n+q=30 \end{equation*}$

(12) $\begin{equation*} 5n+25q=550,\quad\implies n+5q=110 \end{equation*}$

Solving, we find that $n=10$ and $q=20$ . So you have $10$ nickels and $20$ quarters.

Example 6

You have dimes and nickels that amount to $2$ dollars. If the number of nickels is double the number of dimes, how many of each denomination do you have?

Let’s suppose that the numbers of dimes and nickels are $d$ and $n$ , respectively. A dime is worth $10$ cents while a quarter is worth $25$ cents; thus, in terms of $d$ and $n$ , your money’s worth $10d+5n$ , and this must equal $200$ (again we’ve converted $2$ dollars to cents), so $10d+5n=200$ . Also, the number of nickels is double the number of dimes, meaning that $n=2d$ . We now have the linear system:

(13) $\begin{equation*} 10d+5n=200,\quad\textrm{or}~2d+n=40 \end{equation*}$

(14) $\begin{equation*} 2d=n \end{equation*}$

Solving, $d=10$ and $n=20$ . So you have $20$ nickels and $10$ dimes.

Word problems involving distance, speed, and time

The relationship between speed, distance, and time is:

$\boxed{\textrm{speed}=\frac{\textrm{distance}}{\textrm{time}}}$

The above relationship, together with its re-arrangements, will be useful in solving problems relating to distance, speed, and time.

Example 7

A runner has $2$ hours to complete the $15$ km straight line distance from point A to point B, with a point M in between. From A to M, the runner averaged $12$ km/h; from M to B, the runner averaged $6$ km/h. Determine the distance from A to M, and the distance from M to B.

Let the distance from A to M be $x$ , and let the distance from M to B be $y$ . We first have $x+y=15$ , since the distance from A to B is $15$ km. Next, we have to take the total time into consideration. From A to M, the time taken is $\frac{x}{12}$ , whereas $\frac{y}{6}$ is the time taken from M to B. Since the journey took $2$ hours to complete, we have that $\frac{x}{12}+\frac{y}{6}=2$ . Then follows the linear system:

(15) $\begin{equation*} x+y=15 \end{equation*}$

(16) $\begin{equation*} \frac{x}{12}+\frac{y}{6}=2 \end{equation*}$

whose solution is $x=6$ and $y=9$ . So the distance from A to M is $6$ km, and the distance from M to B is $9$ km.

Example 8

A boat traveled $240$ miles downstream in $12$ hours. The return journey took $48$ hours. Find the speed of the boat in still water and the speed of the current.

Ignoring units, let the boat’s speed in still water be $s$ , and let the speed of the current be $c$ . While travelling downstream, the boat is supported by the current, so the “resultant” speed is actually $s+c$ , which must equal $\frac{240}{12}$ . Our first equation is therefore $s+c=20$ . On the return journey, the boat is going against the current, so its “resultant” speed will be $s-c$ , which must equal $\frac{240}{48}$ ; that is, $s-c=5$ . We now have the linear system:

(17) $\begin{equation*} s+c=20 \end{equation*}$

(18) $\begin{equation*} s-c=5 \end{equation*}$

whose solution is $s=12.5$ and $c=7.5$ . Thus, the boat’s speed is $12.5$ miles per hour, while the current flows at $7.5$ miles per hour.

Word problems involving mixtures (weighted averages)

In a sense, many of the previous examples also fall into this “mixture” category.

Example 9

A certain high school mandates that for each subject, students final mark should comprise $t\%$ of their term average and $e\%$ of the final exam mark. In a math course, a student who had a term average of $90$ and an exam mark of $80$ obtained a final mark of $87$ . Determine the values of $t$ and $e$ .

As you’ll recall, this is an example of how “weighted averages” are used to calculate your grades. According to the question, the final mark is obtained by taking $t\%$ of the term average and taking $e\%$ of the exam mark, then adding them together. In our case:

$\Big(\frac{t}{100}\times 90\Big) + \Big(\frac{e}{100}\times 80\Big)=87;\quad\textrm{or,}~9t+8e=870.$

So far we only have one equation. The second information needed to obtain a second equation is somewhat implicit in the question: it is the fact percentages add up to $100$ , so $t+e=100$ . Combining these two equations we obtain the linear system:

(19) $\begin{equation*} 9t+8e=870 \end{equation*}$

(20) $\begin{equation*} t+e=100 \end{equation*}$

whose solution is $t=70$ and $e=30$ .

Example 10

You need to make $10$ litres of a $20\%$ acid solution by mixing $x$ litres of a $10\%$ acid solution with $y$ litres of a $30\%$ acid solution. Determine the values of $x$ and $y$ .

Since you need $10$ litres overall, it follows that $x+y=10$ . Furthermore, $10\%$ of $x$ plus $30\%$ of $y$ should give $20\%$ of $10$ ; that is, $0.1x+0.3y=2$ . The resulting linear system is:

(21) $\begin{equation*} x+y=10 \end{equation*}$

(22) $\begin{equation*} 0.1x+0.3y=2 \end{equation*}$

whose solution is $x=5$ and $y=5$ . So you need $5$ litres of the $10\%$ solution and $5$ litres of the $30\%$ solution. We’re sure you won’t confuse the double use of “solution” in this example — as a chemistry term, and as a mathematical term.

Exercises for the reader

Prove that if the digits of a two-digit number are reversed, then the number either increases or decreases by a multiple of $9$ .
Prove that if the first and third digits of a three-digit number are reversed, then the number either increases or decreases by a multiple of $99$ .