Applications of Linear Systems

Linear systems really pay off in word problems. Whenever a situation has two unknown quantities and gives you two separate facts about them, you can write two equations and solve. This page shows how to turn words into a system for the most common kinds of problems: costs and break-even, money and investments, mixtures, and speeds with a current.

Key ideas

A plan for any word problem

1. Define your variables, with units: “Let $x$ be the amount invested at $3\%$, in dollars.”
2. Write two equations, one for each fact in the problem. Often one equation counts how many (or how much) and the other counts value (money, interest, amount of acid, distance).
3. Choose a method. Use substitution if a variable is already isolated, elimination if both equations are in the form $Ax + By = D$, and a graph when you want to compare two options visually.
4. Solve, and check your answer against the words of the problem, not just your equations. (If you set up an equation wrongly, your answer can satisfy it and still be wrong.)
5. Answer the question in a sentence, with units. Make sure the answer makes sense: you can’t sell $-3$ T-shirts or mix $-250$ mL of solution.

Common set-ups

Type of problem	Equation for “how many”	Equation for “value”
tickets, coins	adult $+$ student $=$ total tickets	price $\times$ number, added up $=$ total money
simple interest (one year)	amount at rate 1 $+$ amount at rate 2 $=$ total invested	$r_1 x + r_2 y =$ total interest, from $I = Prt$ with $t = 1$
mixtures	volume 1 $+$ volume 2 $=$ final volume	concentration $\times$ volume, added up $=$ concentration $\times$ final volume
current or wind	—	$(\text{speed} + c) \times \text{time} = \text{distance}$ and $(\text{speed} - c) \times \text{time} = \text{distance}$

For simple interest, $I = Prt$, where $P$ is the principal, $r$ is the annual rate as a decimal, and $t$ is the time in years. Write $4\%$ as $0.04$.

For a boat in a river with current $c$, going downstream the current helps, so the speed over the ground is $s + c$. Going upstream it works against you, so the speed is $s - c$. Wind does the same for a plane.

Costs, revenue and break-even

A business has fixed costs (paid no matter what, like equipment) and variable costs (paid for each item made). Its revenue is the money it takes in from sales. The break-even point is where cost equals revenue: the business has covered its costs but made no profit yet. Selling more than that gives a profit; selling fewer gives a loss.

Worked examples

Example 1: Break-even point

Maya starts a T-shirt printing business. The printing equipment costs $300, and each shirt costs $7 to make. She sells the shirts for $19 each. How many shirts must she sell to break even?

Solution. Let $n$ be the number of shirts. Let $C$ be the total cost and $R$ the revenue, both in dollars.

$$\begin{aligned} C &= 7n + 300 \\ R &= 19n \end{aligned}$$

At the break-even point, $C = R$. Both equations are solved for the money, so substitute (set them equal):

$$\begin{aligned} 19n &= 7n + 300 \\ 12n &= 300 \\ n &= 25 \end{aligned}$$

At $n = 25$, $R = 19(25) = 475$.

Check: $C = 7(25) + 300 = 175 + 300 = 475$ ✓. Cost and revenue are both $475.

Maya must sell $25$ shirts to break even. Each shirt after that adds $19 - 7 = 12$ dollars of profit.

Below $25$ shirts, cost is higher than revenue (a loss); above $25$, revenue is higher (a profit).

Example 2: Investing at two rates

Liam invests a total of $8000, part in a savings account that pays $3\%$ simple interest per year and the rest in a bond that pays $5\%$ simple interest per year. After one year he has earned $330 in interest. How much did he invest at each rate?

Solution. Let $x$ be the amount at $3\%$ and $y$ the amount at $5\%$, in dollars.

$$\begin{aligned} x + y &= 8000 && \text{total invested} \\ 0.03x + 0.05y &= 330 && \text{total interest} \end{aligned}$$

Multiply the second equation by $100$ to clear the decimals: $3x + 5y = 33\,000$. Multiply the first equation by $3$: $3x + 3y = 24\,000$. Subtract:

$$2y = 9000 \quad\Rightarrow\quad y = 4500$$

Then $x = 8000 - 4500 = 3500$.

Check with the words: $3500 + 4500 = 8000$ ✓, and the interest is $0.03(3500) + 0.05(4500) = 105 + 225 = 330$ ✓.

Liam invested $3500 at $3\%$ and $4500 at $5\%$.

Example 3: Mixing solutions

A science teacher has a $10\%$ acid solution and a $25\%$ acid solution. How much of each should she mix to make $600$ mL of a $15\%$ acid solution?

Solution. Let $x$ be the volume of $10\%$ solution and $y$ the volume of $25\%$ solution, in millilitres.

The volumes add up to $600$ mL. The pure acid in each part adds up to the pure acid in the final mixture, which is $0.15 \times 600 = 90$ mL:

$$\begin{aligned} x + y &= 600 \\ 0.10x + 0.25y &= 90 \end{aligned}$$

Multiply the second equation by $10$: $x + 2.5y = 900$. Subtract the first equation:

$$1.5y = 300 \quad\Rightarrow\quad y = 200$$

Then $x = 600 - 200 = 400$.

Check: $400 + 200 = 600$ ✓, and the acid is $0.10(400) + 0.25(200) = 40 + 50 = 90$ mL ✓.

She should mix $400$ mL of the $10\%$ solution with $200$ mL of the $25\%$ solution. That makes sense: $15\%$ is closer to $10\%$ than to $25\%$, so the mixture needs more of the $10\%$ solution.

Example 4: Paddling with and against the current

A canoeist paddles $36$ km downstream on the Ottawa River in $2$ hours. The return trip upstream takes $3$ hours. Find her paddling speed in still water and the speed of the current.

Solution. Let $s$ be her speed in still water and $c$ the speed of the current, in km/h.

Downstream her speed is $s + c$, and upstream it is $s - c$. Using $\text{speed} = \dfrac{\text{distance}}{\text{time}}$:

$$\begin{aligned} s + c &= \frac{36}{2} = 18 \\ s - c &= \frac{36}{3} = 12 \end{aligned}$$

Add the equations:

$$2s = 30 \quad\Rightarrow\quad s = 15$$

Then $15 + c = 18$, so $c = 3$.

Check: downstream, $(15 + 3) \times 2 = 36$ km ✓; upstream, $(15 - 3) \times 3 = 36$ km ✓.

She paddles at $15$ km/h in still water, and the current flows at $3$ km/h.

Common mistakes

Not defining the variables. “Let $x$ = acid” is too vague: is it a volume? A percent? Write exactly what each variable stands for, with units, before you write any equations.

Writing two equations that say the same thing. You need two different facts. In a mixture problem, one equation is about total volume and the other is about the amount of pure substance.

Forgetting to convert percents. $3\%$ is $0.03$, not $3$. Using $3x + 5y = 330$ in Example 2 would give nonsense.

Adding concentrations. Mixing $10\%$ and $25\%$ solutions doesn’t make a $35\%$ solution. Multiply each concentration by its volume to get the amount of pure acid, and add those amounts.

Mixing up upstream and downstream. Downstream the current adds to your speed ($s + c$); upstream it subtracts ($s - c$). The downstream trip should take less time.

Not checking whether the answer makes sense. Check your answer in the original words, and ask: is it positive? Should it be a whole number? Is it reasonable? A break-even point of $257.1$ loaves means you need to sell $258$.

Practice

1. (Warm-up) The sum of two numbers is $50$. One number is $4$ times the other. Write a system of equations and solve it.

Solution

Let the numbers be $x$ and $y$, with $x = 4y$.

$$\begin{aligned} x + y &= 50 \\ x &= 4y \end{aligned}$$

Substitute: $4y + y = 50$, so $5y = 50$ and $y = 10$. Then $x = 40$.

Check: $40 + 10 = 50$ ✓ and $40 = 4 \times 10$ ✓. The numbers are $40$ and $10$.

2. (Warm-up) A community theatre sold $120$ tickets for a show. Adult tickets cost $15 and child tickets cost $9. Ticket sales came to $1416. How many of each type were sold?

Solution

Let $a$ be the number of adult tickets and $c$ the number of child tickets.

$$\begin{aligned} a + c &= 120 \\ 15a + 9c &= 1416 \end{aligned}$$

From the first equation, $c = 120 - a$. Substitute:

$$\begin{aligned} 15a + 9(120 - a) &= 1416 \\ 15a + 1080 - 9a &= 1416 \\ 6a &= 336 \\ a &= 56 \end{aligned}$$

Then $c = 120 - 56 = 64$.

Check: $15(56) + 9(64) = 840 + 576 = 1416$ ✓. They sold $56$ adult tickets and $64$ child tickets.

3. (Core) A jar holds $40$ coins, all loonies ($1) and toonies ($2). The coins are worth $58 in total. How many of each coin are in the jar?

Solution

Let $l$ be the number of loonies and $t$ the number of toonies.

$$\begin{aligned} l + t &= 40 && \text{number of coins} \\ l + 2t &= 58 && \text{value in dollars} \end{aligned}$$

Subtract the first equation from the second: $t = 18$. Then $l = 40 - 18 = 22$.

Check: $22 + 18 = 40$ ✓ and the value is $22 + 2(18) = 58$ dollars ✓. There are $22$ loonies and $18$ toonies.

4. (Core) Priya invested $12 000, part at $4\%$ simple interest per year and the rest at $2.5\%$ simple interest per year. After one year, the total interest was $405. How much did she invest at each rate?

Solution

Let $x$ be the amount at $4\%$ and $y$ the amount at $2.5\%$, in dollars.

$$\begin{aligned} x + y &= 12\,000 \\ 0.04x + 0.025y &= 405 \end{aligned}$$

Multiply the second equation by $200$: $8x + 5y = 81\,000$. Multiply the first by $5$: $5x + 5y = 60\,000$. Subtract:

$$3x = 21\,000 \quad\Rightarrow\quad x = 7000$$

Then $y = 12\,000 - 7000 = 5000$.

Check: the interest is $0.04(7000) + 0.025(5000) = 280 + 125 = 405$ ✓. She invested $7000 at $4\%$ and $5000 at $2.5\%$.

5. (Core) A bulk food store sells peanuts for $6/kg and cashews for $15/kg. How many kilograms of each should be mixed to make $9$ kg of a mix that sells for $10/kg?

Solution

Let $p$ be the mass of peanuts and $c$ the mass of cashews, in kilograms. The mix is worth $9 \times 10 = 90$ dollars.

$$\begin{aligned} p + c &= 9 \\ 6p + 15c &= 90 \end{aligned}$$

Multiply the first equation by $6$: $6p + 6c = 54$. Subtract it from the second:

$$9c = 36 \quad\Rightarrow\quad c = 4$$

Then $p = 9 - 4 = 5$.

Check: the value is $6(5) + 15(4) = 30 + 60 = 90$ dollars ✓. Mix $5$ kg of peanuts with $4$ kg of cashews.

6. (Core) A small plane flies $1800$ km between two cities with the wind in $3$ hours. The return flight against the same wind takes $4$ hours. Find the speed of the plane in still air and the speed of the wind.

Solution

Let $s$ be the plane’s speed in still air and $w$ the wind speed, in km/h.

$$\begin{aligned} s + w &= \frac{1800}{3} = 600 \\ s - w &= \frac{1800}{4} = 450 \end{aligned}$$

Add: $2s = 1050$, so $s = 525$. Then $w = 600 - 525 = 75$.

Check: $(525 + 75) \times 3 = 1800$ ✓ and $(525 - 75) \times 4 = 1800$ ✓. The plane flies at $525$ km/h in still air, and the wind blows at $75$ km/h.

7. (Core) Phone plan A costs $35 a month plus $5 per gigabyte of data. Plan B costs $50 a month plus $2 per gigabyte.

• (a) Write an equation for the monthly cost $C$ (in dollars) of each plan for $g$ gigabytes.
• (b) For how many gigabytes do the plans cost the same?
• (c) Which plan is cheaper for someone who uses about $8$ GB a month?

Solution

(a) Plan A: $C = 5g + 35$. Plan B: $C = 2g + 50$.

(b) Set the costs equal:

$$5g + 35 = 2g + 50 \quad\Rightarrow\quad 3g = 15 \quad\Rightarrow\quad g = 5$$

Both plans cost $5(5) + 35 = 60$ dollars for $5$ GB. Check: $2(5) + 50 = 60$ ✓.

(c) Above $5$ GB, Plan A costs more, because its cost rises faster ($5 per GB versus $2). For $8$ GB: Plan A costs $5(8) + 35 = 75$ dollars and Plan B costs $2(8) + 50 = 66$ dollars. Plan B is cheaper.

8. (Challenge) A lab technician wants $500$ mL of a $20\%$ salt solution by mixing a $30\%$ solution with a $50\%$ solution. Set up and solve the system. What does the answer tell you?

Solution

Let $x$ be the volume of $30\%$ solution and $y$ the volume of $50\%$ solution, in millilitres. The final mixture contains $0.20 \times 500 = 100$ mL of salt.

$$\begin{aligned} x + y &= 500 \\ 0.30x + 0.50y &= 100 \end{aligned}$$

Multiply the second equation by $10$: $3x + 5y = 1000$. Multiply the first by $3$: $3x + 3y = 1500$. Subtract:

$$2y = -500 \quad\Rightarrow\quad y = -250$$

Then $x = 500 - (-250) = 750$.

The algebra is correct, but a volume can’t be negative. So it’s impossible to make this mixture. That makes sense: mixing a $30\%$ solution with a $50\%$ solution always gives something between $30\%$ and $50\%$, never $20\%$. Checking the answer in context is what caught this.

9. (Challenge) A bakery has fixed costs of $900 a month. Each loaf of bread costs $2.50 to make and sells for $6.

• (a) Write equations for the monthly cost $C$ and revenue $R$ (in dollars) for $n$ loaves.
• (b) How many loaves must the bakery sell each month to break even?
• (c) How many loaves must it sell to make a profit of $500 a month?

Solution

(a) $C = 2.5n + 900$ and $R = 6n$.

(b) Set $R = C$:

$$6n = 2.5n + 900 \quad\Rightarrow\quad 3.5n = 900 \quad\Rightarrow\quad n = \frac{900}{3.5} \approx 257.1$$

The bakery can’t sell part of a loaf. At $257$ loaves, revenue is $6(257) = 1542$ dollars and cost is $2.5(257) + 900 = 1542.50$ dollars, still a small loss. At $258$ loaves, revenue is $1548$ dollars and cost is $1545$ dollars, a profit. So the bakery must sell at least $258$ loaves.

(c) Profit is revenue minus cost, so set $R - C = 500$:

$$6n - (2.5n + 900) = 500 \quad\Rightarrow\quad 3.5n = 1400 \quad\Rightarrow\quad n = 400$$

Check: $R = 6(400) = 2400$ and $C = 2.5(400) + 900 = 1900$, so the profit is $2400 - 1900 = 500$ dollars ✓. The bakery must sell $400$ loaves.