Math 8 · 8.EE.C.8C · A.CED.A.3

Systems in word problems

You know how many there were in all and what they came to in all, but not how many of each kind. How do you find the two amounts?

You need

Solving systems by eliminationMath 8

Where you are 7 in Systems of equations

Algebra9

Systems of equations7

▴ 2 above Systems of linear inequalities4 Solving systems by elimination Systems with a curve Systems in three variables

Systems in word problems

When you reach for it

A museum knows how many people came in on Saturday and how much money it took, but not how many were adults. A shop knows how heavy its trail mix is and what it cost, but not how much of it is hazelnuts. Each time there are two amounts you do not know and two separate facts that tie them together. One fact alone allows many pairs: 230 admissions could be 100 adults and 130 children, or 200 and 30. Two facts usually pin down one pair, and two equations that must hold at once are a system.

Solving is the work of graphing, substitution and elimination. This lesson teaches the two ends: turning a story into two equations, and turning the solution back into an answer the story accepts.

The procedure

Day 1: a quantity equation and a value equation

Sefton’s jar holds only 10-cent and 25-cent coins. There are 32 coins, worth 5.30 dollars in all. How many of each are there?

Name the two unknowns, with their units. Let d be the number of 10-cent coins and q the number of 25-cent coins. Saying what each letter counts is what lets you check that each equation adds like things to like things.

Write the quantity equation. The story says how many coins there are: d + q = 32. Both terms count coins, so adding them is allowed.

Write the value equation. Each kind of coin contributes (how many) \times (value of one):

Coin	Number	Each (cents)	Total (cents)
10-cent	d	10	10d
25-cent	q	25	25q
Jar	32		530

The totals add up to the value of the jar:

10d + 25q = 530

The 5.30 dollars became 530 cents because every term of an equation must be in the same unit. The two equations are two different facts, the Number column and the Total column, so neither can stand in for the other.

Solve by any method. The quantity equation gives d = 32 - q, so substitute:

\begin{aligned} 10(32 - q) + 25q &= 530 \\ 320 + 15q &= 530 \\ q &= 14 \end{aligned}

and then d = 32 - 14 = 18.

Check against the story, not against your equations. 18 + 14 = 32 coins, and 180 + 350 = 530 cents. A check in the story also catches an equation written wrongly.

Mixtures have the same two columns. For a trail mix the Number column holds kilograms and the value of one is the price of a kilogram. For a drink the Number column holds liters of drink, the value of one is the fraction of each liter that is juice, and the total is liters of pure juice: a drink that is 20\% juice (Percent) has 0.20 liters of juice in every liter.

Figure 1The liters of drink add and the liters of juice add, 30 + 10 = 40 and 3 + 5 = 8. The percents do not add.

The liters of drink add and the liters of juice add. The percents never add: a 10\% drink mixed with a 50\% drink cannot give a 60\% drink, any more than 5-dollar raisins and 15-dollar hazelnuts make a 20-dollar mix.

Day 2: two lines of money, and speeds with a current

Break-even. A business pays some costs whatever it sells, its fixed costs, and some for each item. With fixed costs F dollars, each item costing v dollars to make and selling for p dollars, the cost and the income (also called revenue) for x items are

C = F + vx \qquad R = px

Graphed against x, both are straight lines. The break-even point is where they meet: cost equals income. To its left the cost line is higher, a loss; to its right, a profit. Setting R = C gives (p - v)x = F. The number p - v is what each sale brings in above its own cost, and the break-even count is how many sales it takes to pay the fixed costs.

Speeds with a current. Distance is speed times time (Speed, distance and time). A boat moving at k km/h in still water, on a river whose current runs at c km/h, is carried by the water: downstream its speed over the ground is k + c, upstream k - c.

Figure 2The current carries the kayak. Against it the speed over the ground is k - c; with it, k + c.

A wind does the same to a plane: a tailwind adds to its speed, a headwind takes away. Each trip gives speed times time equals distance; dividing by the time gives the two speeds over the ground, and the system

\begin{aligned} k + c &= \text{speed downstream} \\ k - c &= \text{speed upstream} \end{aligned}

has opposite coefficients on c, exactly the case where adding the equations removes a letter.

Judge whether the answer makes sense. The equations do not know that counts are whole numbers, that a weight cannot be negative, or that the current must be slower than the boat. An answer that fails one of these solves the system but not the story (Example 7).

Worked examples

Examples 1 to 3 are Day 1’s work; Examples 4 to 7 are Day 2’s.

Example 1

Admissions at a museum

A museum charges 14 dollars for an adult and 6 dollars for a child. On one Saturday it sold 230 admissions and took 2{,}500 dollars. How many adults and how many children came?

Solution

Let a be the number of adult admissions and c the number of child admissions. The count and the money are the two facts:

\begin{aligned} a + c &= 230 \\ 14a + 6c &= 2500 \end{aligned}

The first equation gives c = 230 - a. Substituting into the second:

\begin{aligned} 14a + 6(230 - a) &= 2500 \\ 14a + 1380 - 6a &= 2500 \\ 8a &= 1120 \end{aligned}

So a = 140 and c = 230 - 140 = 90: 140 adults and 90 children.

Check. 140 + 90 = 230 admissions, and 140 \times 14 = 1960 and 90 \times 6 = 540 dollars, with 1960 + 540 = 2500.

The middle line has a meaning: 230 children would take 1380 dollars, and each adult in place of a child adds 14 - 6 = 8 dollars, so 1120 \div 8 = 140 adults.

Example 2

Trail mix by weight and price

Odalis makes 20 kg of trail mix from hazelnuts at 15 dollars per kg and raisins at 5 dollars per kg. She wants the mix to cost 9 dollars per kg. How much of each should she use?

Solution

Let h be the kilograms of hazelnuts and r the kilograms of raisins.

	kg	Dollars per kg	Dollars
Hazelnuts	h	15	15h
Raisins	r	5	5r
Mix	20	9	180

The 9 is the price of one kilogram, a value of one like the 14 dollars of an adult ticket. All 20 kg cost 20 \times 9 = 180 dollars, and that total is what the two costs add up to:

\begin{aligned} h + r &= 20 \\ 15h + 5r &= 180 \end{aligned}

Substitute r = 20 - h: 15h + 5(20 - h) = 180, so 10h + 100 = 180. So h = 8 and r = 12: 8 kg of hazelnuts and 12 kg of raisins.

Check. 8 + 12 = 20 kg, and 120 + 60 = 180 dollars. The answer also makes sense: 9 is nearer the raisin price than the hazelnut price, so the mix should be mostly raisins, and it is.

Example 3

Mixing two juice drinks

Niall has one drink that is 10\% fruit juice and another that is 50\% fruit juice. How many liters of each should he mix to make 40 liters of a drink that is 20\% fruit juice?

Solution

Let x be the liters of the 10\% drink and y the liters of the 50\% drink. The quantity equation counts liters of drink. The value equation counts liters of pure juice: 0.10x from the first, 0.50y from the second, and 0.20 \times 40 = 8 liters in the mix.

\begin{aligned} x + y &= 40 \\ 0.10x + 0.50y &= 8 \end{aligned}

Multiplying both sides of the second equation by 10 clears the decimals without changing its solutions: x + 5y = 80. Substitute x = 40 - y:

\begin{aligned} (40 - y) + 5y &= 80 \\ 4y &= 40 \end{aligned}

So y = 10 and x = 30: 30 liters of the 10\% drink and 10 liters of the 50\% drink.

Check. 30 + 10 = 40 liters; juice 3 + 5 = 8 liters, and \displaystyle \frac{8}{40} = 0.20, which is 20\%. These are the three bars in the figure above.

Example 4

A food truck breaking even

Mirela’s food truck costs 500 dollars a day to run whatever it sells, and each meal costs 5 dollars in ingredients. She sells each meal for 11 dollars.

(a) Write her daily cost C and income R in terms of the number of meals m.

(b) How many meals must she sell in a day to break even?

Solution

(a) C = 500 + 5m and R = 11m, both in dollars.

Figure 3Cost C = 500 + 5m and income R = 11m. Left of the crossing the cost line is higher and the truck loses money; right of it, it makes money.

(b) Break-even is where the lines meet, the solution of y = 500 + 5m, y = 11m. Both give y, so set the two expressions equal:

\begin{aligned} 11m &= 500 + 5m \\ 6m &= 500 \\ m &= \frac{250}{3} = 83\tfrac{1}{3} \end{aligned}

There is no third of a meal, so look at the whole numbers either side. With 83 meals the income is 913 dollars and the cost 915, a loss of 2. With 84 meals the income is 924 and the cost 920, a profit of 4. She must sell at least 84 meals.

Example 5

A kayak on a river

Piotr paddles his kayak 15 km up a river, against the current, in 2.5 hours. Coming back down the same 15 km, with the current, takes 1.5 hours. Find his speed in still water and the speed of the current.

Solution

Let k be his speed in still water and c the current’s, both in km/h. Speed times time is distance on each trip:

\begin{aligned} 2.5(k - c) &= 15 \\ 1.5(k + c) &= 15 \end{aligned}

Dividing each by its time gives his two speeds over the ground, k - c = 6 and k + c = 10. The coefficients of c are opposites, so add: 2k = 16, so k = 8, and c = 10 - 8 = 2. He paddles at 8 km/h and the current runs at 2 km/h.

Check. Upstream at 6 km/h for 2.5 hours is 15 km; downstream at 10 km/h for 1.5 hours is 15 km. The current is slower than the kayak, as it must be for him to get upstream.

Example 6

A plane with and against the wind

A plane flies 1{,}500 km between two cities. Against the wind the trip takes 3 hours; with the wind, the return takes 2.5 hours. Take the wind as steady and straight along the route, and the plane’s speed through the air as the same on both legs.

(a) Find the plane’s speed in still air and the speed of the wind.

(b) With no wind, how long would the round trip take? Did the wind cost time or save it?

Solution

(a) Let p be the plane’s speed in still air and w the wind’s, in km/h. Then 3(p - w) = 1500 and 2.5(p + w) = 1500. Dividing by the times:

\begin{aligned} p - w &= 500 \\ p + w &= 600 \end{aligned}

Adding, 2p = 1100, so p = 550 and w = 600 - 550 = 50. Check: 500 \times 3 = 1500 and 600 \times 2.5 = 1500.

(b) In still air each leg takes \displaystyle \frac{1500}{550} = \frac{30}{11} hours, so the round trip takes

\frac{60}{11} \approx 5.45 \text{ hours}

about 5 hours 27 minutes. With the wind it took 5.5 hours, longer by \displaystyle 5.5 - \frac{60}{11} = \frac{1}{22} of an hour, a little under 3 minutes. The wind pushes and holds back by the same 50 km/h, but the slow leg lasts longer. Exercise 23 shows that, at a fixed speed through the air, every steady wind along the route lengthens the round trip.

Example 7

Which totals the prices can produce

The museum of Example 1 charges 14 dollars an adult and 6 dollars a child; the trail mix of Example 2 uses hazelnuts at 15 dollars per kg and raisins at 5. For each pair of totals, solve the system and decide whether real counts or amounts can give those totals.

(a) 200 admissions and 2{,}130 dollars.

(b) 200 admissions and 1{,}100 dollars.

Solution

(a) With c = 200 - a, 14a + 6(200 - a) = 2130 gives 8a + 1200 = 2130, so a = 116.25. No count of people is 116.25. The reason shows in the equation: the takings from 200 admissions are 1200 + 8a, which is 1200 plus a multiple of 8, and 930 is not a multiple of 8.

(b) Now 8a + 1200 = 1100, so a = -12.5, with c = 212.5. A count cannot be negative. The least 200 admissions can take is 1200 dollars, all children, so 1100 is out of reach.

(c) The system is h + r = 20 and 15h + 5r = 20 \times 17 = 340. With r = 20 - h, 10h + 100 = 340, so h = 24 and r = -4. A weight cannot be negative. A mix of 15-dollar and 5-dollar ingredients costs between 5 and 15 dollars per kg (Exercise 20), so 17 is out of reach.

In all three the system has exactly one solution. What fails is that the solution is not a count or an amount, so no real admissions or mix gives those totals.

Boundaries

Two unknowns need two separate facts. If the second fact only repeats the first in other words (“230 admissions” and “230 visitors”), the equations describe the same line and there are infinitely many pairs. If the facts contradict each other there are none.

Only totals add. Numbers of things add, and total values add; values of one (prices, percents, coin values) do not. A speed over the ground is the one sum of speeds here, allowed because the boat’s own motion and the river’s happen at once along the same line.

Every term of an equation has the same unit. A time in minutes is turned into hours before it meets a speed in km/h.

The story’s conditions are not in the equations. Counts are whole numbers, amounts are not negative, a mix’s price or percent lies between its ingredients’, and a current is slower than the boat. Each is checked after solving.

Break-even needs a price above the cost of one item. If p = v the two lines are parallel and never meet; if p < v the income line falls further behind. Either way no number of sales pays the fixed costs.

The models assume things stay steady: prices that do not change with quantity, constant speeds and currents, and a wind straight along the route.

Where this goes next

Three kinds of ticket and three facts make a system in three unknowns. When the story says “at least” or “at most” the equations become inequalities and the answer is a region, not a point: that is Systems of linear inequalities, and choosing the best point in such a region is Linear programming. Later, the price of a mix returns as a weighted mean in probability, and a wind blowing across the route is handled with Vectors.

Exercises

Part 1: counting and value

For each, name the two unknowns with their units, write a system, solve it, and check the answer in the story.

A jar holds only 5-cent and 20-cent coins: 40 coins worth 4.40 dollars in all. How many of each are there?
A school play sold 310 tickets and took 3{,}080 dollars. Student tickets cost 8 dollars and adult tickets 12 dollars. How many of each were sold?
A purse holds 27 coins, all of them 1-dollar coins or 50-cent coins, worth 19.50 dollars in all. How many of each are there?
A field holds hens and sheep. Between them they have 30 heads and 84 legs. How many hens and how many sheep are there?

Part 1: mixtures

Cashews cost 18 dollars per kg and peanuts 6 dollars per kg. How much of each makes 15 kg of a mix costing 10 dollars per kg?
One orange drink is 20\% juice and another is 60\% juice. How many liters of each make 50 liters of a drink that is 36\% juice?
How many liters of pure juice (100\% juice) and how many liters of a 25\% juice drink make 12 liters of a drink that is 50\% juice?
Without solving the system, decide whether 9 kg of hazelnuts and 11 kg of raisins make the mix in Example 2.

Part 2: break-even and comparing costs

A club pays a pottery studio 360 dollars to make a mold, then 4 dollars to cast each mug. It sells each mug for 10 dollars. How many must it sell to break even, and what are its cost and its income at that point?
Sefton’s pie stall costs 250 dollars a day to run, each pie costs 3 dollars to make, and he sells pies at 7 dollars each. How many pies must he sell in a day to break even?
One trampoline park charges 20 dollars for grip socks plus 8 dollars a visit; another charges 12 dollars a visit and lends socks free. For how many visits do the two cost the same, and which is cheaper for 7 visits?

Part 2: speeds with a current or a wind

A rowing boat travels 18 km downstream in 2 hours and returns upstream in 3 hours. Find the boat’s speed in still water and the speed of the current.
A plane flies 2{,}100 km with the wind in 3 hours and makes the same trip against the wind in 3.5 hours. Find the plane’s speed in still air and the wind’s speed.
A boat moves at 12 km/h in still water on a river whose current runs at 3 km/h. It travels downstream, turns, and comes back to where it started, and the whole trip takes 4 hours. How long did each leg take, and how far downstream did it go?
Rhiannon rows 4 km downstream in 30 minutes and the same 4 km back upstream in 1 hour. Find her speed in still water and the current’s speed, in km/h.

Part 2: does the answer make sense

Solve each system, then say whether real counts or amounts can give these totals.

The museum of Example 1: 260 admissions and 1{,}500 dollars.
10 liters of a 30\% juice drink, made from a 10\% drink and a 25\% drink.
A boat that moves at 8 km/h downstream and 12 km/h upstream.
Badges at 4 dollars and 7 dollars: 50 badges for 280 dollars.

Reasoning

A mix is made of a kg of one food at A dollars per kg and r kg of another at B dollars per kg, where B < A and both amounts are positive. Explain why its price per kg is between B and A, and use this to explain Example 7(c) and Exercise 17.
Explain why a boat’s speed in still water is half the sum of its downstream and upstream speeds, and the current is half their difference. What would an upstream speed of 0 mean?
Explain why the break-even answer in Example 4 is 84 meals and not 83, even though 83\tfrac{1}{3} is nearer to 83.
A plane flies d km each way at p km/h in still air, with a steady wind of w km/h along the route, where 0 < w < p. Show that the round trip takes longer than it would in still air.

Answers

24 five-cent coins and 16 twenty-cent coins. n + t = 40 and 5n + 20t = 440 cents give 200 + 15t = 440.
160 student tickets and 150 adult tickets. s + a = 310 and 8s + 12a = 3080 give 2480 + 4a = 3080.
12 one-dollar coins and 15 fifty-cent coins. o + h = 27 and 100o + 50h = 1950 cents give 2700 - 50h = 1950.
18 hens and 12 sheep. h + s = 30 and 2h + 4s = 84 give 60 + 2s = 84; the value of one is legs per animal.
5 kg of cashews and 10 kg of peanuts. c + p = 15 and 18c + 6p = 150 give 12c + 90 = 150.
30 liters of the 20\% drink and 20 liters of the 60\% drink. x + y = 50 and 0.2x + 0.6y = 18 give 10 + 0.4y = 18.
4 liters of pure juice and 8 liters of the 25\% drink. j + d = 12 and j + 0.25d = 6; subtracting, 0.75d = 6.
No. The weight is right, 9 + 11 = 20 kg, but the cost is 135 + 55 = 190 dollars, not 180: 9.50 dollars per kg.
60 mugs, where cost and income are both 600 dollars. 10x = 360 + 4x gives 6x = 360.
At least 63 pies. 7x = 250 + 3x gives x = 62.5; 62 pies take 434 dollars against a cost of 436, and 63 take 441 against 439.
5 visits, when both cost 60 dollars; for 7 visits the first park is cheaper, 76 dollars against 84. 20 + 8n = 12n gives 4n = 20.
7.5 km/h in still water and a current of 1.5 km/h. b + c = 9 and b - c = 6.
The plane flies at 650 km/h and the wind blows at 50 km/h. p + w = 700 and p - w = 600.
1.5 hours downstream and 2.5 hours back, and it went 22.5 km downstream. Speeds 15 and 9 km/h: t + u = 4 and 15t = 9u give 24t = 36.
6 km/h in still water and a current of 2 km/h. In 0.5 hours and 1 hour, k + c = 8 and k - c = 4.
No: a = -7.5. 14a + 6(260 - a) = 1500 gives 8a = -60; even 260 children take 1560 dollars.
No: \displaystyle x = -\frac{10}{3} and \displaystyle y = \frac{40}{3}. x + y = 10 and 0.1x + 0.25y = 3 give 0.15y = 2; an amount cannot be negative.
No: b = 10 and c = -2. A current running downstream makes the downstream speed the larger; c = -2 would mean the water flows the other way.
No: the 7-dollar badges would number \displaystyle s = \frac{80}{3}. 4(50 - s) + 7s = 280 gives 3s = 80, and a count is a whole number.
The price per kg is \displaystyle \frac{Aa + Br}{a + r}. Since B < A and r > 0, Aa + Br < A(a + r), so the price is below A; likewise Aa + Br > B(a + r), so it is above B. So no mix of 15- and 5-dollar ingredients costs 17 dollars per kg, and no mix of 10\% and 25\% drinks is 30\% juice.
Adding k + c = D and k - c = U gives 2k = D + U; subtracting gives 2c = D - U. An upstream speed of 0 means the current is as fast as the boat: it holds still against the bank.
The profit 6m - 500 grows with m and is 0 only at m = 83\tfrac{1}{3}, so every whole number up to 83 gives a loss and 84 is the first that does not.
With the wind the round trip takes \displaystyle \frac{d}{p - w} + \frac{d}{p + w} = \frac{2dp}{p^2 - w^2} hours; in still air, \displaystyle \frac{2d}{p} = \frac{2dp}{p^2}. Since 0 < w < p, the denominator p^2 - w^2 is positive and smaller than p^2, so the first fraction is larger.