Solving word problems with simultaneous equations

A farm keeps chickens and cows, $50$ animals in all. Nam counts $140$ legs altogether. How many chickens and how many cows are there on the farm?

The problem has two unknown numbers and two conditions. That is the sign that we should form a pair of simultaneous equations.

Three steps to solve a problem

Explore: suppose that…

Suppose all $50$ animals were chickens. Then there would be $50\cdot2=100$ legs, which is $140-100=40$ legs short. Each time we swap a chicken for a cow, the number of legs goes up by $2$ . So we need $40\div2=20$ swaps: there are $20$ cows and $30$ chickens.

The "suppose" trick is clever but hard to find for harder problems. Forming simultaneous equations is a general method that works for every problem.

The solution is where the two lines meet

Number problems and motion problems

In digit problems, a two-digit number with tens digit $a$ and units digit $b$ is written $\overline{ab}=10a+b$ .

A motion problem on a river has more data, so we work through the four steps.

Example A boat downstream and upstream

A motor boat travels $48$ km downstream in $2$ hours and $48$ km upstream in $3$ hours. Find the speed of the boat in still water and the speed of the current.

Solution.

Understand. Going downstream the current pushes the boat along; going upstream it holds the boat back.

Plan. Let $x$ be the speed of the boat in still water and $y$ the speed of the current (km/h, $x>y>0$ ). The downstream speed is $x+y$ and the upstream speed is $x-y$ .

Solve. The downstream speed is $48\div2=24$ km/h and the upstream speed is $48\div3=16$ km/h:

$\begin{cases} x+y=24\\ x-y=16 \end{cases}\Rightarrow 2x=40,\ x=20,\ y=4.$

Look back. $x=20>y=4>0$ , so the conditions hold. The boat travels at $20$ km/h in still water and the current flows at $4$ km/h.

Percentage problems

When a quantity $x$ increases by $15\%$ , the increase is $0.15x$ and the new value is $1.15x$ (not $x+15$ ).

Example Two production teams

Last month two production teams made $800$ products. This month team I made $15\%$ more than last month and team II made $20\%$ more, so together they made $945$ products. How many products did each team make last month?

Solution. Let $x$ and $y$ be the numbers of products made last month by team I and team II ( $x$ , $y$ positive whole numbers). Together the extra products come to $945-800=145$ :

$\begin{cases} x+y=800\\ 0.15x+0.2y=145 \end{cases}\iff\begin{cases} x+y=800\\ 3x+4y=2900 \end{cases}\Rightarrow y=2900-3\cdot800=500,\ x=300.$

So last month team I made $300$ products and team II made $500$ .

Check. $1.15\cdot300+1.2\cdot500=345+600=945$ .

Key terms

English	Tiếng Việt
solving word problems with simultaneous equations	giải bài toán bằng cách lập hệ phương trình
choose the unknowns	chọn ẩn
condition on the unknowns	điều kiện của ẩn
speed	vận tốc
downstream / upstream	xuôi dòng / ngược dòng
exceed (a previous level)	vượt mức
percentage concentration	nồng độ phần trăm
solution (mixture)	dung dịch

Summary

Step 1: choose two unknowns, state their conditions, write the quantities in terms of them, form two equations.
Step 2: solve the equations (substitution or elimination).
Step 3: check against the conditions, then answer, with units.
$\overline{ab}=10a+b$ ; downstream $x+y$ , upstream $x-y$ ; an increase of $p\%$ means multiplying by $1+\dfrac{p}{100}$ .

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

There are $36$ chickens and rabbits, with $100$ legs altogether. How many chickens and how many rabbits are there?

What you need and how to start

What you need: a chicken has $2$ legs and a rabbit has $4$ .

Steps:

Let $x$ be the number of chickens and $y$ the number of rabbits.
Form the equations from the number of animals and the number of legs, then solve.

Show solution

$x+y=36$ and $2x+4y=100$ , that is $x+2y=50$ . Subtracting: $y=14$ , $x=22$ .

Answer. $22$ chickens and $14$ rabbits.

Exercise 2

Class 9B visits a museum: $42$ people, teachers and students together. A student ticket costs $20$ thousand đồng, a teacher ticket costs $40$ thousand đồng, and the tickets cost $920$ thousand đồng in total. How many students and how many teachers are there?

What you need and how to start

What you need: the number of people and the cost give two equations.

Steps:

Let $x$ be the number of students and $y$ the number of teachers.
Form the equations, divide the cost equation by $20$ , then solve.

Show solution

$x+y=42$ and $20x+40y=920$ , that is $x+2y=46$ . Subtracting: $y=4$ , $x=38$ .

Answer. $38$ students and $4$ teachers.

Exercise 3

In a two-digit number, the tens digit is twice the units digit. If the two digits are swapped, the new number is $36$ less than the original number. Find the number.

What you need and how to start

What you need: $\overline{ab}=10a+b$ ; $(10a+b)-(10b+a)=9(a-b)$ .

Steps:

Form $a=2b$ and $9(a-b)=36$ .
Substitute $a=2b$ into the second equation.

Show solution

$a-b=4$ and $a=2b$ give $b=4$ , $a=8$ . Check: $84-48=36$ .

Answer. $84$ .

Exercise 4

A rectangular garden has perimeter $56$ m. If its length is increased by $2$ m and its width is decreased by $1$ m, its area stays the same. Find the length and the width of the garden.

What you need and how to start

What you need: perimeter $2(x+y)$ ; area $xy$ . When you expand $(x+2)(y-1)=xy$ , the $xy$ terms cancel.

Steps:

Let the length be $x$ and the width $y$ (m, $x\ge y>1$ ).
Form $x+y=28$ and $(x+2)(y-1)=xy$ , and simplify the second equation.
Solve the equations.

Show solution

$(x+2)(y-1)=xy\iff -x+2y-2=0\iff -x+2y=2$ . Adding $x+y=28$ : $3y=30$ , $y=10$ , $x=18$ . Check: $20\cdot9=180=18\cdot10$ .

Answer. Length $18$ m, width $10$ m.

Exercise 5

How many grams of a $10\%$ salt solution and how many grams of a $30\%$ salt solution must be mixed to make $200$ g of a $16\%$ salt solution?

What you need and how to start

What you need: mass of salt $=$ mass of solution $\times$ concentration; the total mass of solution and the total mass of salt do not change when you mix.

Steps:

Let $x$ and $y$ be the grams of the $10\%$ and the $30\%$ solutions.
Form the equations from the total mass and the total salt ( $16\%$ of $200$ g is $32$ g).

Show solution

$x+y=200$ and $0.1x+0.3y=32$ , that is $x+3y=320$ . Subtracting: $2y=120$ , $y=60$ , $x=140$ .

Answer. $140$ g of the $10\%$ solution and $60$ g of the $30\%$ solution.

Exercise 6 Review

Solve the chickens-and-rabbits problem of exercise 1 again by forming one equation in one unknown (as in Grade 8), then compare with the method of simultaneous equations.

What you need and how to start

What you need: if there are $x$ chickens, there are $36-x$ rabbits.

Steps:

Write the number of legs in terms of $x$ .
Solve the linear equation in one unknown.

Show solution

$2x+4(36-x)=100\iff 144-2x=100\iff x=22$ , so there are $14$ rabbits. This is really the substitution method: we wrote $y=36-x$ as soon as we chose the unknown.

Answer. $22$ chickens and $14$ rabbits.

Exercise 7

Last month two teams sewed $1000$ shirts. This month team A sewed $10\%$ more and team B sewed $5\%$ fewer than last month, so together they sewed $1040$ shirts.

a) Let $x$ and $y$ be the numbers of shirts sewn last month by team A and team B. Form the simultaneous equations.

b) Solve them to find $x$ and $y$ .

c) Which team sewed more this month, and by how many shirts?

What you need and how to start

What you need: $10\%$ more means multiplying by $1.1$ ; $5\%$ fewer means multiplying by $0.95$ .

Steps:

Form $x+y=1000$ and $1.1x+0.95y=1040$ .
Multiply the second equation by $20$ , then substitute $x=1000-y$ .
Work out each team's shirts this month.

Show solution

a) $\begin{cases} x+y=1000\\ 1.1x+0.95y=1040 \end{cases}$

b) Multiplying by $20$ : $22x+19y=20800$ . Substituting $x=1000-y$ : $22000-3y=20800$ , $y=400$ , $x=600$ .

c) Team A: $1.1\cdot600=660$ shirts; team B: $0.95\cdot400=380$ shirts; the difference is $280$ shirts.

Answer. b) $x=600$ , $y=400$ ; c) team A sewed $280$ more shirts.

Exercise 8 Challenge

Hà cycles from home to school. At $15$ km/h she arrives $6$ minutes before the start of lessons; at $12$ km/h she arrives $3$ minutes late. Find the distance from home to school and the time Hà needs to arrive exactly on time.

What you need and how to start

What you need: distance $=$ speed $\times$ time; in hours, $6$ minutes $=0.1$ h and $3$ minutes $=0.05$ h.

Steps:

Let $s$ be the distance (km) and $t$ the time to arrive exactly on time (hours), $t>0.1$ .
Form $s=15(t-0.1)$ and $s=12(t+0.05)$ .
Solve by substitution.

Show solution

$15(t-0.1)=12(t+0.05)\iff 15t-1.5=12t+0.6\iff 3t=2.1\iff t=0.7$ . Then $s=15\cdot0.6=9$ . Check: $12\cdot0.75=9$ .

Answer. The distance is $9$ km; the time is $0.7$ hours $=42$ minutes.

a) $x+y=30$ .	True	False
b) $4x+2y=84$ .	True	False
c) There are $12$ rabbits.	True	False
d) There are $6$ more chickens than rabbits.	True	False

Three steps to solve a problem

Explore: suppose that…

Number problems and motion problems

Percentage problems

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer