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Linear Equation

Unit 11 · CDC textbook pages 214–235

Hide the stories, facts, photos, activities and interactives, for revising before an exam
In this unit 7 sections
  1. 11.0 Review
  2. 11.1 Simultaneous linear equations with two variables
  3. 11.2 Methods of solving simultaneous linear equations with two variables
  4. Exercise 11.1
  5. 11.2.3 Word problems related to simultaneous equations
  6. How the exam asks it
  7. Exercise 11.2

About 1,600 years ago a Chinese maths book, the Sunzi Suanjing, set this puzzle: pheasants and rabbits are shut in one cage. Counted from above there are 35 heads; counted from below, 94 feet. How many pheasants are there, and how many rabbits?

Each fact alone has many answers: 35 heads could be 20 pheasants and 15 rabbits, or 30 and 5. Only one pair fits both facts at once. Two facts about two unknown numbers, true together, are called simultaneous equations, and this unit gives you three ways to find the pair that fits. By Back to the cage you will solve Sunzi’s puzzle in two lines, and see the trick he used.

Try it

Pheasants and rabbits

Each point is a guess: pheasants across, rabbits up. Drag the dot until both the heads and the feet are right.

0102030400102030pheasantsrabbits35 heads94 feet
Your guess
10 pheasants, 10 rabbits
Heads
20
Feet
60

Neither is right yet. Drag the dot (or use the arrow keys) until there are 35 heads.

The puzzle: MacTutor History of Mathematics, “Chinese problems”.

11.0 Review

A non-veg meal at a hotel costs Rs 200. Ronish needs Rs 50 more than he has to buy one. If he has Rs x, then
x+50=200 x + 50 = 200
Discuss with your friends:

  • a.What are the variables and the constants in this statement?
  • b.How much money does Ronish have to begin with?
  • c.Which sign joins the left-hand side to the right-hand side?
  • d.What is the power of x?

An equation in which every variable has power 1 (no x2 x^2 , no xy, no 1x \frac{1}{x} ) is a linear equation, or an equation of the first degree. “Linear” because, as you will see, its graph is a straight line. x + 50 = 200 has one variable and exactly one answer, x = 150. This unit is about linear equations with two variables, x and y.

Which of these is a linear equation?

Pick an answer to see why.

11.1 Simultaneous linear equations with two variables

Take the equation x + y = 3. On its own it has many answers: x = 3, y = 0; x = 2, y = 1; x = 1.5, y = 1.5; x = 5, y = −2… Every pair that works is a point (x, y), and all those points lie on one straight line. So one linear equation in two variables has a whole line of answers. To pin down one pair you need a second fact.

A linear equation is of the first degree, so its graph is a straight line. Two such equations about the same two variables, which must be true at the same time, are called simultaneous linear equations in two variables. Their solution is the pair of values of x and y that satisfies both: on a graph, the point where the two lines cross.

Is x = 4, y = 1 a solution of the simultaneous equations x + y = 5 and x − y = 3?

Pick an answer to see why.

11.2 Methods of solving simultaneous linear equations with two variables

That is the graphical method, which you used in earlier grades. A graph shows what a solution means, but reading it is slow, and a crossing at (1.37, 2.6) can’t be read exactly. The next two methods are algebra, and give the exact pair every time.

11.2.1 Substitution method

The idea: one equation tells you what one letter is in terms of the other. Put (substitute) that into the second equation, and it has only one letter left. The method is also called the replacement method.

Substitution, step by step:

  1. From one equation, write one letter in terms of the other (pick the letter whose coefficient is 1, if there is one: no fractions).
  2. Substitute it into the other equation and solve for the remaining letter.
  3. Put that value back to find the first letter.
  4. Check the pair in both equations: left-hand side (LHS) = right-hand side (RHS) in each.

11.2.2 Elimination method

The elimination method finds one variable by removing (eliminating) the other:

  1. Multiply one or both equations by numbers so that one letter has the same coefficient in both.
  2. If those terms have the same sign, subtract the equations; if opposite signs, add them. That letter disappears.
  3. Solve for the letter that is left, put it back into either equation for the other, and check in both.

Choose which letter to eliminate by the smaller numbers: in Activity 4, making the y’s equal (×2 and ×3) is easier than the x’s (×5 and ×4).

Try it

Make one letter vanish

Choose a number for each equation, then subtract or add. Try to make the x’s cancel, then the y’s.

(i) × 14x + 3y = 100 (ii) × 15x + 2y = 90 subtract−x + y = 10  

Both letters are still there. Choose the numbers so that the two x’s, or the two y’s, are the same, then subtract.

To eliminate y from 3x + 4y = 18 and 2x − 3y = 1, what should you do?

Pick an answer to see why.

Exercise 11.1

Question 1

Solve these simultaneous linear equations by the substitution method.

a.

3x + 5y = 31 and 2x − y = 12

Answer
b.

5x + 6y = 27 and 3x + 4y = 17

Answer
c.

3x − 2y = 11 and x + 3y = 11

Answer
d.

3x − 2y = 8 and x + 2y = 8

Answer
e.

4x − 3y + 1 = 0 and 3x + 2y − 12 = 0

Answer
f.

5x − y + 1 = 0 and 2x − 5y + 51 = 0

Answer
g.

9x − 8y = 12 and 2x + 3y = 17

Answer
h.

y = 5x − 23 and 3x − 2y = 4

Answer
i.

2x − y = 7 and x + y = 5

Answer
j.

3x + 2y = 15 and 5x − 3y − 25 = 0

Answer

Question 2

Solve these simultaneous linear equations by the elimination method, and check the result.

a.

x + y = 16 and x − y = −4

Answer
b.

3x − 2y = 4 and 5x − y = 23

Answer
c.

5x − 2y = 2 and 2x + 3y = 16

Answer
d.

2x + 5y + 7 = 0 and 2x − 2y = 14

Answer
e.

9x − 8y = 12 and 2x + 3y = 17

Answer
f.

3x + 4y = 17 and 5x + 6y = 27

Answer
g.

7x + 8y = −1 and 10x + 15y = −5

Answer
h.

4x − 16 = 3y and 5y = 12 − 3x

Answer
i.

3x − 3y + 6 = 0 and 4y − 2 − 2x = 0

Answer
j.

3x = 4y + 18 and 5x = 7y + 31

Answer

Question 3

Solve these pairs of simultaneous linear equations.

a.

3x + 4y = 2 and 5x + 3y + 4 = 0

Answer
b.

2x + 5y = 120 and 8x − 9y + 100 = 0

Answer
c.

6x5+7y5=1 \dfrac{6x}{5} + \dfrac{7y}{5} = 1 and 7x3+8y3=2 \dfrac{7x}{3} + \dfrac{8y}{3} = 2

Hint
Answer
d.

x2+y3=6 \dfrac{x}{2} + \dfrac{y}{3} = 6 and 3x8+1=2y3 \dfrac{3x}{8} + 1 = \dfrac{2y}{3}

Hint
Answer

Most uses of simultaneous equations start as words. The work is in turning each fact into an equation; the solving is the part you already know.

The words In algebra
the sum of x and y x + y
one more than three times their sum 3(x + y) + 1
x is 5 times y x = 5y
two less than twice y 2y − 2
ages a years from now x + a and y + a
ages a years ago x − a and y − a
a two-digit number with tens digit x and ones digit y 10x + y
the number with its digits swapped (interchanged, or reversed) 10y + x
a fraction with numerator x and denominator y xy \dfrac{x}{y}

The routine:

  1. Say what each letter stands for, with units (“let 1 kg of apples cost Rs x”).
  2. Write one equation for each fact.
  3. Solve by substitution or elimination.
  4. Check the answer against the words of the question (not only your equations, which could be wrong), and answer in a sentence.

Let there be p pheasants and r rabbits. A pheasant has 2 feet and a rabbit 4.
p+r=35(heads)2p+4r=94(feet) \begin{aligned} p + r &= 35 \quad \text{(heads)} \\ 2p + 4r &= 94 \quad \text{(feet)} \end{aligned}
Multiply the heads equation by 2 and subtract it from the feet equation: 2r = 24, so r = 12 and p = 23. There are 23 pheasants and 12 rabbits. Check: 23 + 12 = 35 heads, 46 + 48 = 94 feet ✓.

How the exam asks it

Grade 9’s final exam is set to CDC’s specification grid for Compulsory Mathematics, Grades 9–10. There, Algebra is 15 marks in three questions. In the grid’s Grade 9 model paper, simultaneous equations are question 10, a word problem about two prices:

Part Level Marks What it asks
a. understanding 1 write the statement as linear equations
b. application 2 solve them: the two prices
c. higher ability 2 use the prices in a new situation (here, after a price rise)

Write the letters’ meanings and the equations even in part b: the first mark there is for a correct method. Here is a question of our own in that shape.

Exercise 11.2

Question 1

a.

How do you write one more than three times the sum of two positive integers x and y?

Answer
b.

Anisha is x years old and Tashi is y years old now. Write their ages 10 years from now.

Answer
c.

Two brothers are 15 and 11 years old now. Write their ages m years ago.

Answer
d.

How do you write a two-digit number with x in the tens place and y in the ones place? What is the number with its digits swapped (the reversed number)?

Answer

Question 2

2.

The sum of two positive integers is 128 and their difference is 16. Find the numbers.

Answer

Question 3

3.

One number is 5 times the other, and their sum is 72. Find the numbers.

Answer

Question 4

4.

Two acute angles of a triangle add up to 105° and differ by 15°. Find them.

Answer

Question 5

5.

Bishal and Sumita bought the same number of chocolates at the same price. After Bishal ate 5 and Sumita 12, Bishal had 2 fewer than twice as many as Sumita. How many did each have at the start?

Answer

Question 6

6.

Five years ago a father was 5 times as old as his son. Now their ages add up to 46 years. Find their present ages.

Answer

Question 7

7.

15 years ago a father was 4 times as old as his daughter. In 5 years he will be twice as old as she will be. Find their present ages.

Answer

Question 8

8.

In one year a mother will be 4 times as old as her son. Two years ago she was three times as old as he will be four years from now. Find their present ages.

Answer

Question 9

9.

Three years ago the ages of Rajan and Nimesh were in the ratio 4 : 3. Three years from now they will be in the ratio 11 : 9. Find their present ages.

Answer

Question 10

10.

Ten years ago the ratio of a father’s age to his son’s was 11 : 3. In five years the father will be 10 years more than twice his son’s age then. In how many years will the son be as old as his father is now?

Answer

Question 11

11.

The numerator and denominator of a fraction add up to three less than twice the denominator. If 1 is taken from both the numerator and the denominator, the numerator becomes half the denominator. Find the fraction.

Answer

Question 12

12.

A two-digit number is three times the sum of its digits. If 45 is added to the number, its digits swap places. Find the number.

Answer

Question 13

13.

A two-digit number is larger than the number formed by swapping its digits. The two numbers add up to 121 and differ by 27. Find the number.
The textbook prints it without saying which of the two numbers is larger, and then its numbers don’t fit one answer only: 74 and 47 both work. Saying that the number is the larger, as here, leaves 74.

Answer

Question 14

14.

A number between 10 and 100 has digits that add up to 9. Seven times the number equals four times the number with its digits swapped. Compare its two digits as a percentage.

Answer

Question 15

15.

5 kg of onions and 7 kg of sugar cost Rs 810 in all, and 5 kg of onions cost the same as 2 kg of sugar. Find the price of 1 kg of each.

Answer

Question 16

16.

A rural municipality gives a school a rectangular piece of land whose length and breadth are in the ratio 11 : 4. Fencing it with one round of wire at Rs 100 a meter costs Rs 75,000. Find the length and breadth.

Answer

Question 17

17.

A school’s rectangular playground is one third as wide as it is long, and its perimeter is 32 m. Find its length and breadth.

Answer

Question 18

18.

Sambridi goes to Nepal Rastra Bank to change Rs 2,000 into new Rs 50 and Rs 100 notes. The bank gives her 25 notes in all. How many of each does she get?

Answer
Where these facts come from
  • Sunzi’s pheasants and rabbits (the opening, Back to the cage, Sunzi’s trick): MacTutor History of Mathematics, “Chinese problems” (35 heads, 94 feet; 23 pheasants and 12 rabbits; halve the feet and take away the heads), and Wikipedia, “Sunzi Suanjing” (written in the 3rd to 5th centuries CE), read November 2026.
  • The Nine Chapters (the story): Wikipedia, “The Nine Chapters on the Mathematical Art” (Chapter 8, fangcheng: problems in two to five unknowns, solved by a method indistinguishable from Gaussian elimination; rules for positive and negative numbers; composed no later than 93 CE), and “Gaussian elimination” (Gauss’s notation for it, 1810; named after him only in the 1950s), read November 2026.
  • The CT scanner (the fact in 11.1): Wikipedia, “Algebraic reconstruction technique” (an iterative solver of a system of linear equations for image reconstruction, first used for it by Gordon, Bender and Herman in 1970), read November 2026.
  • Counting rods (the photo’s caption): Wikipedia, “Counting rods” (red rods positive, black negative; in Japan, sangi laid on a grid-marked counting board), read November 2026. The photo: Wikimedia Commons (CC BY-SA 4.0), credited under the picture.
  • The exam’s question (For the exam, How the exam asks it): CDC, Specification Grid, Compulsory Mathematics, Grades 9–10 (Algebra 15 marks in three questions) and its Grade 9 model question 10 (1U, 2A, 2HA); our question is our own.
  • Every answer: worked out here and checked with a computer algebra program (SymPy).

A companion to your textbook, not a replacement for it: study from the book your school uses, and where the two differ, follow your teacher. Found a mistake? Please tell us.