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Simultaneous Equations

Simultaneous equations are two or more algebraic equations that share common variables and are solved at the same time (that is, simultaneously). For example, equations x + y = 5 and x - y = 6 are simultaneous equations as they have the same unknown variables x and y and are solved simultaneously to determine the value of the variables. We can solve simultaneous equations using different methods such as substitution method, elimination method, and graphically.

In this article, we will explore the concept of simultaneous equations and learn how to solve them using different methods of solving. We shall discuss the simultaneous equations rules and also solve a few examples based on the concept for a better understanding.

What are Simultaneous Equations?

Simultaneous equations are two or more algebraic equations with the same unknown variables and the same value of the variables satisfies all such equations. This implies that the simultaneous equations have a common solution. Some of the examples of simultaneous equations are:

• 2x - 4y = 4, 5x + 8y = 3
• 2a - 3b + c = 9, a + b + c = 2, a - b - c = 9
• 3x - y = 5, x - y = 4
• a 2 + b 2 = 9, a 2 - b 2 = 16

We can solve such a set of equations using different methods. Let us discuss different methods to solve simultaneous equations in the next section.

Solving Simultaneous Equations

We use different methods to solve simultaneous equations. Some of the common methods are:

• Substitution Method
• Elimination Method
• Graphical Method

Simultaneous equations can have no solution, an infinite number of solutions, or unique solutions depending upon the coefficients of the variables. We can also use the method of cross multiplication and determinant method to solve linear simultaneous equations in two variables . We can add/subtract the equations depending upon the sign of the coefficients of the variables to solve them.

To solve simultaneous equations, we need the same number of equations as the number of unknown variables involved. We shall discuss each of these methods in detail in the upcoming sections with examples to understand their applications properly.

Simultaneous Equations Rules

To solve simultaneous equations, we follow certain rules first to simplify the equations. Some of the important rules are:

• Simplify each side of the equation first by removing the parentheses, if any.
• Combine the  like terms .
• Isolate the variable terms on one side of the equation.
• Then, use the appropriate method to solve for the variable.

Solving Simultaneous Equations Using Substitution Method

Now that we have discussed different methods to solve simultaneous equations. Let us solve a few examples using the substitution method to understand it better. Consider a system of equations x + y = 4 and 2x - 3y = 9. Now, we will find the value of one variable in terms of another variable using one of the equations and substitute it into the other equation. We have

x + y = 4 --- (1)

2x - 3y = 9 --- (2)

From (1), we have

x = 4 - y --- (3)

Substituting this in (2), we get

2(4 - y) - 3y = 9

⇒ 8 - 2y - 3y = 9

⇒ 8 - 5y = 9

Isolating the variable term to one side of the equation, we have

⇒ -5y = 9 - 8

⇒ y = 1/(-5)

Substituting the value of y in (3), we have

x = 4 - (-1/5)

= (20 + 1)/5

Answer: So, the solution of the simultaneous equations x + y = 4 and 2x - 3y = 9 is x = 21/5 and y = -1/5.

Solving Simultaneous Equations By Elimination Method

To solve simultaneous equations by the elimination method, we eliminate a variable from one equation using another to find the value of the other variable. Let us solve an example to understand find the solution of simultaneous equations using the elimination method. Consider equations 2x - 5y = 3 and 3x - 2y = 5. We have

2x - 5y = 3 --- (1)

and 3x - 2y = 5 --- (2)

Here, we will eliminate the variable y, so we find the LCM of the coefficients of y. LCM (5, 2) = 10. So, multiply equation (1) by 2 and equation (2) by 5. So, we have

[ 2x - 5y = 3 ] × 2

⇒ 4x - 10y = 6 --- (3)

[ 3x - 2y = 5 ] × 5

⇒ 15x - 10y = 25 --- (4)

Now, subtracting equation (3) from (4), we have

(15x - 10y) - (4x - 10y) = 25 - 6

⇒ 15x - 10y - 4x + 10y = 19

⇒ (15x - 4x) + (-10y + 10y) = 19

⇒ 11x + 0 = 19

⇒ x = 19/11

Now, substituting this value of x in (1), we have

2(19/11) - 5y = 3

⇒ 38/11 - 5y = 3

⇒ 5y = 38/11 - 3

⇒ 5y = (38 - 33) / 11

⇒ y = 5/(11×5)

So, the solution of the simultaneous equations 2x - 5y = 3 and 3x - 2y = 5 using the elimination method is x = 19/11 and y = 1/11.

Solving Simultaneous Equations Graphically

In this section, we will learn to solve the simultaneous equations using the graphical method. We will plot the lines on the coordinate plane and then find the point of intersection of the lines to find the solution. Consider simultaneous equations x + y = 10 and x - y = 4. Now, find two points (x, y) satisfying for each equation such that the equation holds.

For x + y = 10, we have

So, we have coordinates (0, 10) and (10, 0). Plot them and join the points and plot the line x + y = 10.

For equation x - y = 4, we have

So, we have coordinates (0, -4) and (4, 0). Plot them and join the points and plot the line x - y = 4.

Now, as we have plotted the two lines, find their intersecting point. The two lines x + y = 10 and x - y = 4 intersect each other at (7, 3). So, we have found the solution of the simultaneous equations x + y = 10 and x - y = 4 graphically which is x = 7 and y = 3.

Important Notes on Simultaneous Equations

• Simultaneous equations are two or more algebraic equations that share common variables and are solved at the same time.
• Simultaneous equations can be solved using different methods such as substitution method, elimination method, and graphically.
• We can also use the cross multiplication and determinant method to solve simultaneous linear equations in two variables.

☛ Related Articles:

• Solutions of a Linear Equation
• Simultaneous Linear Equations

Simultaneous Equations Examples

Example 1: Solve the simultaneous equations 2x - y = 5 and y - 4x = 1 using the appropriate method.

Solution: To solve 2x - y = 5 and y - 4x = 1, we will use the elimination method as it is easy to eliminate the variable y by adding the two equations. So, we have

2x - y = 5 --- (1)

y - 4x = 1 --- (2)

Adding (1) and (2), we get

(2x - y) + (y - 4x) = 5 + 1

⇒ 2x - y + y - 4x = 6

Substitute this value of x in (1)

2(-3) - y = 5

⇒ -6 - y = 5

⇒ y = -6 - 5

Answer: Solution of simultaneous equations 2x - y = 5 and y - 4x = 1 is x = -3 and y = -11.

Example 2: Find the solution of the simultaneous equations 2x - 4y + z = 2, x + 5y - 3z = 7, 3x + 2y - z = 10 using the substitution method.

Solution: We have

2x - 4y + z = 2 --- (1)

x + 5y - 3z = 7 --- (2)

3x + 2y - z = 10 --- (3)

z = 2 - 2x + 4y

Substituting this value of z in (2) and (3),

x + 5y - 3(2 - 2x + 4y) = 7

⇒ x + 5y - 6 + 6x - 12y = 7

⇒ 7x - 7y = 13 --- (4)

3x + 2y - (2 - 2x + 4y) = 10

3x + 2y - 2 + 2x - 4y = 10

⇒ 5x - 2y = 12 --- (5)

Now, solving the two-variable equations (4) and (5), multiply (4) by 2 and (5) by 7, we have

[7x - 7y = 13 ] × 2 and [5x - 2y = 12 ] × 7

⇒ 14x - 14y = 26 and 35x - 14y = 84

Now, subtracting the above two equations, we have

(14x - 14y) - (35x - 14y)= 26 - 84

⇒ 14x - 35x - 14y + 14y = -58

⇒ -21x = -58

⇒ x = 58/21 --- (A)

Substitute the value of x in (5)

5(58/21) - 2y = 12

⇒ 290/21 - 2y = 12

⇒ 2y = 290/21 - 12

= (290 - 252)/21

⇒ y = 19/21 --- (B)

Substituting the values of x and y in z = 2 - 2x + 4y, we have

z = 2 - 2(58/21) + 4(19/21)

= (42 - 116 + 76)/21

= 2/21 --- (C)

From (A), (B), (C), we have x = 58/21, y = 19/21, and z = 2/21

Answer: Solution is x = 58/21, y = 19/21, and z = 2/21.

Example 3: Find the solution of simultaneous equations x - y = 10 and 2x + y = 9.

Solution: We will solve the given equations using the elimination method.

Adding x - y = 10 and 2x + y = 9, we have

(x - y) + (2x + y) = 10 + 9

⇒ x + 2x - y + y = 19

So, we have

19/3 - y = 10

⇒ y = 19/3 - 10

= (19 - 30)/3

Answer: The solution of x - y = 10 and 2x + y = 9 is x = 19/3 and y = -11/3.

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Simultaneous Equations Questions

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FAQs on Simultaneous Equations

Simultaneous equations are two or more algebraic equations that share common variables and are solved at the same time (that is, simultaneously).

How to Solve Simultaneous Equations?

What is the substitution method in simultaneous equations.

According to the substitution method, we obtain the value of one variable in terms of another and then substitute that into another equation to find the value of the other variable.

What is the Rule for Simultaneous Equations?

Some of the important rules of simultaneous equations are:

• Combine the like terms.

What are Linear Simultaneous Equations?

Linear simultaneous equations refer to simultaneous equations where the degree of the variables is one.

How to Solve 3 Simultaneous Equations?

We can solve 3 simultaneous equations using various methods such as:

It also depends upon the number of variables involved.

What are the Three Methods to Solve Simultaneous Equations?

The three methods to solve simultaneous equations are:

Solving Simultaneous Equations: Worksheets with Answers

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System of Linear Equations (Simultaneous Equations)

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• Hemang Agarwal
• Andrew Ellinor
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• Yuxuan Seah
• Mahindra Jain
• Aditya Virani
• A Former Brilliant Member
• Abhishu Brahmecha
• Lii Gang Hah
• Chung Kevin

A system of linear equations is a collection of linear equations which involve the same set of variables. As an example,

\[ \begin{align} x+2y & =2 \\ -x+y & =1 \end{align} \]

is a system of equations that has two variables \(x\) and \(y.\) The solution to a linear system is an assignment of numbers to the variables that satisfy every equation in the system. For a given system, we could have one solution, no solutions or infinitely many solutions.

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Number line.

• x+y+z=25,\:5x+3y+2z=0,\:y-z=6
• x+2y=2x-5,\:x-y=3
• 5x+3y=7,\:3x-5y=-23
• x^2+y=5,\:x^2+y^2=7
• xy+x-4y=11,\:xy-x-4y=4
• 3-x^2=y,\:x+1=y
• xy=10,\:2x+y=1
• substitution\:x+2y=2x-5,\:x-y=3
• elimination\:x+2y=2x-5,\:x-y=3
• How to solve linear Simultaneous equations with two variables by graphing?
• To solve linear simultaneous equations with two variables by graphing, plot both equations on the same set of axes. The coordinates of the points at which the two lines intersect are the solutions to the system.
• What are Simultaneous Equations?
• Simultaneous equations are a set of equations that are solved at the same time. These equations are used to define the relationships between variables and can have multiple solutions, a single solution, or no solution.
• What are Simultaneous Equations used for?
• Simultaneous equations can be used to solve a wide range of problems in finance, science, engineering, and other fields. They are often used to find the values of variables that make multiple equations or expressions true at the same time.
• What are the methods for solving Simultaneous Equations?
• The common methods for solving simultaneous equations are Graphing, Substitution, and Elimination. The choice of method depends on the specific equations and the desired solution.

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• Middle School Math Solutions – Simultaneous Equations Calculator Solving simultaneous equations is one small algebra step further on from simple equations. Symbolab math solutions...

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Simultaneous Equations Questions

Simultaneous equations questions with solutions are given here for practice and understanding how to determine the solution of given equations. Simultaneous equations are the system of two or more independent equations which satisfies a common solution. Simultaneous equations could have,

• Linear Equation
• Quadratic Equation
• Polynomial Equations

There are several methods of solving simultaneous equations, and they are:

Learn more about Simultaneous Equations .

Video Lesson on Consistent and Inconsistent Simultaneous Equations

Simultaneous Equations Questions with Solutions

Let us solve a few simultaneous equations questions.

Question 1:

Solve the following:

2x + 3y = –2

5x + 4y + 2 = 0

2x + 3y = –2 ….(i)

5x + 4y = –2 ….(ii)

Multiply (i) by 5 and (ii) 2 on both sides, subtracting (ii) from (i), we get

10x + 15y = –10

10x + 8y = –4

(–)_ (–)___(+)___

7y = –6

⇒ y = –6/7

Substituting the value of y in (i), we get

∴ the solution of the given simultaneous equation is (2/7, –6/7)

Question 2:

Solve the following equations:

2a – b = 4 and 2b – 3a = 2

2a – b = 4 …..(i)

2b – 3a = 2

⇒ –3a + 2b = 2 ….(ii)

a = ½ (b + 4)

Putting this value in (ii)

–3/2(b + 4) + 2b = 2

⇒ –3b – 12 + 4b = 4

⇒ b = 16

Thus, a = ½(16 + 4) = 10

∴ the solution of the given simultaneous equation is a = 10 and b = 16.

Question 3:

Solve the following simultaneous equations:

7m – 9n = 3 and 2m – 3n = 1

7m – 9n = 3 ….(i)

2m – 3n = 1 ….(ii)

Multiplying equation (ii) by 3 on both sides, then subtracting from (i), we get,

(7m – 6m) + ( –9n + 9n) = 3 – 3

⇒ m = 0

Consequently, n = –⅓

∴ the solution of the given simultaneous equation is m = 0 and n = –⅓.

Question 4:

a 2 + 2b = 9 and b – a = 3

a 2 + 2b = 9 …(i)

b – a = 3 …(ii)

From (ii) b = 3 + a

Substituting the value of b in (i), we get

a 2 + 2(3 + a) = 9

⇒ a 2 + 2a + 6 = 9

⇒ a 2 + 2a + – 3 = 0

⇒ (a + 3)(a – 1) = 0

⇒ a = 1 and –3

When a = 1, b = 4, and when a = –3, b = 0

∴ the solution of the given simultaneous equation is a = 1, b = 4 and a = –3, b = 0.

Also, learn about solving simultaneous equations by Cramer’s rule .

Question 5:

a 2 – b = 14 and 2b – 4 = 12a

a 2 – b = 14 ….(i)

2b – 4 = 12a

⇒ 12a – 2b = –4

⇒ 6a – b = –2

⇒ b = 6a + 2

a 2 – 6a – 2 = 14

⇒ a 2 – 6a – 16 = 0

⇒ (a – 8)(a + 2) = 0

⇒ a = 8 and –2

When a = 8, b = 50 and when a = –2, b = 10

∴ the solution of the given simultaneous equation is a = 8, b = 50 and a = –5, b = 11.

Question 6:

8q + p + r = 0

q + 2p + r = 0

q + p – 1600 = 0

8q + p + r = 0 …(i)

q + 2p + r = 0 …(ii)

q + p – 1600 = 0 …(iii)

Subtracting (ii) from (i), we get

7q – p = 0 ….(iv)

Solving (iii) and (iv), we get,

q = 200 and p = 1400

And from (i), we get r = –3000

Question 7:

(m + n – 8)/2 = (m + 2n – 14)/3 = (3m + n – 12)/11

Taking the first two equations:

(m + n – 8)/2 = (m + 2n – 14)/3

⇒ 3m + 3n – 24 = 2m + 4n – 28

⇒ m – n = – 28 + 24

⇒ m – n = –4 ….(i)

Taking the last two equations:

(m + 2n – 14)/3 = (3m + n – 12)/11

⇒ 11m + 22n – 154 = 9m + 3n – 36

⇒ 2m + 19n = 154 – 36

⇒ 2m + 19n = 118 ….(ii)

Multiplying both sides of (i) by 2 and then subtracting from (ii), we get

19n + 2n = 118 + 8

⇒ 21n = 126

⇒ n = 126/21 = 6

∴ the solution of the given simultaneous equation is m = 2 and n = 6.

The other methods of solving simultaneous equations are:

• Gaussian Elimination
• Jacobi and Gauss-Seidel Iterative Methods

Question 8:

A man can row 16 km downstream and 8 km upstream in 6 hours. He can row 6 km upstream and 24 km downstream in 6 hours. Find the speed of man in still water.

Let the speed of a man rowing a boat in still water be x km/hr and the speed of the stream be y km/hr.

Speed of rowing in downstream = x + y = 1/u (let)

Speed of rowing in upstream = x – y = 1/v

16/(x + y) + 8/(x – y) = 6

⇒ 16u + 8v = 6

⇒ 8u + 4v = 3 ….(i)

And 24/(x + y) + 6/(x – y) = 6

⇒ 24u + 6v = 6

⇒ 4u + v = 1 ….(ii)

Solving (i) and (ii) we get, u = ⅛ ⇒ x + y = 8 and v = ½ ⇒ x – y = 2

Thus, we get x = 5km/hr and y = 3km/hr.

∴ The speed of the man rowing the boat in still water is 5 km/hr.

Question 9:

If (sin 𝜃, cos 𝜃) satisfies the system of equations mx + ny + a + b = a – b and nx + my + 2b = 0, then find the value of 𝜃 where 0≤ 𝜃 ≤ 90 o .

Given equations,

mx + ny + a + b = a – b

⇒ mx + ny + 2b = 0 …..(i)

And nx + my + 2b = 0 ….(ii)

Multiply (i) by n and (ii) by m on both sides, subtracting (ii) from (i), we get,

(n 2 – m 2 )y + 2nb – 2mb = 0

y = –2/(m + n)

Similarly, x = –2/(m + n)

That is, x = y ⇒ sin 𝜃 = cos 𝜃, for 𝜃 = 45 o , the value of sin 𝜃 is equal to the value of cos 𝜃.

Question 10:

A sum of ₹ 400 was distributed among the students of a class. Each boy received ₹ 8, and each girl received ₹ 4. If each girl had received ₹ 10, then each boy would have received ₹ 5. Find the total number of students in the class.

Let the number of boys be x, and the number of girls be y. Then

8x + 4y = 400 and 5x + 10y = 400

⇒ 2x + y = 100 and x + 2y = 80

Multiplying the second equation by 2 on both sides, then subtracting from the first equation, we get,

Simultaneous Linear Equations

To remember the process of framing simultaneous linear equations from mathematical problems

●  To remember how to solve simultaneous equations by the method of comparison and method of elimination

●  To acquire the ability to solve simultaneous equations by the method of substitution and method of cross-multiplication

●  To know the condition for a pair of linear equations to become simultaneous equations

●  To acquire the ability to solve mathematical problems framing simultaneous equations We know that if a pair of definite values of two unknown quantities satisfies simultaneously two distinct linear equations in two variables, then those two equations are called simultaneous equations in two variables. We also know the method of framing simultaneous equations and two methods of solving these simultaneous equations. 

We have already learnt that linear equation in two variable x and y is in the form ax + by + c = 0.

Where a, b, c are constant (real number) and at least one of a and b is non-zero. 

The graph of linear equation ax + by + c = 0 is always a straight line.

Every linear equation in two variables has an infinite number of solutions. Here, we will learn about two linear equations in 2 variables. (Both equations having to same variable i.e., x, y) Simultaneous linear equations: Two linear equations in two variables taken together are called simultaneous linear equations.

The solution of system of simultaneous linear equation is the ordered pair (x, y) which satisfies both the linear equations. Necessary steps for forming and solving simultaneous linear equations Let us take a mathematical problem to indicate the necessary steps for forming simultaneous equations: In a stationery shop, cost of 3 pencil cutters exceeds the price of 2 pens by $2. Also, total price of 7 pencil cutters and 3 pens is $43. Follow the steps of instruction along with the method of solution. Step I: Indentify the unknown variables; assume one of them as x and the other as y

Here two unknown quantities (variables) are:

Price of each pencil cutter = $x

Price of each pen = $y

Step II: Identify the relation between the unknown quantities.

Price of 3 pencil cutter =$3x

Price of 2 pens = $2y

Therefore, first condition gives: 3x – 2y = 2

Step III: Express the conditions of the problem in terms of x and y

Again price of 7 pencil cutters = $7x

Price of 3 pens = $3y

Therefore, second condition gives: 7x + 3y = 43

Simultaneous equations formed from the problems:

3x – 2y = 2 ----------- (i)

7x + 3y = 43 ----------- (ii)

For examples: (i) x + y = 12 and x – y = 2 are two linear equation (simultaneous equations). If we take x = 7 and y = 5, then the two equations are satisfied, so we say (7, 5) is the solution of the given simultaneous linear equations. (ii) Show that x = 2 and y = 1 is the solution of the system of linear equation x + y = 3and 2x + 3y = 7 Put x = 2 and y = 1 in the equation x + y = 3

L.H.S. = x + y = 2 + 1 = 3, which is equal to R.H.S. In 2ⁿᵈ equation , 2x + 3y = 7, put x = 2 and y = 1 in L.H.S.

L.H.S. = 2x + 3y = 2 × 2 + 3 × 1 = 4 + 3 = 7, which is equal to R.H.S.

Thus, x = 2 and y = 1 is the solution of the given system of equations.

Worked-out problems on solving simultaneous linear equations: 1. x + y = 7         ………… (i)

  3x - 2y = 11      ………… (ii) Solution: The given equations are:

x + y = 7      ………… (i)

3x - 2y = 11      ………… (ii) From (i) we get y = 7 – x

Now, substituting the value of y in equation (ii), we get;

3x - 2 (7 - x) = 11

or, 3x - 14 + 2x = 11

or, 3x + 2x - 14 = 11

or, 5x - 14 = 11

or, 5x -14 + 14 = 11 + 14 [add 14 in both the sides]

or, 5x = 11 + 14

or, 5x = 25

or, 5x/5 = 25/5 [divide by 5 in both the sides]

or, x = 5 Substituting the value of x in equation (i), we get;

Put the value of x = 5

or, 5 + y = 7

or, 5 – 5 + y = 7 – 5

or, y = 7 – 5

or, y = 2 Therefore, (5, 2) is the solution of the system of equation x + y = 7 and 3x – 2y = 11

2. Solve the system of equation 2x – 3y = 1 and 3x – 4y = 1. Solution: The given equations are:

2x – 3y = 1      ………… (i)

3x – 4y = 1      ………… (ii)

From equation (i), we get;

2x = 1 + 3y

or, x = ¹/₂(1 + 3y) Substituting the value of x in equation (ii), we get;

or, 3 × ¹/₂(1 + 3y) – 4y = 1

or, ³/₂ + ⁹/₂y - 4y = 1

or, (9y – 8y)/2 = 1 - ³/₂

or, ¹/₂y = (2 – 3)/2

or, ¹/₂y = \(\frac{-1}{2}\)

or, y = \(\frac{-1}{2}\) × \(\frac{2}{1}\)

Substituting the value of y in equation (i) 

2x – 3 × (-1) = 1

or, 2x + 3 = 1

or, 2x = 1 - 3 or, 2x = -2

or, x = -2/2

or, x = -1 Therefore, x = -1 and y = -1 is the solution of the system of equation

2x – 3y = 1 and 3x – 4y = 1 .

●   Simultaneous Linear Equations

Comparison Method

Elimination Method

Substitution Method

Cross-Multiplication Method

Solvability of Linear Simultaneous Equations

Pairs of Equations

Word Problems on Simultaneous Linear Equations

Practice Test on Word Problems Involving Simultaneous Linear Equations

●   Simultaneous Linear Equations - Worksheets

Worksheet on Simultaneous Linear Equations

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4.1: Simultaneous linear equations and symmetry

• Last updated
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• Page ID 23457

• Alexandre Borovik & Tony Gardiner
• University of Manchester & University of Birmingham via Open Book Publishers

Problem 92 Dad took our new baby to the clinic to be weighed. But the baby would not stay still and caused the needle on the scales to wobble. So Dad held the baby still and stood on the scales, while nurse read off their combined weight: 78kg. Then nurse held the baby, while Dad read off their combined weight: 69kg. Finally Dad held the nurse, while the baby read off their combined weight: 137kg. How heavy was the baby?

The situation described in Problem 92 is representative of a whole class of problems, where the given information incorporates a certain symmetry, which the solver would be wise to respect. Hence one should hesitate before applying systematic brute force (as when using the information from one weighing to substitute for one of the three unknown weights – a move which effectively reduces the number of unknowns, but which fails to respect the symmetry in the data).

A similar situation arises in certain puzzles like the following.

Problem 93 Numbers are assigned (secretly) to the vertices of a polygon. Each edge of the polygon is then labelled with the sum of the numbers at its two end vertices.

(a) If the polygon is a triangle ABC , and the labels on the three sides are c (on AB ), b (on AC ), and a (on BC ), what were the numbers written at each of the three vertices?

(b) If the polygon is a quadrilateral ABCD, and the labels on the four sides are w (on AB), x (on BC), y (on CD ), and z (on DA ), what numbers were written at each of the four vertices?

(c) the polygon is a pentagon ABCDE, and the labels on the five sides are d (on AB ), e (on BC ), a (on CD ), b (on DE ), and c (on EA ), what numbers were written at each of the five vertices?

In case any reader is inclined to dismiss such problems as “artificial puzzles”, it may help to recall two familiar instances (Problems 94 and 96 ) which give rise to precisely the above situation.

Problem 94 In the triangle ABC with sides of lengths a (opposite A ), b (opposite B ), and c (opposite C ), we want to locate the three points where the incircle touches the three sides - at point P (on BC ), Q (on CA ), and R (on AB ). To this end, let the two tangents to the incircle from A (namely AQ and AR) have length x, the two tangents from B (namely BP and BR) have length y, and the two tangents from C (namely CP and CQ) have length z . Find the values of x, y, z in terms of a, b, c.

The second instance requires us first to review the basic properties of midpoints in terms of vectors.

(a) Write down the coordinates of the midpoint M of the line segment joining Y = (a, b) and Z = (c, d). Justify your answer.

(b) Position a general triangle XYZ so that the vertex X lies at the origin (0,0). Suppose that Y then has coordinates (a, b) and Z has coordinates (c, d). Let M be the midpoint of XY , and N be the midpoint of XZ . Prove the Midpoint Theorem, namely that

“ MN is parallel to YZ and half its length”.

(c) Given any quadrilateral ABCD, let P be the midpoint of AB , let Q be the midpoint of BC , let R be the midpoint of CD , and let S be the midpoint of DA . Prove that PQRS is always a parallelogram.

(a) Suppose you know the position vectors p, q, r corresponding to the midpoints of the three sides of a triangle. Can you reconstruct the vectors x, y, z corresponding to the three vertices?

(b) Suppose you know the vectors p, q, r, s corresponding to the midpoints of the four sides of a quadrilateral. Can you reconstruct the vectors w, x, y, z corresponding to the four vertices?

(c) Suppose you know the vectors p, q, r, s, t corresponding to the midpoints of the five sides of a pentagon. Can you reconstruct the vectors v, w, x, y, z corresponding to the five vertices?

The previous five problems explore a common structural theme - namely the link between certain sums (or averages) and the original, possibly unknown, data. However this algebraic link was in every case embedded in some practical, or geometrical, context. The next few problems have been stripped of any context, leaving us free to focus on the underlying structure in a purely algebraic, or arithmetical, spirit.

Problem 97 Solve the following systems of simultaneous equations.

(a) (i) x + y = 1, y + z = 2, x + z = 3

(ii) uv = 2, vw = 4, uw = 8

(b) (i) x + y = 2, y + z = 3, x + z = 4

(ii) uv = 6, vw = 10, uw = 15

(iii) uv = 6, vw = 10, uw = 30

(iv) uv = 4, vw = 8, uw = 16

Problem 98 Use what you know about solving two simultaneous linear equations in two unknowns to construct the general positive solution to the system of equations:

u a v b = m ,   u c v d = n .

Interpret your result in the language of Cramer’s Rule. (Gabriel Cramer (1704–1752)).

(a) For which values b , c does the following system of equations have a unique solution?

x + y + z = 3 ,   x y + y z + z x = b ,   x 2 + y 2 + z 2 = c

(b) For which values a, b, c does the following system of equations have a unique solution?

x + y + z = a ,   x y + y z + z x = b ,   x 2 + y 2 + z 2 = c

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