ICSE Class 8 Maths Chapter 19 Simultaneous Equations With Problems PDF Download

Chapter 19: Simultaneous Equations

19.1 Introduction

1. Simultaneous Equations

Two equations, each containing two same variables (unknowns), are called simultaneous equations.

For example:

(i) 3x + 2y = 7 and x - y = 8 (Here; x and y are variables)

(ii) 5a - 4b = 18 and 2a + b = 4 (Here; a and b are variables) and so on.

2. Solution

Consider the linear equation in two variables to be 3x - 4y = 7.

Any pair of values of x and y which satisfy this equation is called its solution.

For example, if x = 5 and y = 2; then:

3x - 4y = 7 implies 3 \times 5 - 4 \times 2 = 7

implies 15 - 8 = 7; which is true.

Therefore x = 5 and y = 2 is a solution of 3x - 4y = 7.

Now, if x = 9 and y = 5:

3x - 4y = 7 implies 3 \times 9 - 4 \times 5 = 7; which is true

Therefore x = 9 and y = 5 is also a solution of equation 3x - 4y = 7

On trying more, we shall be able to get an infinite number of solutions of the equation 3x - 4y = 7.

In the same way, it can easily be shown that every linear equation in two variables has an infinite number of solutions.

Example:

Is x = 3 and y = 2 a solution of equation x + 2y = 7?

x + 2y = 7 implies 3 + 2 \times 2 = 7

implies 3 + 4 = 7; which is true.

Is x = 3 and y = 2 a solution of equation 3x - 4y = 1?

3x - 4y = 1 implies 3 \times 3 - 4 \times 2 = 1

implies 9 - 8 = 1; which is true.

Therefore x = 3 and y = 2 is a solution of simultaneous equations x + 2y = 7 and 3x - 4y = 1.

Now, if we try to get more pairs of values of x and y to satisfy the equations x + 2y = 7 and 3x - 4y = 1, we find it impossible.

The solution of two simultaneous equations is unique.

Teacher's Note

Simultaneous equations help solve real-world problems like finding prices of items when given total costs, or determining distances when given speeds and times.

19.2 Solving Simultaneous Equations Algebraically

To solve simultaneous equations means; to find the values of variables used in them.

For example:

(i) To solve the equations 3x - 2y = 8 and x + 4y = 10 means to find the values of variables x and y.

(ii) To solve the equations 8m + 7n = 5 and 2m - 5n = 15 means to find the values of variables m and n.

There are mainly two methods of solving simultaneous equations algebraically:

(i) Substitution method (ii) Addition or subtraction method

Example 1

Solve equations x + 3y = 3 and 4x - 5y = 29 by substitution method.

Solution

Steps: 1. Take one of the two given equations (Generally, a simpler equation is taken)

Let us take equation x + 3y = 3

2. Find the value of one variable (x or y) in terms of the other variable.

(Here we find x in terms of y)

Therefore x + 3y = 3 implies x = 3 - 3y

3. Substitute the value of x in the other equation.

Substituting in equation 4x - 5y = 29, we get:

4(3 - 3y) - 5y = 29

implies 12 - 12y - 5y = 29

implies -17y = 29 - 12 = 17

Therefore y = 17 / -17 = -1

4. Substitute y = -1 in the result of step 2.

i.e. x = 3 - 3y = 3 - 3(-1) = 6

The solution of the given equations is x = 6 and y = -1 (Ans.)

If instead of finding x in terms of y (step 1), we find y in terms of x and substitute in the other equation, then also the solution of the equations will be the same. Let us try:

x + 3y = 3 implies 3y = 3 - x implies y = \frac{3 - x}{3}

Substituting in 4x - 5y = 29; we get:

4x - 5\left(\frac{3 - x}{3}\right) = 29

implies \frac{12x - 15 + 5x}{3} = 29

implies 17x - 15 = 29 \times 3 = 87

implies 17x = 87 + 15 = 102 implies x = \frac{102}{17} = 6

Therefore y = \frac{3 - x}{3} = \frac{3 - 6}{3} = \frac{-3}{3} = -1

Required solution is: x = 6 and y = -1 (Ans.)

Example 2

Solve equations x + 3y = 3 and 4x - 5y = 29 by addition or subtraction method.

Solution

Steps: 1. Multiply one or both the equations, if necessary, by a suitable number or numbers so that either the coefficients of x or coefficients of y in both the equations are numerically equal.

By inspection we find that if the first equation x + 3y = 3 be multiplied by 4, the coefficients of x in both the equations will be the same.

Multiplying x + 3y = 3 by 4, we get 4x + 12y = 12.

2. Add or subtract one equation from the other so that the terms with equal numerical coefficients, cancel mutually. Then solve the resulting equation.

Thus,

4x + 12y = 12

4x - 5y = 29

- + - (Subtracting)

17y = -17

implies y = -17/17 = -1

3. Substitute y = -1 in any of the two given equations.

Substituting y = -1, in x + 3y = 3, we get:

x + 3(-1) = 3

implies x - 3 = 3

implies x = 3 + 3 = 6

Therefore x = 6 and y = -1 (Ans.)

Example 3

Solve by substitution method: 3x - 2y = -7 and 5x - 3y = 13.

Solution

Steps: 1. 3x - 2y = -7

2. 3x = 2y - 7 implies x = \frac{2y - 7}{3}

3. 5x - 3y = 13 implies 5\left(\frac{2y - 7}{3}\right) - 3y = 13

i.e. \frac{10y - 35 - 9y}{3} = 13 implies y - 35 = 39

implies y = 39 + 35 = 74

4. x = \frac{2y - 7}{3} = \frac{2 \times 74 - 7}{3} = 47

Therefore x = 47 and y = 74 (Ans.)

Example 4

Solve by addition or subtraction method: 9x + 4y = 5 and 4x - 5y = 9.

Solution

Steps: 1. Multiply the first equation by 5 and second equation by 4.

45x + 20y = 25 [(9x + 4y = 5) \times 5]

16x - 20y = 36 [(4x - 5y = 9) \times 4]

2. 61x = 61 (Adding)

implies x = 61/61 = 1

3. Substituting x = 1 in equation 9x + 4y = 5, we get:

9 \times 1 + 4y = 5 implies 4y = 5 - 9 = -4

implies y = -4/4 = -1

Therefore x = 1 and y = -1 (Ans.)

When students are not asked to solve the given simultaneous equations by a particular method, the solution is generally done by addition or subtraction method.

Teacher's Note

The addition and subtraction method is often preferred in practical applications because it systematically eliminates variables and reduces computational errors compared to substitution.

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