ICSE Class 8 Maths Chapter 17 Simultaneous Linear Equations PDF Download

Chapter 17: Simultaneous Linear Equations

Linear Equation In Two Variables

An equation of the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are real numbers and \(a\) and \(b\) both are non-zero, is called a linear equation in two variables \(x\) and \(y\).

For example:

\(x + y - 5 = 0\) is a linear equation in the two variables (unknown) \(x\) and \(y\). Note that \(x = 2\) and \(y = 3\) satisfies this linear equation.

Simultaneous Linear Equations

Let us consider two linear equations in two variables.

\(x + y - 5 = 0\)

\(2x - y - 1 = 0\)

These two equations are said to form a pair of simultaneous linear equations in two variables \(x\) and \(y\).

A solution to a pair of simultaneous linear equations in two variables is an ordered pair of real numbers which satisfy both the equations.

In the above example, \(x = 2, y = 3\) is a solution to the pair of simultaneous linear equations. We can check this by substituting \(x = 2, y = 3\) in each of these two equations.

The process of finding an ordered pair of real numbers which satisfy both the equations is called solving a pair of simultaneous equations.

There are two methods to solve a pair of simultaneous linear equations:

(i) Substitution method (ii) Addition-subtraction (or Elimination) method.

Substitution Method

Procedure to solve a pair of simultaneous linear equations in two variables:

(i) Solve one of the given equations for one of the variables.

(ii) Substitute that value of the variable in the other equation.

(iii) Solve the resulting single variable equation and substitute this value into either of the two original equations and solve it to find the value of the other variable.

Remark

The solution may be checked by substituting in both the original equations.

Teacher's Note

Understanding simultaneous equations helps us solve real-world problems involving multiple unknown quantities, such as calculating costs, speeds, or ages when several conditions must be satisfied together.

Example 1

Solve the following pair of simultaneous linear equations: \(2x + 7y = 30\), \(x - 3y = 2\).

Solution.

Given equations are:

\(2x + 7y = 30\) ... (i)

\(x - 3y = 2\) ... (ii)

We can use any equation to express one variable in terms of another. To avoid fractions, we can express \(x\) in terms of \(y\) by using equation (ii).

\(x - 3y = 2 \Rightarrow x = 3y + 2\)

Substituting this value of \(x\) in equation (i), we get

\(2(3y + 2) + 7y = 30\)

\(\Rightarrow 6y + 4 + 7y = 30\)

\(\Rightarrow 13y = 30 - 4 \Rightarrow 13y = 26 \Rightarrow y = 2.\)

Substituting this value of \(y\) in equation (ii), we get

\(x - 3 \times 2 = 2 \Rightarrow x - 6 = 2 \Rightarrow x = 8.\)

Hence, the solution is \(x = 8, y = 2\).

Teacher's Note

The substitution method is like detective work - we solve for one unknown and use that clue to find the other, similar to how we might find a friend's location by asking about their usual hangouts.

Example 2

Solve \(4x + 3y = 14\), \(9x - 5y = 55\) by substitution method.

Solution.

Given equations are:

\(4x + 3y = 14\) ... (i)

\(9x - 5y = 55\) ... (ii)

Using equation (i), we get

\(4x + 3y = 14 \Rightarrow x = \frac{14 - 3y}{4}.\)

Substituting this value of \(x\) in equation (ii), we get

\(9 \times \frac{14 - 3y}{4} - 5y = 55\)

\(\Rightarrow 9(14 - 3y) - 20y = 220\) (Multiplying by 4)

\(\Rightarrow 126 - 27y - 20y = 220\)

\(\Rightarrow -47y = 220 - 126 \Rightarrow -47y = 94\)

\(\Rightarrow y = -2.\)

Substituting this value of \(y\) in equation (i), we get

\(4x + 3 \times (-2) = 14 \Rightarrow 4x - 6 = 14\)

\(\Rightarrow 4x = 20 \Rightarrow x = 5\)

Hence, the solution is \(x = 5, y = -2\).

Remark

An alternative method of solving a pair of simultaneous linear equations that is perhaps superior to substitution for systems such as

\(4x + 3y = 14, 9x - 5y = 55\)

is known as addition-subtraction method or elimination method. In example 2, we noticed that solving these equations by using substitution method involves fractions regardless of our choice of variable and choice of equation. The addition-subtraction method enables us to avoid fractions.

Teacher's Note

When fractions appear in substitution, the elimination method becomes more practical - it's like combining two recipes to eliminate an ingredient you don't want to measure.

Addition-Subtraction Method

Procedure:

- Multiply one or both equations (if necessary) to transform them so that addition or subtraction will eliminate one variable.

- Solve the resulting single variable equation and substitute this value into either of the two original equations, and solve to find the value of the second variable.

Example 3

Solve \(4x + 3y = 14, 9x - 5y = 55\) by addition-subtraction method.

Solution.

Given equations are:

\(4x + 3y = 14\) ... (i)

\(9x - 5y = 55\) ... (ii)

To make coefficients of \(y\) equal (numerically), multiplying equation (i) by 5 and equation (ii) by 3, we get

\(20x + 15y = 70\) ... (iii)

\(27x - 15y = 165\) ... (iv)

Adding (iii) and (iv), we get

\(47x = 235 \Rightarrow x = 5.\)

Substituting \(x = 5\) in equation (i), we get

\(4 \times 5 + 3y = 14 \Rightarrow 20 + 3y = 14\)

\(\Rightarrow 3y = -6 \Rightarrow y = -2.\)

Hence, the solution is \(x = 5, y = -2\).

Teacher's Note

The elimination method is like balancing scales - we adjust the weights (multiply equations) strategically so that one side cancels out, leaving us with a simple solution.

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