ICSE Class 8 Maths Chapter 16 Linear Equations and Inequations

Linear Equations And Inequations

In the previous classes, you have studied mathematical open sentences like \(x + 3 = 5\) or \(2x - 1 = 9\) which may or may not be true depending upon the value of \(x\). You have learnt how to find truth set or solution set of such equations. You also studied word problems where you first have to write equations corresponding to the given problem stated in words and then find the solution set. In this chapter, we shall strengthen these concepts and introduce the idea of inequations.

Equations

An (algebraic) equation is a statement that two expressions are equal. It may involve one or more than one variables (literals). Thus, \(2x - 1 = 9\), \(x^2 - 3x + 2 = 0\) are equations in one variable and \(5x - 7y = -3\), \(x^2 + 3xy + 5y^2 = 4\) are equations in two variables.

Linear Equation

An equation of the type \(ax + b = 0\), where \(a \neq 0\), is called a linear equation in the variable \(x\).

Thus, \(2x - 1 = 9\) is a linear equation in the variable \(x\).

Solution Or Root

Any value (or values) of the variable (or variables) which when substituted in an equation makes its both sides equal is called a solution (or root) of the equation.

Thus, a number \(\alpha\) is a root (or solution) of the equation \(ax + b = 0\) if and only if \(a\alpha + b = 0\).

For example, when we substitute \(x = 5\) in the equation \(2x - 1 = 9\), we get \(2 \times 5 - 1 = 9\) i.e. \(9 = 9\), which is true. Therefore, 5 is a solution (or root) of the equation \(2x - 1 = 9\).

Solving An Equation

To solve an equation is to find all its solutions (or roots), and the process of finding all the solutions is called solving the equation.

Permissible Rules For Solving Equations

We can add the same number or expression to both sides of an equation.

We can subtract the same number or expression from both sides of an equation.

We can multiply both sides of an equation by the same non-zero number or expression.

We can divide both sides of an equation by the same non-zero number or expression.

A term may be transposed from one side of the equation to the other side, but its sign will change.

Solving Linear Equations In One Variable

For solving a linear equation in one variable, proceed as under:

(i) Simplify both sides by removing group symbols and collecting like terms.

(ii) Remove fractions (or decimals) by multiplying both sides by an appropriate factor (L.C.M. of fractions or a power of 10 in case of decimals).

(iii) Isolate all variable terms on one side and all constants on the other side. Collect like terms.

(iv) Make the coefficient of the variable 1.

Remark

The solution may be checked (verified) by substituting in the original equation.

Example 1

Solve the following equations:

(i) \(3 - 4x = 2x + 25\)

(ii) \(5 - 3(5x + 2) = 4(7 - 3x) + 1\)

Solution

(i) Given \(3 - 4x = 2x + 25\)

\(\Rightarrow -4x - 2x = 25 - 3\)

\(\Rightarrow -6x = 22 \Rightarrow x = -\frac{22}{6} = -\frac{11}{3}\)

(ii) Given \(5 - 3(5x + 2) = 4(7 - 3x) + 1\)

\(\Rightarrow 5 - 15x - 6 = 28 - 12x + 1\)

\(\Rightarrow -15x - 1 = -12x + 29\)

\(\Rightarrow -15x + 12x = 29 + 1\)

\(\Rightarrow -3x = 30 \Rightarrow x = -10\)

Example 2

Solve the following equations:

(i) \(\frac{3x}{5} - \frac{x}{3} = \frac{x}{6} + 1\frac{1}{2}\)

(ii) \(\frac{3y - 1}{5} - \frac{1 + y}{2} = 3 - \frac{y - 1}{4}\)

Solution

(i) Given \(\frac{3x}{5} - \frac{x}{3} = \frac{x}{6} + \frac{3}{2}\)

Multiplying both sides by 30, L.C.M. of 5, 3, 6 and 2, we get

\(18x - 10x = 5x + 45\)

\(\Rightarrow 18x - 10x - 5x = 45\)

\(\Rightarrow 3x = 45 \Rightarrow x = 15\)

(ii) Given \(\frac{3y - 1}{5} - \frac{1 + y}{2} = 3 - \frac{y - 1}{4}\)

Multiplying both sides by 20, L.C.M. of 5, 2 and 4, we get

\(4(3y - 1) - 10(1 + y) = 20 \times 3 - 5(y - 1)\)

\(\Rightarrow 12y - 4 - 10 - 10y = 60 - 5y + 5\)

\(\Rightarrow 2y - 14 = -5y + 65\)

\(\Rightarrow 2y + 5y = 65 + 14\)

\(\Rightarrow 7y = 79 \Rightarrow y = \frac{79}{7} = 11\frac{2}{7}\)

Cross-Multiplication

If \(\frac{a}{b} = \frac{c}{d}\), then \(ad = bc\) or \(bc = ad\).

This is known as cross-multiplication.

Consider the equation \(\frac{ax + b}{cx + d} = \frac{m}{n} \Rightarrow (ax + b) \times n = (cx + d) \times m\) (By cross-multiplication)

Example 3

Solve the following equations:

(i) \(\frac{3}{x + 8} = \frac{4}{6 - x}\)

(ii) \(\frac{x + 1}{x - 1} = \frac{2x + 3}{2x - 5}\)

Solution

(i) Given \(\frac{3}{x + 8} = \frac{4}{6 - x}\)

\(\Rightarrow 4(x + 8) = 3(6 - x)\) (By cross-multiplication)

\(\Rightarrow 4x + 32 = 18 - 3x\)

\(\Rightarrow 4x + 3x = 18 - 32\)

\(\Rightarrow 7x = -14 \Rightarrow x = -2\)

(ii) Given \(\frac{x + 1}{x - 1} = \frac{2x + 3}{2x - 5}\)

\(\Rightarrow (x - 1)(2x + 3) = (x + 1)(2x - 5)\) (By cross-multiplication)

\(\Rightarrow 2x^2 + 3x - 2x - 3 = 2x^2 - 5x + 2x - 5\)

\(\Rightarrow 2x^2 + x - 3 = 2x^2 - 3x - 5\)

\(\Rightarrow 2x^2 + x - 2x^2 + 3x = -5 + 3\)

\(\Rightarrow 4x = -2 \Rightarrow x = -\frac{1}{2}\)

Example 4

If \(p = x + 1\) and \(\frac{4p - 3}{2} - \frac{3x + 2}{5} = \frac{3}{2}\), find \(x\).

Solution

Given \(p = x + 1\) ...(i)

and \(\frac{4p - 3}{2} - \frac{3x + 2}{5} = \frac{3}{2}\) ...(ii)

Substituting the value of \(p\) from (i) in (ii), we get

\(\frac{4(x + 1) - 3}{2} - \frac{3x + 2}{5} = \frac{3}{2} \Rightarrow \frac{4x + 1}{2} - \frac{3x + 2}{5} = \frac{3}{2}\)

Multiplying both sides by 10, L.C.M. of 2, 5 and 2, we get

\(5(4x + 1) - 2(3x + 2) = 15\)

\(\Rightarrow 20x + 5 - 6x - 4 = 15\)

\(\Rightarrow 14x + 1 = 15 \Rightarrow 14x = 15 - 1\)

\(\Rightarrow 14x = 14 \Rightarrow x = 1\)

Exercise 16.1

Solve the following (1 to 9) equations:

1. (i) \(4x - 8 = 7 - x\) (ii) \(3x - 7 = 3(5 - x)\)

2. (i) \(3(2x - 1) = 5 - (3x - 2)\) (ii) \(5x - 2[x - 3(x - 5)] = 6\)

3. (i) \(\frac{x - 1}{3} = \frac{x + 2}{6} + 3\) (ii) \(\frac{x + 7}{3} = 1 + \frac{3x - 2}{5}\)

4. (i) \(\frac{y + 1}{3} - \frac{y - 1}{2} = \frac{1 - 2y}{3}\) (ii) \(\frac{p}{3} + \frac{p}{4} = 55 - \frac{p + 40}{5}\)

5. (i) \(0.3(6 - x) = 0.4(x + 8)\) (ii) \(\frac{1}{2}(x + 1.7) - \frac{1}{3}(x - 2.3) = 1\)

6. (i) \(\frac{3x + 2}{x - 6} = 5\) (ii) \(\frac{1}{x - 3} + \frac{3}{5} = \frac{1}{7}\)

7. (i) \(\frac{5}{x} = \frac{7}{x - 4}\) (ii) \(\frac{2x - 3}{2x - 1} = \frac{3x - 1}{3x + 1}\)

8. (i) \(\frac{2x + 5}{x - 1} - \frac{5x}{x - 1} = x\) (ii) \(\frac{1}{5}\left(\frac{1}{3x} - 5\right) = \frac{1}{3}\left(3 - \frac{1}{x}\right)\)

9. (i) \(\frac{4}{x - 3} + \frac{2}{x - 2} = \frac{6}{x}\) (ii) \(\frac{3}{2x - 1} + \frac{4}{2x + 1} = \frac{7}{2x}\)

10. If \(x = p + 1\), find the value of \(p\) from the equation \(\frac{1}{2}(5x - 30) - \frac{1}{3}(1 + 7p) = \frac{1}{4}\)

11. Solve \(\frac{x + 3}{3} - \frac{x - 2}{2} = 1\). Hence find \(p\) if \(\frac{1}{x} + p = 1\).

12. Solve \(\frac{2x + 1}{10} - \frac{3 - 2x}{15} = \frac{x - 2}{6}\). Hence find the value of \(y\) if \(\frac{2}{x} + \frac{5}{y} = 5\).

Teacher's Note

Understanding how to solve linear equations helps students develop critical thinking when breaking down real-world problems like calculating distances, budgeting expenses, or determining fair pricing in everyday transactions.

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