ICSE Class 8 Maths Algebra Chapter 08 Linear Equations and Inequations

Linear Equations and Inequations

Equations

An algebraic equation represents the equality of two mathematical expressions involving at least one variable (literal).

Examples (i) \(7 - 3x = 4\) (ii) \(2x + 5 = 9\) (iii) \(x^2 + 6x = 7\) (iv) \(y = 0\)

Linear Equations

A linear equation in one variable, say x, is an equation in which the exponent of x is 1. The general form of a linear equation is \(ax + b = 0\).

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

Solution of an Equation

The solution or root of an equation is a number, which when substituted for the variable in the equation makes the left-hand side (LHS) of the equation equal to the right-hand side (RHS).

Examples (i) The solution of the equation \(x - 2 = 6\) is 8 because \(8 - 2 = 6\).

(ii) 5 is not the solution of the equation \(2x + 3 = 3x - 1\) because \(2 \times 5 + 3 \neq 3 \times 5 - 1\).

Laws of Equality

1. If the same number or quantity is added to or subtracted from both sides of an equation, the two sides remain equal. We can express this symbolically as follows.

If \(x = y\) then \(x + k = y + k\) and \(x - c = y - c\).

2. If both sides of an equation are multiplied or divided by the same nonzero quantity, the sides remain equal.

If \(x = y\) then \(ax = ay\) where \(a \neq 0\) and \(\frac{x}{c} = \frac{y}{c}\) where \(c \neq 0\).

Transposition

Any term on one side of an equation can be shifted to the other side by changing the sign of the term. This process is called transposition.

Example: If \(a + b = c - d\) then \(a + b - c = -d\) or \(a + b - c + d = 0\).

Note: An equation remains unchanged if all the terms on the LHS are shifted to the RHS and all the terms on RHS are shifted to LHS.

Solving Linear Equations

To solve a linear equation, proceed step by step.

Steps 1. Simplify both sides of the equation. Use the distributive law (if necessary) to separate terms containing the variable and the constant terms.

2. If the equation involves fractions, multiply both sides by the LCM of the denominators to clear the fractions.

3. If decimals are present, multiply both sides by a suitable power of 10 to eliminate the decimals.

4. Collect all the terms containing the variable on one side of the equation (generally, LHS) and all constant terms on the other side.

5. Divide both sides of the equation by the resulting coefficient of the variable.

Verification of the Solution

Substitute the value of the variable on both sides of the equation. If the values of both sides are equal, the solution of the equation is correct.

Example

Solve \(2 - 3(2x + 3) = 9 - 2x\) and verify the solution.

Solution

The equation is \(2 - (6x + 9) = 9 - 2x\).

Simplifying, \(2 - 6x - 9 = 9 - 2x\) or \(-6x - 7 = 9 - 2x\).

Collecting terms containing x on the LHS and constants on the RHS, \(-6x + 2x = 9 + 7\) or \(-4x = 16\).

Dividing both sides by the coefficient of x, \(\frac{-4x}{-4} = \frac{16}{-4}\) or \(x = -4\).

Verification

Substituting \(x = -4\) on the LHS, we get \(2 - 3(2x + 3) = 2 - 3(-8 + 3) = 2 + 15 = 17\).

Substituting \(x = -4\) on the RHS, we get \(9 - 2x = 9 - 2 \times (-4) = 9 + 8 = 17\).

∴ LHS = RHS. So, the solution \(x = -4\) is correct.

Teacher's Note

Linear equations are used daily in budgeting and shopping, where we calculate unknown prices or quantities by setting up equations based on known relationships and totals.

Solved Examples

Example 1

Solve \(3(z + 1) - 2(4z - 3) = 4z\).

Solution

The equation is \(3z + 3 - (8z - 6) = 4z\).

∴ \(3z + 3 - 8z + 6 = 4z\) or \(-5z + 9 = 4z\) or \(9 = 4z + 5z\) or \(4z + 5z = 9\) or \(9z = 9\).

∴ \(z = \frac{9}{9} = 1\).

Example 2

Solve \(3x - \frac{2(x + 3)}{3} = 16 - \frac{x + 2}{2}\).

Solution

Multiplying both sides by the LCM of 3 and 2, that is, 6,

\(6 \times \left[3x - \frac{2(x + 3)}{3}\right] = 6 \times \left[16 - \frac{x + 2}{2}\right]\).

∴ \(6 \times 3x - 6 \times \frac{2}{3}(x + 3) = 6 \times 16 - 6 \times \frac{x + 2}{2}\)

or \(18x - 4(x + 3) = 96 - 3(x + 2)\) or \(18x - 4x - 12 = 96 - 3x - 6\)

or \(14x - 12 = 90 - 3x\) or \(14x + 3x = 90 + 12\) or \(17x = 102\).

∴ \(x = \frac{102}{17} = 6\).

Example 3

Solve \(\frac{1}{4} + \frac{9}{x} = 1\) or \(\left(0.25 + \frac{9}{x} = 1\right)\).

Solution

The given equation is \(\frac{1}{4} + \frac{9}{x} = 1\).

∴ \(\frac{9}{x} = 1 - \frac{1}{4}\) or \(\frac{9}{x} = \frac{3}{4}\).

By cross multiplication, \(9 \times 4 = 3 \times x\) or \(3 \times x = 9 \times 4\).

∴ \(x = \frac{9 \times 4}{3} = 12\).

Example 4

Solve \(\frac{4}{x - 2} = \frac{9}{x + 8}\).

Solution

The given equation is \(\frac{4}{x - 2} = \frac{9}{x + 8}\).

By cross multiplication,

\(4 \times (x + 8) = 9 \times (x - 2)\) or \(4x + 32 = 9x - 18\)

or \(32 + 18 = 9x - 4x\) or \(5x = 50\).

∴ \(x = \frac{50}{5} = 10\).

Example 5

Solve \(\frac{3x + 7}{5x + 16} = \frac{3x - 2}{5x - 2}\)

Solution

The given equation is \(\frac{3x + 7}{5x + 16} = \frac{3x - 2}{5x - 2}\).

By cross multiplication, \((3x + 7)(5x - 2) = (3x - 2)(5x + 16)\)

or \(15x^2 + 35x - 6x - 14 = 15x^2 - 10x + 48x - 32\)

or \(15x^2 + 29x - 14 = 15x^2 + 38x - 32\)

or \(15x^2 + 38x - 15x^2 - 29x = 32 - 14\).

∴ \(9x = 18\) or \(x = \frac{18}{9} = 2\).

Example 6

Solve \(\frac{1}{x + 3} - \frac{x}{x^2 - 9} = \frac{-2}{3 - x}\)

Solution

The given equation is

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

[∴ \(x^2 - 9 = (x + 3)(x - 3)\) and \(3 - x = -(x - 3)\)]

Multiplying both sides by the LCM of the denominators or \((x + 3)(x - 3)\),

\((x + 3)(x - 3)\left[\frac{1}{x + 3} - \frac{x}{(x + 3)(x - 3)}\right] = \frac{-2(x + 3)(x - 3)}{x - 3}\)

or \(x - 3 - x = -2(x + 3)\) or \(-3 = -2(x + 3)\) or \(3 = 2x + 6\) or \(2x = 3 - 6 = -3\).

Thus, \(x = -\frac{3}{2}\).

Example 7

Solve \(\frac{x + 6}{x + 5} - \frac{2x - 1}{x - 4} + \frac{x + 4}{x - 2} = 0\).

Solution

Multiplying both sides by the LCM of the denominators, i.e., \((x + 5)(x - 4)(x - 2)\),

\((x + 6)(x^2 - 4 + 2)x + 4 \times 2) - (2x - 1)(x^2 + (5 - 2)x - 5 \times 2) + (x + 4)(x + 5)(x - 4) = 0\)

or \((x + 6)(x^2 - 6x + 8) - (2x - 1)(x^2 + 3x - 10) + (x + 5)(x^2 - 16) = 0\)

or \((x + 6)(x^2 - 6x + 8) - (2x - 1)(x^2 + 3x - 10) + (x + 5)(x^2 - 16) = 0\)

or \(x(x^2 - 6x + 8) + 6(x^2 - 6x + 8) - 2x(x^2 + 3x - 10) + x^2 + 3x - 10 + x(x^2 - 16) + 5(x^2 - 16) = 0\)

or \(x^3 - 6x^2 + 8x + 6x^2 - 36x + 48 - 2x^3 - 6x^2 + 20x + x^2 + 3x - 10 + x^3 - 16x + 5x^2 - 80 = 0\)

or \(8x - 36x + 48 + 20x + 3x - 10 - 16x - 80 = 0\)

or \(-21x - 42 = 0\) or \(21x = -42\).

∴ \(x = \frac{-42}{21} = -2\)

Teacher's Note

Solving complex equations teaches logical problem-solving that applies to real-world scenarios like calculating loan repayment schedules or adjusting recipe ingredients proportionally.

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