ICSE Class 9 Maths Chapter 07 Linear Equations

Linear Equations

Points To Remember

Linear Equation: An equation which involves only one variable with highest power, is called a linear equation.

Solving the Equation

To solve an equation means to find the value of the variable which satisfies the equation.

Rules for solving a Linear Equation:

The equality of a linear equation is not changed when

(i) the same number is added or subtracted from both sides of the equation.

(ii) both sides of the equation are multiplied or divided by the same non zero number.

Transposition:

When any term of the equation is taken from one side to the other, this process is called transposition.

Cross multiplication

If \(\frac{a}{b} = \frac{c}{d}\) then \(a \times d = b \times c\)

This process is called the cross multiplication.

Word problems can be solved by means of equation by representing the unknown quantity as x, y, z etc and then solve the equation so formed by the condition or conditions given in the problem.

Teacher's Note

Linear equations are fundamental in everyday situations like calculating costs, distances, and time. When you split a restaurant bill or figure out how many hours to work for a certain amount, you are using linear equations.

Exercise 7 (A)

Solve:

Q. 1. \(5x - 16 = 19 - 2x\)

Sol. \(5x - 16 = 19 - 2x\)

\(\Rightarrow 5x + 2x = 19 + 16\)

(By Transposition)

\(\Rightarrow 7x = 35\)

\(\therefore x = \frac{35}{7} = 5\) Ans.

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

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

\(\Rightarrow \frac{6x - x}{2} = \frac{3}{2}\)

\(\Rightarrow \frac{5x}{2} = \frac{3}{2}\)

\(\therefore x = \frac{3}{5}\) Ans.

Q. 3. \(3\frac{3}{4}x = 5x - 2\frac{1}{2}\)

Sol. \(3\frac{3}{4}x = 5x - 2\frac{1}{2}\)

\(\Rightarrow \frac{15}{4}x = 5x - \frac{5}{2}\)

Multiplying by 4, the LCM of 4 and 2

\(\frac{15}{4}x \times 4 = 5x \times 4 - \frac{5}{2} \times 4\)

\(\Rightarrow 15x = 20x - 10\)

\(\Rightarrow 15x - 20x = -10\)

(By transposition)

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

\(\therefore x = \frac{-10}{-5} = 2\) Ans.

Q. 4. \(3x - 5 = \frac{2x}{3} + 9\)

Sol. \(3x - 5 = \frac{2x}{3} + 9\)

\(\Rightarrow 3x \times 3 - 5 \times 3 = \frac{2x}{3} \times 3 + 9 \times 3\)

(Multiplying by 3)

\(\Rightarrow 9x - 15 = 2x + 27\)

\(\Rightarrow 9x - 2x = 27 + 15\)

(By transposition)

\(\Rightarrow 7x = 42\)

\(\therefore x = \frac{42}{7} = 6\) Ans.

Q. 5. \((x - 4)(x + 4) = (x + 4)(x - 7) + 33\)

Sol. \((x - 4)(x + 4) = (x + 4)(x - 7) + 33\)

\(\Rightarrow x^2 - 4x + 4x - 16\)

\(= x^2 - 7x + 4x -28 + 33\)

\(\Rightarrow x^2 - 16 = x^2 - 3x + 5\)

\(\Rightarrow x^2 - x^2 + 3x = 5 + 16\)

(By transposition)

\(\Rightarrow 3x = 21\)

\(\therefore x = \frac{21}{3} = 7\) Ans.

Q. 6. \((x - 2)(x + 3) = (x^2 - 4)\)

Sol. \((x - 2)(x + 3) = x^2 -4\)

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

\(\Rightarrow x^2 + x - 6 = x^2 - 4\)

\(\Rightarrow x^2 + x - x^2 = -4 + 6\)

(By transposition)

\(\Rightarrow x = 2\) Ans.

Q. 7. \(\frac{(y - 4)}{5} + \frac{(y + 2)}{2} = 10\)

Sol. \(\frac{y - 4}{5} + \frac{y + 2}{2} = 10\)

Multiplying by 10, the LCM of 5 and 2

\(\frac{(y - 4)}{5} \times 10 + \frac{(y + 2)}{2} \times 10 = 10 \times 10\)

\(\Rightarrow 2(y - 4) + 5(y + 2) = 100\)

\(\Rightarrow 2y - 8 + 5y + 10 = 100\)

\(\Rightarrow 2y + 5y = 100 + 8 - 10\)

(By transposition)

\(\Rightarrow 7y = 98\)

\(\Rightarrow y = \frac{98}{7} = 14\)

\(\therefore y = 14\) Ans.

Teacher's Note

Solving equations step by step teaches logical thinking and problem-solving skills that help in debugging code, understanding recipes, or planning budgets.

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