Vardaan Watermark
Vardaan Learning Institute
Created by Team Vardaan with Powered by VARDAAN COMET
ICSE Class 9 Master Editorial Notes

Simultaneous Linear Equations (In Two Variables)

Complete ICSE Class 9 Syllabus (M.L. Aggarwal Reference) + Exhaustive 100/100 Practice Kit

Topic 1: Foundations & Method of Substitution

Substitution Algorithm
  1. From any one equation, express one variable (say $y$) in terms of the other ($x$).
  2. Substitute this expression for $y$ into the OTHER equation to get a single-variable equation in $x$.
  3. Solve for $x$, then back-substitute to find $y$.
Topic 1: Method of Substitution Practice Kit (Every Question Type)
Problem 1.1 (Standard Integer Substitution) 3 MARKS / ICSE

Solve by substitution: $$2x + y = 7 \quad \text{and} \quad 3x - 2y = 7$$

Step-by-Step Solution:
1. From Eq 1: $y = 7 - 2x$.
2. Substitute into Eq 2: $3x - 2(7 - 2x) = 7 \implies 3x - 14 + 4x = 7 \implies 7x = 21 \implies \mathbf{x = 3}$.
3. $y = 7 - 2(3) = \mathbf{1}$. $$\mathbf{x = 3, \quad y = 1}$$
Substitution x = 3, y = 1
Fig 5.1: Integer Solution
Problem 1.2 (Fractional Coefficients) 3 MARKS / ICSE

Solve by substitution: $$\frac{x}{3} + \frac{y}{4} = 11 \quad \text{and} \quad \frac{5x}{6} - \frac{y}{3} = -7$$

Step-by-Step Solution:
1. Multiply Eq 1 by 12: $4x + 3y = 132 \implies y = \frac{132 - 4x}{3}$.
2. Multiply Eq 2 by 6: $5x - 2y = -42$.
3. Substitute $y$: $5x - 2\left(\frac{132 - 4x}{3}\right) = -42 \implies 15x - 264 + 8x = -126 \implies 23x = 138 \implies \mathbf{x = 6}$.
4. $y = \frac{132 - 24}{3} = \mathbf{36}$. $$\mathbf{x = 6, \quad y = 36}$$
LCM Clearing x = 6, y = 36
Fig 5.2: Fraction Solution
Problem 1.3 (Decimal Coefficients) 3 MARKS / ICSE

Solve for $x$ and $y$: $$0.4x + 0.3y = 1.7 \quad \text{and} \quad 0.7x - 0.2y = 0.8$$

Step-by-Step Solution:
1. Multiply both equations by 10 to clear decimals: $$4x + 3y = 17 \quad \text{and} \quad 7x - 2y = 8$$ 2. From 1st: $3y = 17 - 4x \implies y = \frac{17 - 4x}{3}$.
3. Substitute in 2nd: $7x - 2\left(\frac{17 - 4x}{3}\right) = 8 \implies 21x - 34 + 8x = 24 \implies 29x = 58 \implies \mathbf{x = 2}$.
4. $y = \frac{17 - 8}{3} = \mathbf{3}$. $$\mathbf{x = 2, \quad y = 3}$$
Decimal Clear x = 2, y = 3
Fig 5.3: Decimal Solution

Topic 2: Method of Elimination by Equating Coefficients

Equating Coefficients Algorithm
  1. Multiply equations by constants so coefficients of one variable become equal in magnitude.
  2. Add if signs are opposite; subtract if signs are identical.
ICSE M.L. Aggarwal Symmetry Shortcut ($ax + by = c, bx + ay = d$)
When coefficients of $x$ and $y$ are interchanged:
1. Add equations $\implies (a+b)(x+y) = c+d \implies x + y = A$.
2. Subtract equations $\implies (a-b)(x-y) = c-d \implies x - y = B$.
3. Add/subtract the simplified equations instantaneously!
Topic 2: Elimination Practice Kit (Every Question Type)
Problem 2.1 (Standard Elimination) 3 MARKS / ICSE

Solve by equating coefficients: $$3x + 4y = 10 \quad \text{and} \quad 2x - 2y = 2$$

Step-by-Step Solution:
1. Multiply 2nd equation by 2: $4x - 4y = 4$.
2. Add to 1st equation: $(3x + 4y) + (4x - 4y) = 10 + 4 \implies 7x = 14 \implies \mathbf{x = 2}$.
3. Substitute $x = 2$ in 2nd: $2(2) - 2y = 2 \implies 4 - 2y = 2 \implies 2y = 2 \implies \mathbf{y = 1}$. $$\mathbf{x = 2, \quad y = 1}$$
Elimination x = 2, y = 1
Fig 5.4: Basic Elimination
Problem 2.2 (Symmetry Trick Problem 1) 4 MARKS / ICSE

Solve for $x$ and $y$: $$37x + 43y = 123 \quad \text{and} \quad 43x + 37y = 117$$

Step-by-Step Solution:
1. Add equations: $80x + 80y = 240 \implies \mathbf{x + y = 3} \quad \text{--- (Eq 3)}$.
2. Subtract Eq 1 from Eq 2: $6x - 6y = -6 \implies \mathbf{x - y = -1} \quad \text{--- (Eq 4)}$.
3. Add Eq 3 & Eq 4: $2x = 2 \implies \mathbf{x = 1}$.
4. Substitute $x = 1$: $1 + y = 3 \implies \mathbf{y = 2}$. $$\mathbf{x = 1, \quad y = 2}$$
Symmetry Trick x = 1, y = 2
Fig 5.5: Symmetric Pair 1
Problem 2.3 (Symmetry Trick Problem 2) 4 MARKS / ICSE

Solve for $x$ and $y$: $$103x + 97y = 497 \quad \text{and} \quad 97x + 103y = 503$$

Step-by-Step Solution:
1. Add equations: $200x + 200y = 1000 \implies \mathbf{x + y = 5} \quad \text{--- (Eq 3)}$.
2. Subtract 2nd from 1st: $6x - 6y = -6 \implies \mathbf{x - y = -1} \quad \text{--- (Eq 4)}$.
3. Add Eq 3 & Eq 4: $2x = 4 \implies \mathbf{x = 2}$.
4. Substitute $x = 2$: $2 + y = 5 \implies \mathbf{y = 3}$. $$\mathbf{x = 2, \quad y = 3}$$
Symmetry Trick x = 2, y = 3
Fig 5.6: Symmetric Pair 2

Topic 3: Method of Cross-Multiplication (ICSE Classical Method)

Cross-Multiplication Master Formula

For $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$:

$$\mathbf{\frac{x}{b_1 c_2 - b_2 c_1} = \frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1}}$$
Topic 3: Cross-Multiplication Practice Kit (Every Question Type)
Problem 3.1 (Standard Cross-Multiplication) 3 MARKS / ICSE

Solve by cross-multiplication: $$2x + 3y - 17 = 0 \quad \text{and} \quad 3x - 2y - 6 = 0$$

Step-by-Step Solution:
$a_1=2, b_1=3, c_1=-17$; $a_2=3, b_2=-2, c_2=-6$. $$\frac{x}{3(-6) - (-2)(-17)} = \frac{y}{(-17)(3) - (-6)(2)} = \frac{1}{2(-2) - 3(3)}$$ $$\frac{x}{-18 - 34} = \frac{y}{-51 + 12} = \frac{1}{-4 - 9} \implies \frac{x}{-52} = \frac{y}{-39} = \frac{1}{-13}$$ $$x = \frac{-52}{-13} = \mathbf{4}, \qquad y = \frac{-39}{-13} = \mathbf{3}$$
Cross-Mult x = 4, y = 3
Fig 5.7: Standard Cross-Mult
Problem 3.2 (Literal Coefficients $a, b$) 4 MARKS / ICSE

Solve for $x$ and $y$: $$ax + by = a^2 \quad \text{and} \quad bx + ay = b^2$$

Step-by-Step Solution:
Rewrite: $ax + by - a^2 = 0$ and $bx + ay - b^2 = 0$. $$\frac{x}{b(-b^2) - a(-a^2)} = \frac{y}{(-a^2)(b) - (-b^2)(a)} = \frac{1}{a(a) - b(b)}$$ $$\frac{x}{a^3 - b^3} = \frac{y}{ab(b - a)} = \frac{1}{a^2 - b^2}$$ $$x = \frac{a^3 - b^3}{a^2 - b^2} = \mathbf{\frac{a^2 + ab + b^2}{a + b}}, \qquad y = \mathbf{\frac{-ab}{a + b}}$$
Literal Proof Literal Terms
Fig 5.8: Literal Cross-Mult

Topic 4: Reducible Linear Forms & Binomial Denominators

Substitution Strategies
Topic 4: Reciprocal & Reducible Practice Kit (Every Question Type)
Problem 4.1 (Simple Reciprocal $1/x, 1/y$) 3 MARKS / ICSE

Solve for $x$ and $y$: $$\frac{2}{x} + \frac{3}{y} = 13 \quad \text{and} \quad \frac{5}{x} - \frac{4}{y} = -2$$

Step-by-Step Solution:
Let $u = 1/x, v = 1/y \implies 2u + 3v = 13$ and $5u - 4v = -2$.
Multiply 1st by 4, 2nd by 3: $8u + 12v = 52$ and $15u - 12v = -6$.
Add: $23u = 46 \implies \mathbf{u = 2} \implies \mathbf{x = 1/2}$.
Substitute $u=2$: $4 + 3v = 13 \implies 3v = 9 \implies \mathbf{v = 3} \implies \mathbf{y = 1/3}$. $$\mathbf{x = 1/2, \quad y = 1/3}$$
Reciprocal x=1/2, y=1/3
Fig 5.9: Simple Reciprocal
Problem 4.2 (Binomial Denominators $x+y$ and $x-y$) 4 MARKS / ICSE

Solve for $x$ and $y$: $$\frac{10}{x+y} + \frac{2}{x-y} = 4 \quad \text{and} \quad \frac{15}{x+y} - \frac{5}{x-y} = -2$$

Step-by-Step Solution:
Let $u = \frac{1}{x+y}, v = \frac{1}{x-y} \implies 10u + 2v = 4$ and $15u - 5v = -2$.
Multiply 1st by 5, 2nd by 2: $50u + 10v = 20$ and $30u - 10v = -4$.
Add: $80u = 16 \implies \mathbf{u = 1/5} \implies x + y = 5$.
Substitute $u=1/5$: $10(1/5) + 2v = 4 \implies 2v = 2 \implies \mathbf{v = 1} \implies x - y = 1$.
Solve $x+y=5$ and $x-y=1 \implies 2x = 6 \implies \mathbf{x = 3}, \mathbf{y = 2}$. $$\mathbf{x = 3, \quad y = 2}$$
Binomial u,v x = 3, y = 2
Fig 5.10: Binomial Denominator
Problem 4.3 (Advanced Binomial $3x+y$ and $3x-y$) 5 MARKS / ICSE

Solve for $x$ and $y$: $$\frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} \quad \text{and} \quad \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8}$$

Step-by-Step Solution:
Let $u = \frac{1}{3x+y}, v = \frac{1}{3x-y} \implies u + v = 3/4$ and $\frac{u}{2} - \frac{v}{2} = -1/8 \implies u - v = -1/4$.
Add equations: $2u = 3/4 - 1/4 = 2/4 = 1/2 \implies \mathbf{u = 1/4} \implies 3x + y = 4$.
Subtract equations: $2v = 3/4 + 1/4 = 1 \implies \mathbf{v = 1/2} \implies 3x - y = 2$.
Solve $3x+y=4$ and $3x-y=2 \implies 6x = 6 \implies \mathbf{x = 1}, \mathbf{y = 1}$. $$\mathbf{x = 1, \quad y = 1}$$
Advanced u,v x = 1, y = 1
Fig 5.11: Advanced Binomial

Topic 5: Consistency, Dependency & Unknown Parameters ($k, a, b$)

Consistency Criteria
Ratio Condition Graphical Meaning Algebraic Nature
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ Intersecting Lines Unique Solution (Consistent)
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ Coincident Lines Infinitely Many Solutions (Dependent)
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ Parallel Lines No Solution (Inconsistent)
Topic 5: Parameter Finding Practice Kit
Problem 5.1 (Infinite Solutions Parameter $k$) 4 MARKS / ICSE

Find the value of $k$ for which the system has infinitely many solutions: $$kx + 3y = k - 3 \quad \text{and} \quad 12x + ky = k$$

Step-by-Step Solution:
For infinite solutions: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies \frac{k}{12} = \frac{3}{k} = \frac{k-3}{k}$.
1. From $\frac{k}{12} = \frac{3}{k} \implies k^2 = 36 \implies k = \pm 6$.
2. Test $k = 6$: $\frac{6}{12} = \frac{3}{6} = \frac{6-3}{6} = \frac{1}{2}$ (Valid!).
3. Test $k = -6$: $\frac{-6}{12} = -\frac{1}{2}$, but $\frac{-6-3}{-6} = \frac{3}{2}$ (Invalid!). $$\mathbf{k = 6}$$
Infinite Criteria k = 6
Fig 5.12: Infinite Parameter
Problem 5.2 (Dual Unknowns $a$ and $b$) 4 MARKS / ICSE

Find $a$ and $b$ for which the system has infinitely many solutions: $$(2a - 1)x + 3y = 5 \quad \text{and} \quad 3x + (b - 1)y = 2$$

Step-by-Step Solution:
Condition: $\frac{2a-1}{3} = \frac{3}{b-1} = \frac{5}{2}$.
1. $\frac{2a-1}{3} = \frac{5}{2} \implies 4a - 2 = 15 \implies 4a = 17 \implies \mathbf{a = 17/4 = 4.25}$.
2. $\frac{3}{b-1} = \frac{5}{2} \implies 5b - 5 = 6 \implies 5b = 11 \implies \mathbf{b = 11/5 = 2.2}$. $$\mathbf{a = 17/4, \quad b = 11/5}$$
Dual Parameters a=17/4, b=11/5
Fig 5.13: Dual Parameters

6. ICSE Examination Master Formula Sheet

Master Checklist
Created by Team Vardaan with
Powered by VARDAAN COMET