Login / Register
Follow Us
0
0
Register Free!
Register Free!
Online Mock Tests & Exam Preparation | Target Exams
Select Medium / माध्यम चुनें:
Eng (English) Hindi (हिन्दी)
ICSE • Class 9 • Mathematics • Ch 5
Estimated Time: 45 Mins
Study Progress: In Progress

Simultaneous (Linear) Equations

In ICSE Class 9 Mathematics, "Simultaneous (Linear) Equations" provides the indispensable algebraic framework for finding unique solution pairs $(x, y)$ that concurrently satisfy two distinct first-degree linear equations in two variables. A general system of two linear equations in two variables is represented canonically as $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. Geometrically, each equation describes an infinite straight line in the Cartesian plane; a unique simultaneous solution exists if and only if the two lines intersect at exactly one point, governed analytically by the non-proportionality condition $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. The ICSE curriculum rigorously trains students in four fundamental algebraic solution methods: (1) The Method of Substitution, where one variable is isolated in terms of the other and substituted into the second equation; (2) The Method of Elimination by Equating Coefficients, where equations are multiplied by suitable non-zero constants so that adding or subtracting eliminates one variable; (3) The Method of Cross-Multiplication, using the determinant rule $\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$; and (4) Equations reducible to linear form via clever algebraic substitutions (such as setting $u = \frac{1}{x}$ and $v = \frac{1}{y}$, or adding and subtracting symmetric systems like $ax + by = c$ and $bx + ay = d$). The chapter culminates in modeling complex real-world word problems involving two-digit numbers and digit reversals, speed-distance-time with upstream and downstream river currents, fractional adjustments, geometry of rectangles, and commercial cost calculations.

Take Chapter CBT Test View All Chapters
Overview Core Concepts Formulas & Identities Summary Self Check Take MCQ Test

The Air Traffic Collision Mystery: How Simultaneous Equations Keep 100,000 Airplanes from Crashing Every Single Day

At any given moment, there are over 10,000 commercial aircraft cruising through the skies at 900 kilometers per hour. High above the Atlantic Ocean, two jetliners—one departing New York for London and another flying from Paris to Chicago—are traveling along paths defined by two linear trajectories across the globe. How do air traffic controllers and flight computers guarantee that these two giant metal birds will never occupy the exact same point $(x, y)$ in airspace at the same time $t$? They solve Simultaneous Linear Equations! If the two flight lines are parallel (same slope, no solution), they will never meet. If the lines intersect at an exact point $(x_0, y_0)$, the system computes the exact time each jet reaches that intersection to enforce mandatory safety separation. In our daily life, whether calculating the speed of a rowing boat battling against a river current or deciphering a secret two-digit number, simultaneous equations turn tangled real-world mysteries into clean, deterministic solutions! Let us master the mathematical engines of coordination.

Why This Chapter Matters

Simultaneous linear equations form the absolute cornerstone of coordinate geometry, linear programming, Kirchhoff's laws in electrical circuit theory, economic market equilibrium (supply vs demand), and engineering optimization.

Before You Begin (Prerequisites)

  • Solving simple linear equations in one variable.
  • Basic algebraic manipulation, LCM of coefficients, and fraction simplification.

What You Will Learn (Core Objectives)

  • Understand the algebraic and geometric meaning of a simultaneous solution pair $(x, y)$.
  • Master the Method of Elimination by equating coefficients with precision.
  • Apply the Method of Substitution systematically to solve linear systems.
  • Execute the Cross-Multiplication Method using determinant memory schemes.
  • Solve non-linear rational systems reducible to linear form using auxiliary variables.
  • Formulate and solve real-world word problems (two-digit numbers, upstream/downstream, fractions, age).

Chapter Roadmap & Progression

1 1. Standard Form and Consistency Co...
2 2. The Four Primary Algebraic Solut...
3 3. Equations Reducible to Linear Fo...
4 4. Real-World Word Problem Archetyp...
5 5. Worked ICSE Problem Archetypes

Complete Concept Guide (100% Curriculum Coverage)

1. Standard Form and Consistency Conditions

Theoretical Foundations
A. Canonical System Form:
$$a_1x + b_1y + c_1 = 0 \quad \text{and} \quad a_2x + b_2y + c_2 = 0$$

where $a_1, b_1, c_1, a_2, b_2, c_2 \in \mathbb{R}$ and $a_1^2 + b_1^2 \neq 0, \; a_2^2 + b_2^2 \neq 0$.

B. Geometric Criteria for Solutions:
Condition on CoefficientsGeometric Behavior of LinesNature and Number of Solutions
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$Intersecting lines at a single pointUnique Solution (Consistent & Independent)
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$Parallel lines (never intersect)No Solution (Inconsistent)
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$Coincident lines (identical line)Infinitely Many Solutions (Consistent & Dependent)

2. The Four Primary Algebraic Solution Methods

Solution Techniques
1. Method of Elimination by Equating Coefficients:
  1. Multiply one or both equations by suitable non-zero integers so that the absolute values of the coefficients of one variable (either $x$ or $y$) become identical.
  2. If the signs of these coefficients are opposite, add the two equations; if the signs are the same, subtract one equation from the other to eliminate that variable.
  3. Solve the resulting single-variable linear equation.
  4. Substitute this value back into either original equation to obtain the remaining variable.
2. Method of Substitution:

Express $y$ in terms of $x$ (or $x$ in terms of $y$) from equation (1): $y = \frac{-c_1 - a_1x}{b_1}$. Substitute this expression into equation (2), eliminating $y$ completely.

3. Method of Cross-Multiplication:

Using the cyclic determinant scheme:

$$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$$ $$\mathbf{x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}, \quad y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}} \quad (\text{provided } a_1b_2 - a_2b_1 \neq 0)$$
4. Symmetric System Shortcut (When $a_1 = b_2$ and $b_1 = a_2$):

Given $ax + by = m$ and $bx + ay = n$:

  • Add equations: $(a+b)(x+y) = m+n \implies x+y = \frac{m+n}{a+b}$.
  • Subtract equations: $(a-b)(x-y) = m-n \implies x-y = \frac{m-n}{a-b}$.
  • Solve the ultra-simple system of $(x+y)$ and $(x-y)$ by adding and subtracting!

3. Equations Reducible to Linear Form

Reducible Systems
Fractional Reciprocal Variables:

When variables appear in the denominator, such as:

$$\frac{a_1}{x} + \frac{b_1}{y} = c_1 \quad \text{and} \quad \frac{a_2}{x} + \frac{b_2}{y} = c_2$$

Substitute auxiliary variables: let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The system transforms into standard linear form:

$$a_1u + b_1v = c_1 \quad \text{and} \quad a_2u + b_2v = c_2$$

After solving for $u$ and $v$, recover $x$ and $y$ via $x = \frac{1}{u}$ and $y = \frac{1}{v}$.

4. Real-World Word Problem Archetypes

Word Problem Modeling
A. Two-Digit Number Problems:
  • Let tens digit be $x$ and units digit be $y$. Original number value $= 10x + y$.
  • Number obtained by reversing the digits $= 10y + x$.
  • Sum of original and reversed number $= (10x + y) + (10y + x) = 11(x + y)$.
  • Difference of original and reversed number $= (10x + y) - (10y + x) = 9(x - y)$.
B. Upstream and Downstream Boat Problems:
  • Let speed of boat in still water $= x\text{ km/h}$, speed of stream $= y\text{ km/h}$ ($x > y$).
  • Downstream speed (boat aided by stream) $= (x + y)\text{ km/h}$.
  • Upstream speed (boat resisted by stream) $= (x - y)\text{ km/h}$.
  • Time equation: $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.

5. Worked ICSE Problem Archetypes

Exemplary Solutions
Problem 1: Solve the following system of linear equations using elimination:
$$3x + 4y = 10 \quad \text{--- (1)}$$ $$2x - 2y = 2 \quad \text{--- (2)}$$

Solution:

Multiply Equation (2) by $2$ to make the coefficient of $y$ equal to $-4$:

$$2 \times (2x - 2y) = 2 \times 2 \implies 4x - 4y = 4 \quad \text{--- (3)}$$

Add Equation (1) and Equation (3) to eliminate $y$:

$$(3x + 4y) + (4x - 4y) = 10 + 4$$ $$7x = 14 \implies x = 2$$

Substitute $x = 2$ into Equation (1):

$$3(2) + 4y = 10 \implies 6 + 4y = 10 \implies 4y = 4 \implies y = 1$$ $$\mathbf{\text{Solution: } x = 2, \quad y = 1}$$
Problem 2: Solve the symmetric system:
$$37x + 43y = 123 \quad \text{--- (1)}$$ $$43x + 37y = 117 \quad \text{--- (2)}$$

Solution:

Step 1 (Add equations):

$$(37 + 43)x + (43 + 37)y = 123 + 117 \implies 80x + 80y = 240 \implies x + y = 3 \quad \text{--- (3)}$$

Step 2 (Subtract (1) from (2)):

$$(43 - 37)x + (37 - 43)y = 117 - 123 \implies 6x - 6y = -6 \implies x - y = -1 \quad \text{--- (4)}$$

Step 3 (Solve (3) and (4)):

Adding (3) and (4): $2x = 2 \implies x = 1$.

Subtracting (4) from (3): $2y = 4 \implies y = 2$.

$$\mathbf{\text{Solution: } x = 1, \quad y = 2}$$

Key Formulas, Identities & Theorems

General Linear System
$$a_1x + b_1y + c_1 = 0, \quad a_2x + b_2y + c_2 = 0$$
Canonical standard form.
Unique Solution Condition
$$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$
Lines intersect at exactly one point.
Cross-Multiplication Formula
$$x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}, \quad y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$$
Valid when denominator is non-zero.
Boat Speed Modeling
$$v_{\text{down}} = x + y, \quad v_{\text{up}} = x - y$$
x is still water speed, y is river current speed.

Mathematics: Geometric Intersecting Lines & Cross-Multiplication Scheme

Simultaneous Linear Equations: Geometry & Cross-Multiplication Geometric Interpretation: Unique Solution X Y L₁: 3x + 4y = 10 L₂: 2x - 2y = 2 P (2, 1) x=2 y=1 Unique Intersection Point = Unique Solution Cross-Multiplication Method Scheme x y 1 b₁ c₁ a₁ b₁ b₂ c₂ a₂ b₂ Master Formula: x / (b₁c₂ - b₂c₁) = y / (c₁a₂ - c₂a₁) = 1 / (a₁b₂ - a₂b₁) ⇒ x = (b₁c₂ - b₂c₁) / (a₁b₂ - a₂b₁) ⇒ y = (c₁a₂ - c₂a₁) / (a₁b₂ - a₂b₁)

Chapter Summary & 10 Key Takeaways

Takeaway 1
A system of simultaneous linear equations consists of two first-degree equations with two shared unknown variables.
Takeaway 2
Geometrically, the solution pair (x, y) represents the exact point of intersection of two straight lines.
Takeaway 3
A unique solution exists if and only if a1/a2 ≠ b1/b2 (lines intersect).
Takeaway 4
No solution exists if a1/a2 = b1/b2 ≠ c1/c2 (lines are parallel and inconsistent).
Takeaway 5
Infinitely many solutions exist if a1/a2 = b1/b2 = c1/c2 (lines are coincident and dependent).
Takeaway 6
Elimination by equating coefficients involves multiplying equations to cancel one variable upon addition or subtraction.
Takeaway 7
The cross-multiplication method uses the cyclic determinant formula x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1).
Takeaway 8
Symmetric systems ax + by = m and bx + ay = n are solved instantly by adding and subtracting equations.
Takeaway 9
Equations with variables in denominators are solved by substituting auxiliary variables u = 1/x and v = 1/y.
Takeaway 10
Downstream speed is (x + y) and upstream speed is (x - y), where x is boat speed in still water and y is stream speed.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Solve the following system of linear equations using the method of cross-multiplication:
$$2x + 3y - 17 = 0$$
$$3x - 2y - 6 = 0$$
Reveal Answer & Explanation
Answer: • Write the system in standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 2, b_1 = 3, c_1 = -17$
$a_2 = 3, b_2 = -2, c_2 = -6$
• Apply the cross-multiplication formula:
$$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$$
• Compute each denominator:
$$b_1c_2 - b_2c_1 = (3)(-6) - (-2)(-17) = -18 - 34 = -52$$
$$c_1a_2 - c_2a_1 = (-17)(3) - (-6)(2) = -51 - (-12) = -51 + 12 = -39$$
$$a_1b_2 - a_2b_1 = (2)(-2) - (3)(3) = -4 - 9 = -13$$
• Set up the equalities:
$$\frac{x}{-52} = \frac{y}{-39} = \frac{1}{-13}$$
• Solve for $x$ and $y$:
$$x = \frac{-52}{-13} = \mathbf{4}$$
$$y = \frac{-39}{-13} = \mathbf{3}$$
Identify a1=2, b1=3, c1=-17 and a2=3, b2=-2, c2=-6. Compute b1c2 - b2c1 = -52 and a1b2 - a2b1 = -13.
2
Solve for $x$ and $y$:
$$\frac{10}{x+y} + \frac{2}{x-y} = 4$$
$$\frac{15}{x+y} - \frac{5}{x-y} = -2$$
Reveal Answer & Explanation
Answer: • Let $u = \frac{1}{x+y}$ and $v = \frac{1}{x-y}$.
• The system transforms into linear form in $u$ and $v$:
$$10u + 2v = 4 \implies 5u + v = 2 \quad \text{--- (1)}$$
$$15u - 5v = -2 \quad \text{--- (2)}$$
• From (1), $v = 2 - 5u$. Substitute into (2):
$$15u - 5(2 - 5u) = -2 \implies 15u - 10 + 25u = -2$$
$$40u = 8 \implies u = \frac{8}{40} = \frac{1}{5}$$
• Find $v$: $v = 2 - 5\left(\frac{1}{5}\right) = 2 - 1 = 1$.
• Substitute back $u$ and $v$ into $x$ and $y$:
$$u = \frac{1}{x+y} = \frac{1}{5} \implies x + y = 5 \quad \text{--- (3)}$$
$$v = \frac{1}{x-y} = 1 \implies x - y = 1 \quad \text{--- (4)}$$
• Add (3) and (4): $2x = 6 \implies \mathbf{x = 3}$.
• Subtract (4) from (3): $2y = 4 \implies \mathbf{y = 2}$.
Set u = 1/(x+y) and v = 1/(x-y). Solve for u = 1/5, v = 1, then solve x + y = 5 and x - y = 1.
3
The sum of the digits of a two-digit number is $9$. If $27$ is added to the number, its digits are reversed. Find the original number.
Reveal Answer & Explanation
Answer: • Let the tens digit be $x$ and the units digit be $y$.
Original number $= 10x + y$.
Number with reversed digits $= 10y + x$.
• Condition 1: Sum of the digits is $9$:
$$x + y = 9 \quad \text{--- (1)}$$
• Condition 2: Adding 27 reverses the digits:
$$(10x + y) + 27 = 10y + x$$
$$10x - x + y - 10y = -27 \implies 9x - 9y = -27 \implies x - y = -3 \quad \text{--- (2)}$$
• Add Equation (1) and Equation (2):
$$2x = 6 \implies x = 3$$
• Substitute $x = 3$ into (1):
$$3 + y = 9 \implies y = 6$$
• Therefore, the original number is $10(3) + 6 = \mathbf{36}$.
• Check: $3 + 6 = 9$, and $36 + 27 = 63$ (digits reversed). Correct!
Set original number as 10x + y and reversed as 10y + x. System: x + y = 9 and 9x - 9y = -27.
4
A boat travels $30\text{ km}$ upstream and $44\text{ km}$ downstream in $10\text{ hours}$. In $13\text{ hours}$, it can cover $40\text{ km}$ upstream and $55\text{ km}$ downstream. Determine the speed of the boat in still water and that of the stream.
Reveal Answer & Explanation
Answer: • Let speed of boat in still water $= x\text{ km/h}$, speed of stream $= y\text{ km/h}$.
Upstream speed $= (x - y)\text{ km/h}$, Downstream speed $= (x + y)\text{ km/h}$.
• Let $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$.
• Time equations:
$$30u + 44v = 10 \quad \text{--- (1)}$$
$$40u + 55v = 13 \quad \text{--- (2)}$$
• Multiply (1) by $4$ and (2) by $3$ to eliminate $u$:
$$120u + 176v = 40 \quad \text{--- (3)}$$
$$120u + 165v = 39 \quad \text{--- (4)}$$
• Subtract (4) from (3):
$$11v = 1 \implies v = \frac{1}{11}$$
• Substitute $v = \frac{1}{11}$ into (1):
$$30u + 44\left(\frac{1}{11}\right) = 10 \implies 30u + 4 = 10 \implies 30u = 6 \implies u = \frac{6}{30} = \frac{1}{5}$$
• Thus:
$$x - y = \frac{1}{u} = 5 \quad \text{--- (5)}$$
$$x + y = \frac{1}{v} = 11 \quad \text{--- (6)}$$
• Adding (5) and (6): $2x = 16 \implies \mathbf{x = 8\text{ km/h}}$ (Speed of boat in still water).
• Subtracting (5) from (6): $2y = 6 \implies \mathbf{y = 3\text{ km/h}}$ (Speed of stream).
Set u = 1/(x-y) and v = 1/(x+y). Solve to get u = 1/5 (upstream=5) and v = 1/11 (downstream=11).
5
For what value of $k$ will the following system of linear equations have no solution?
$$kx + 3y = k - 2$$
$$12x + ky = k$$
Reveal Answer & Explanation
Answer:

• The condition for a system to have no solution (parallel lines) is:

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$


• Here, $a_1 = k, b_1 = 3, c_1 = -(k-2)$ and $a_2 = 12, b_2 = k, c_2 = -k$.
• Setting $\frac{a_1}{a_2} = \frac{b_1}{b_2}$:

$$\frac{k}{12} = \frac{3}{k} \implies k^2 = 36 \implies k = \pm 6$$


• Check the inequality $\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$:
- If $k = 6$: $\frac{3}{6} = \frac{1}{2}$ and $\frac{-(6-2)}{-6} = \frac{4}{6} = \frac{2}{3}$. Since $\frac{1}{2} \neq \frac{2}{3}$, $k = 6$ produces no solution.
- If $k = -6$: $\frac{3}{-6} = -\frac{1}{2}$ and $\frac{-(-6-2)}{-(-6)} = \frac{-8}{6} = -\frac{4}{3} \neq -\frac{1}{2}$.
- For $k = -6$: $\frac{-6}{12} = \frac{3}{-6} = -\frac{1}{2} \neq -\frac{4}{3}$.
• Thus, for $k = -6$, the equations also have no solution. Specifically, for distinct lines with no intersection: $k = -6$ (and $k = 6$).


Equate k/12 = 3/k to get k^2 = 36, then verify the constant ratio.
6
Solve by substitution: $2x + y = 7$ and $4x - 3y + 1 = 0$.
Reveal Answer & Explanation
Answer:

• From the first equation, isolate $y$:

$$y = 7 - 2x \quad \text{--- (1)}$$


• Substitute $y = 7 - 2x$ into the second equation:

$$4x - 3(7 - 2x) + 1 = 0$$


$$4x - 21 + 6x + 1 = 0 \implies 10x - 20 = 0 \implies 10x = 20 \implies x = 2$$


• Substitute $x = 2$ back into (1):

$$y = 7 - 2(2) = 7 - 4 = 3$$


• Solution: $\mathbf{x = 2, \quad y = 3}$.


Isolate y = 7 - 2x from first equation and substitute into 4x - 3y + 1 = 0.
7
A fraction becomes $\frac{1}{3}$ when $1$ is subtracted from the numerator, and it becomes $\frac{1}{4}$ when $8$ is added to its denominator. Find the fraction.
Reveal Answer & Explanation
Answer: • Let the fraction be $\frac{x}{y}$.
• Condition 1: $\frac{x - 1}{y} = \frac{1}{3} \implies 3(x - 1) = y \implies 3x - y = 3$   --- (1)
• Condition 2: $\frac{x}{y + 8} = \frac{1}{4} \implies 4x = y + 8 \implies 4x - y = 8$   --- (2)
• Subtract Equation (1) from Equation (2):
$$(4x - y) - (3x - y) = 8 - 3 \implies x = 5$$
• Substitute $x = 5$ into Equation (1):
$$3(5) - y = 3 \implies 15 - y = 3 \implies y = 12$$
• Therefore, the fraction is $\mathbf{\frac{5}{12}}$.
• Check: $\frac{5-1}{12} = \frac{4}{12} = \frac{1}{3}$, and $\frac{5}{12+8} = \frac{5}{20} = \frac{1}{4}$. Perfect!
Set fraction as x/y. System: 3x - y = 3 and 4x - y = 8. Subtract to find x = 5.
8
Solve the symmetric system of equations: $217x + 131y = 913$ and $131x + 217y = 827$.
Reveal Answer & Explanation
Answer:

• Step 1 (Add both equations):

$$(217 + 131)x + (131 + 217)y = 913 + 827$$


$$348x + 348y = 1740 \implies x + y = \frac{1740}{348} = 5 \quad \text{--- (1)}$$


• Step 2 (Subtract the second from the first):

$$(217 - 131)x + (131 - 217)y = 913 - 827$$


$$86x - 86y = 86 \implies x - y = 1 \quad \text{--- (2)}$$


• Step 3 (Solve the simplified system):
Adding (1) and (2): $2x = 6 \implies \mathbf{x = 3}$.
Subtracting (2) from (1): $2y = 4 \implies \mathbf{y = 2}$.
• Solution: $\mathbf{x = 3, \quad y = 2}$.


Add equations to get x + y = 5. Subtract to get x - y = 1. Then x = 3, y = 2.
Finished Studying This Chapter?
READY TO PRACTICE?

Timed CBT Practice Tests (Exam Simulator)

Put your concepts to the test with official curriculum-aligned Foundation and Advanced practice tests. Get instant accuracy scores, time metrics, and step-by-step verified explanations.

Foundation Test
15 MCQs • Core Concepts
Advanced Test
15 MCQs • HOTS & Traps

All Class 9 Mathematics Chapters

AI Study Friend

Instant Concept Tutor

Have a doubt in Simultaneous (Linear) Equations? Ask our AI study tutor for rapid explanations or customized quizzes.