Introduction and Pair of Linear Equations

What is a Linear Equation in Two Variables?

Let's start with something familiar. An equation like $2x + 3y = 12$ involves two unknowns, $x$ and $y$. This is a linear equation in two variables — 'linear' because each variable appears to the power 1, and 'two variables' because there are two unknowns.

The general form is:

$$ax + by + c = 0$$

where $a$, $b$, $c$ are real numbers and $a$ and $b$ are not both zero (i.e. $a^2 + b^2 \neq 0$).

Key Point: A single linear equation in two variables has infinitely many solutions — each solution is a pair $(x, y)$. For example, $(0, 4)$, $(6, 0)$, $(3, 2)$ all satisfy $2x + 3y = 12$.

Think of it this way: with two unknowns but only one equation, there isn't enough information to pin down a single answer — many pairs work.

[Board Important] Each solution of a linear equation corresponds to a point on its straight-line graph. That's why a line has infinitely many points.

A Pair (System) of Linear Equations

To nail down a unique answer, we usually need two equations together. Two linear equations in the same two variables form a pair of linear equations (also called a system):

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

A solution of the pair is a pair $(x, y)$ that satisfies both equations at the same time.

Example

Consider the pair: $x + y = 10$ and $x - y = 4$. The pair $(7, 3)$ works in both ($7+3 = 10$ ✓ and $7-3 = 4$ ✓), so $(7, 3)$ is the solution.

Key Point: A solution of a pair must satisfy both equations simultaneously. This is stricter than satisfying just one, which is why a pair usually has a single solution.

[Board Important] The whole chapter is about finding the pair $(x, y)$ that satisfies both equations — graphically or algebraically.

Forming Equations from Word Problems

Many board questions describe a real-life situation in words and ask you to form a pair of linear equations. The skill is to name the unknowns and translate each sentence into an equation.

Worked outline

'The cost of 2 pens and 3 pencils is ₹16, and the cost of 4 pens and 1 pencil is ₹22.'

1. Let the cost of one pen be ₹$x$ and one pencil be ₹$y$.
2. First sentence: $2x + 3y = 16$.
3. Second sentence: $4x + y = 22$.

These two equations are the required pair.

Key Point: Always begin by clearly stating what $x$ and $y$ represent (with units). Then convert each condition into one equation.

[Board Important] A common error is mixing up which quantity is $x$ and which is $y$. Write the 'Let …' statement first — it earns marks and prevents confusion.

Types of Solutions — A Preview

When we solve a pair of linear equations, exactly one of three things happens. Geometrically the two equations are two straight lines, and they can relate in three ways:

Lines	Solutions	Name
Intersect at one point	Exactly one (unique)	Consistent (independent)
Coincide (same line)	Infinitely many	Consistent (dependent)
Parallel (never meet)	No solution	Inconsistent

Key Point: A pair is consistent if it has at least one solution (one or infinitely many), and inconsistent if it has no solution.

We will explore each case in detail — graphically in the next section, and through the ratios of coefficients later.

[Board Important] Remember the vocabulary: 'consistent' = has a solution; 'inconsistent' = no solution. These exact words appear in board questions.

Solved Examples

Example 1: Check a solution

Is $(2, 3)$ a solution of the pair $x + y = 5$ and $2x - y = 1$?

Solution:

1. First equation: $2 + 3 = 5$. ✓
2. Second equation: $2(2) - 3 = 4 - 3 = 1$. ✓
3. Both are satisfied.

Final Answer: Yes, $(2, 3)$ is a solution.

Takeaway: A solution of a pair must satisfy both equations.

Example 2: Find some solutions of one equation

Find any two solutions of $3x + y = 9$.

Solution:

1. Put $x = 0$: $y = 9$, giving $(0, 9)$.
2. Put $x = 3$: $9 + y = 9 \Rightarrow y = 0$, giving $(3, 0)$.

Final Answer: $(0, 9)$ and $(3, 0)$ (others exist).

Takeaway: Choose convenient values of $x$ and solve for $y$.

Example 3: Form equations (cost problem)

Form a pair of equations: '5 oranges and 3 apples cost ₹35; 2 oranges and 4 apples cost ₹28.'

Solution:

1. Let one orange cost ₹$x$ and one apple cost ₹$y$.
2. $5x + 3y = 35$.
3. $2x + 4y = 28$.

Final Answer: $5x + 3y = 35$ and $2x + 4y = 28$.

Takeaway: Define the variables, then translate each sentence.

Example 4: Form equations (ages)

Form equations: 'The sum of the ages of a father and son is 45 years. Five years ago, the father was six times as old as the son.'

Solution:

1. Let the father's age be $x$ years and the son's age be $y$ years.
2. $x + y = 45$.
3. Five years ago: $(x - 5) = 6(y - 5)$, i.e. $x - 6y + 25 = 0$.

Final Answer: $x + y = 45$ and $x - 6y + 25 = 0$.

Takeaway: For 'years ago', subtract from both ages before forming the relation.

Example 5: Identify general-form coefficients

Write $2x = 5 - 3y$ in the form $ax + by + c = 0$ and state $a$, $b$, $c$.

Solution:

1. Move all terms to one side: $2x + 3y - 5 = 0$.
2. So $a = 2$, $b = 3$, $c = -5$.

Final Answer: $a = 2, b = 3, c = -5$.

Takeaway: Rearrange to standard form before reading off coefficients.

Example 6: How many solutions does one equation have?

How many solutions does the single equation $x + 2y = 6$ have?

Solution:

1. For each value of $x$, there is a value of $y$ (e.g. $(0,3), (2,2), (6,0)$).
2. There are infinitely many such pairs.

Final Answer: Infinitely many solutions.

Takeaway: A single linear equation in two variables has infinitely many solutions.

Example 7: Form equations (two-digit number)

Form equations: 'A two-digit number has digit-sum 9. On reversing the digits, the new number exceeds the original by 27.'

Solution:

1. Let the tens digit be $x$ and the units digit be $y$. Number $= 10x + y$.
2. Digit sum: $x + y = 9$.
3. Reversed number $= 10y + x$; condition: $(10y + x) - (10x + y) = 27 \Rightarrow 9y - 9x = 27 \Rightarrow y - x = 3$.

Final Answer: $x + y = 9$ and $y - x = 3$.

Takeaway: A two-digit number with digits $x$ (tens) and $y$ (units) equals $10x + y$.

Example 8: Consistent or not (by inspection)

The pair $x + y = 5$ and $x + y = 8$ — does it have a solution?

Solution:

1. Both ask for $x + y$ to equal two different numbers (5 and 8) at once.
2. This is impossible, so there is no solution.

Final Answer: No solution — the pair is inconsistent.

Takeaway: If two equations demand contradictory values, the system is inconsistent.

Example 9: Form equations (fraction problem)

Form equations: 'A fraction becomes $\dfrac{1}{2}$ when 1 is subtracted from the numerator, and becomes $\dfrac{1}{3}$ when 8 is added to the denominator.'

Solution:

1. Let the fraction be $\dfrac{x}{y}$.
2. $\dfrac{x - 1}{y} = \dfrac{1}{2} \Rightarrow 2(x-1) = y \Rightarrow 2x - y = 2$.
3. $\dfrac{x}{y + 8} = \dfrac{1}{3} \Rightarrow 3x = y + 8 \Rightarrow 3x - y = 8$.

Final Answer: $2x - y = 2$ and $3x - y = 8$.

Takeaway: Cross-multiply each fraction condition to clear denominators.

Example 10: Verify with both equations

Verify whether $(4, 1)$ solves $x + y = 5$ and $x - y = 3$.

Solution:

1. $x + y = 4 + 1 = 5$. ✓
2. $x - y = 4 - 1 = 3$. ✓
3. Both hold.

Final Answer: Yes, $(4, 1)$ is the solution.

Takeaway: Substitute the candidate pair into both equations to confirm.