Pair of Linear Equations in Two Variables - Algebraic Methods & Consistency Conditions

An equation of the form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers such that $a$ and $b$ are not both zero ($a^2 + b^2 \neq 0$), is called a linear equation in two variables $x$ and $y$. Geometrically, every solution $(x, y)$ of a linear equation corresponds to a point on the line representing it in the Cartesian plane.

In practical situations—such as determining unknown costs, rates of work, speeds, or age relationships—we often encounter situations involving two unknown quantities governed by two simultaneous conditions. This leads to a system or pair of linear equations in two variables.

The general form of a pair of linear equations in two variables is: $a_1x + b_1y + c_1 = 0$ $a_2x + b_2y + c_2 = 0$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers, and $a_1^2 + b_1^2 \neq 0$, $a_2^2 + b_2^2 \neq 0$.

While graphical methods provide an intuitive visualization of how two lines behave, they have limitations when coordinates are non-integers, fractions, or irrational numbers like $\sqrt{3}$. Hence, mastering algebraic methods (Substitution and Elimination) alongside understanding the conditions for consistency is essential for precision and efficient problem-solving in Class 10 Board examinations.

Conditions for Consistency

When two linear equations in two variables are plotted on a Cartesian plane, three geometric possibilities arise:

1. The two lines intersect at a single point.
2. The two lines are parallel and never meet.
3. The two lines coincide (overlap entirely).

Algebraically, we do not need to draw graphs to determine which scenario applies. We can predict the geometric behavior and the number of solutions simply by comparing the ratios of the coefficients $a_1, a_2, b_1, b_2$ and constant terms $c_1, c_2$.

Ratio Comparison Table

Detailed Analysis of Consistency Conditions

1. Unique Solution (Consistent System)

• Condition: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
• Explanation: The slopes of the two lines are different. Because their directions differ, they are guaranteed to cross each other at exactly one point $(x, y)$. This unique pair satisfies both equations simultaneously.
• Example: For $2x + 3y - 9 = 0$ and $4x + 5y - 18 = 0$: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{5}$ Since $\frac{1}{2} \neq \frac{3}{5}$, the system has a unique solution and is consistent.

2. Infinitely Many Solutions (Consistent & Dependent System)

• Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
• Explanation: One equation is simply a non-zero scalar multiple of the other. Graphically, both equations represent the exact same line. Every point lying on line 1 also lies on line 2, resulting in infinitely many common solutions.
• Example: For $2x + 3y - 9 = 0$ and $4x + 6y - 18 = 0$: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-9}{-18} = \frac{1}{2}$ Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines coincide, offering infinitely many solutions.

3. No Solution (Inconsistent System)

• Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
• Explanation: The slopes of the lines are equal (as indicated by equal coefficient ratios for $x$ and $y$), but their $y$-intercepts are different (as indicated by the unequal constant ratio). The lines run parallel alongside each other and will never intersect. Thus, no common pair $(x, y)$ satisfies both equations.
• Example: For $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$: $\frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-4}{-12} = \frac{1}{3}$ Here, $\frac{1}{2} = \frac{1}{2} \neq \frac{1}{3}$. The system has no solution and is inconsistent.

Algebraic Methods of Solving Linear Equations

When a pair of linear equations is consistent and possesses a unique solution, we employ algebraic methods to calculate the exact values of $x$ and $y$.

Method 1: The Substitution Method

The substitution method operates by expressing one variable in terms of the other from one equation, and then substituting this expression into the second equation. This reduces a system of two equations with two unknowns into a single linear equation in one variable.

Step-by-Step Algorithm for Substitution

1. Choose one equation (preferably the one where a coefficient of a variable is $+1$ or $-1$) and express one variable in terms of the other. For example, express $y$ in terms of $x$: $y = g(x)$.
2. Substitute this value of $y$ into the other equation. This produces a linear equation containing only $x$.
3. Solve the single-variable equation for $x$.
4. Substitute the calculated value of $x$ back into the expression obtained in Step 1 to find $y$.
5. Verify your solution by checking if $(x, y)$ satisfies both original equations.

Special Cases During Substitution:

• If while solving you get a statement with no variables that is true (e.g., $9 = 9$), the system has infinitely many solutions.
• If you get a statement with no variables that is false (e.g., $0 = 5$), the system has no solution.

Method 2: The Elimination Method

The elimination method involves multiplying one or both equations by suitable non-zero constants so that the coefficients of one variable become equal in absolute value. Adding or subtracting the resulting equations then "eliminates" that variable.

Step-by-Step Algorithm for Elimination

1. Write both equations in the standard form $ax + by = c$ or $ax + by + c = 0$.
2. Choose a variable to eliminate ($x$ or $y$).
3. Multiply one or both equations by suitable non-zero constants such that the coefficients of the chosen variable become equal in absolute value (the Least Common Multiple/LCM of the original coefficients).
4. Add or Subtract:
   • If the signs of the coefficients of the chosen variable are opposite, add the two equations.
   • If the signs are the same, subtract one equation from the other.
5. Solve the resulting single-variable equation for the remaining variable.
6. Substitute this value into either of the original equations to solve for the eliminated variable.

Method Comparison: Substitution vs. Elimination

Real-World Applications

1. Speed, Distance, and Time (Upstream and Downstream)

In water transportation, a boat moves with its own speed in still water ($x$ km/h), while the river current moves at a speed ($y$ km/h).

• Upstream Speed (against current): $(x - y)\text{ km/h}$
• Downstream Speed (with current): $(x + y)\text{ km/h}$

Using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$, real-world scenarios convert into a pair of linear equations.

2. Fixed and Variable Cost Structure

Many daily services use split-pricing models:

• Taxis charge a fixed fee plus a rate per kilometer.
• Public libraries charge a fixed fee for initial days plus a daily penalty fee thereafter.

Let $x$ be the fixed charge and $y$ be the variable rate per unit. Formulating two user instances yields a pair of equations that are easily solved via elimination.

3. Number Theory and Digit Reversal

Two-digit numbers are represented using place value:

• If tens digit $= x$ and units digit $= y$, then the number $= 10x + y$.
• Reversing the digits yields the new number $= 10y + x$.

Problems involving sums or differences of digits and numbers create linear systems.

Step-by-Step Solved Examples

Example 1: Finding an Unknown Parameter for Consistency

Problem: Find the value of $k$ for which the following system of linear equations has a unique solution: $kx + 2y = 5$ $3x + y = 1$

Solution:

Step 1: Write both equations in standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. $kx + 2y - 5 = 0 \quad \text{--- (Equation 1)}$ $3x + 1y - 1 = 0 \quad \text{--- (Equation 2)}$

Step 2: Identify coefficients and constant terms:

• $a_1 = k, \quad b_1 = 2, \quad c_1 = -5$
• $a_2 = 3, \quad b_2 = 1, \quad c_2 = -1$

Step 3: Apply the condition for a unique solution: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Step 4: Substitute the coefficient values into the inequality: $\frac{k}{3} \neq \frac{2}{1}$

Step 5: Cross-multiply to solve for $k$: $k \neq 3 \times 2 \implies k \neq 6$

Final Answer: The given system of linear equations will have a unique solution for all real values of $k$ except $k = 6$.

Example 2: Solving System via Substitution Method

Problem: Solve the following pair of linear equations by the Substitution Method: $7x - 15y = 2 \quad \text{--- (Equation 1)}$ $x + 2y = 3 \quad \text{--- (Equation 2)}$

Solution:

Step 1: Select Equation 2 because the coefficient of $x$ is $1$, making it easy to isolate $x$: $x + 2y = 3 \implies x = 3 - 2y \quad \text{--- (Equation 3)}$

Step 2: Substitute $x = 3 - 2y$ into Equation 1: $7(3 - 2y) - 15y = 2$

Step 3: Expand and solve the single-variable equation for $y$: $21 - 14y - 15y = 2$ $21 - 29y = 2$ $-29y = 2 - 21$ $-29y = -19 \implies y = \frac{19}{29}$

Step 4: Substitute $y = \frac{19}{29}$ back into Equation 3 to find $x$: $x = 3 - 2\left(\frac{19}{29}\right)$ $x = 3 - \frac{38}{29}$ $x = \frac{3 \times 29 - 38}{29} = \frac{87 - 38}{29} = \frac{49}{29}$

Step 5: Verification: Substitute $x = \frac{49}{29}, y = \frac{19}{29}$ into Equation 2: $\frac{49}{29} + 2\left(\frac{19}{29}\right) = \frac{49 + 38}{29} = \frac{87}{29} = 3 \quad \text{(Verified)}$

Final Answer: The required solution is $x = \frac{49}{29}, y = \frac{19}{29}$.

Example 3: Solving System via Elimination Method

Problem: Solve the following pair of linear equations using the Elimination Method: $9x - 4y = 2000 \quad \text{--- (Equation 1)}$ $7x - 3y = 2000 \quad \text{--- (Equation 2)}$

Solution:

Step 1: Choose a variable to eliminate. Let us eliminate $y$.

• Coefficient of $y$ in Eq 1 = $-4$
• Coefficient of $y$ in Eq 2 = $-3$
• LCM of $4$ and $3$ = $12$

Step 2: Multiply Equation 1 by $3$ and Equation 2 by $4$ to make $y$-coefficients equal: $\text{Eq 1} \times 3 \implies 27x - 12y = 6000 \quad \text{--- (Equation 3)}$ $\text{Eq 2} \times 4 \implies 28x - 12y = 8000 \quad \text{--- (Equation 4)}$

Step 3: Since both coefficients of $y$ have the same sign ($-12$), subtract Equation 3 from Equation 4: $(28x - 12y) - (27x - 12y) = 8000 - 6000$ $28x - 27x - 12y + 12y = 2000$ $x = 2000$

Step 4: Substitute $x = 2000$ into Equation 1 to find $y$: $9(2000) - 4y = 2000$ $18000 - 4y = 2000$ $18000 - 2000 = 4y$ $16000 = 4y \implies y = \frac{16000}{4} = 4000$

Step 5: Verification: Substitute $x = 2000, y = 4000$ into Equation 2: $7(2000) - 3(4000) = 14000 - 12000 = 2000 \quad \text{(Verified)}$

Final Answer: The solution is $x = 2000, y = 4000$.

Example 4: Word Problem Involving Two-Digit Numbers

Problem: The sum of a two-digit number and the number obtained by reversing the digits is $66$. If the digits of the number differ by $2$, find the number. How many such numbers are there?

Solution:

Step 1: Let the tens digit be $x$ and the units digit be $y$.

• Original Number $= 10x + y$
• Reversed Number $= 10y + x$

Step 2: Express Condition 1 algebraically: $\text{Original Number} + \text{Reversed Number} = 66$ $(10x + y) + (10y + x) = 66$ $11x + 11y = 66$ Divide by 11: $x + y = 6 \quad \text{--- (Equation 1)}$

Step 3: Express Condition 2 algebraically: The digits differ by $2$, which means either $x - y = 2$ or $y - x = 2$.

• Case I: $x - y = 2 \quad \text{--- (Equation 2)}$ Add Equation 1 and Equation 2: $(x + y) + (x - y) = 6 + 2$ $2x = 8 \implies x = 4$ Substitute $x = 4$ into Eq 1: $4 + y = 6 \implies y = 2$ $\text{Number} = 10(4) + 2 = 42$

• Case II: $y - x = 2 \quad \text{or} \quad -x + y = 2 \quad \text{--- (Equation 3)}$ Add Equation 1 and Equation 3: $(x + y) + (-x + y) = 6 + 2$ $2y = 8 \implies y = 4$ Substitute $y = 4$ into Eq 1: $x + 4 = 6 \implies x = 2$ $\text{Number} = 10(2) + 4 = 24$

Final Answer: There are two such numbers: $42$ and $24$.

Common Student Mistakes to Avoid

1. Inconsistent Formats for Constants $c_1$ and $c_2$

• Mistake: Checking ratio conditions when one equation is written as $a_1x + b_1y = c_1$ and the other as $a_2x + b_2y + c_2 = 0$.
• Correct Approach: Always align both equations into identical forms—either both standard ($a x + b y + c = 0$) or both with constants on the right-hand side ($a x + b y = c$)—before extracting coefficients.

2. Sign Errors During Subtraction in Elimination

• Mistake: When subtracting $(28x - 12y = 8000)$ from $(27x - 12y = 6000)$, failing to distribute the negative sign across all terms, resulting in wrong arithmetic like $-12y - 12y = -24y$.
• Correct Approach: Enclose the subtracted equation inside parentheses and explicitly flip the sign of every term: $-(28x - 12y = 8000) \implies -28x + 12y = -8000$.

3. Digit Value Representation Errors

• Mistake: Expressing a two-digit number simply as $xy$ (which algebraically means $x \times y$).
• Correct Approach: Use place value expanded notation: a two-digit number with tens digit $x$ and unit digit $y$ must strictly be written as $(10x + y)$.

4. Ignoring Absolute Difference in Word Problems

• Mistake: Assuming "the difference between digits is 2" means only $x - y = 2$.
• Correct Approach: Unless specified which digit is larger, you must analyze both cases: $x - y = 2$ and $y - x = 2$.

Practice Questions for Self-Assessment

Practice Question 1

Determine the value of $k$ for which the following system of linear equations has infinitely many solutions: $2x + 3y = 7$ $(k+1)x + (2k-1)y = 4k-1$

Practice Question 2

Solve the following pair of linear equations by the Substitution Method: $\sqrt{2}x + \sqrt{3}y = 0$ $\sqrt{3}x - \sqrt{8}y = 0$

Practice Question 3

A fraction becomes $\frac{9}{11}$ if $2$ is added to both the numerator and the denominator. If $3$ is added to both the numerator and the denominator, it becomes $\frac{5}{6}$. Find the fraction using the Elimination Method.

Solutions to Practice Questions

Solution to Practice Question 1

Write both equations in standard form $ax + by - c = 0$:

1. $2x + 3y - 7 = 0 \implies a_1 = 2, b_1 = 3, c_1 = -7$
2. $(k+1)x + (2k-1)y - (4k-1) = 0 \implies a_2 = k+1, b_2 = 2k-1, c_2 = -(4k-1)$

For infinitely many solutions: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Equating Ratio 1 and Ratio 2: $\frac{2}{k+1} = \frac{3}{2k-1}$ $2(2k - 1) = 3(k + 1)$ $4k - 2 = 3k + 3$ $4k - 3k = 3 + 2 \implies k = 5$

Verification with Ratio 3: $\frac{c_1}{c_2} = \frac{-7}{-(4(5)-1)} = \frac{7}{20-1} = \frac{7}{19}$ Checking $\frac{a_1}{a_2} = \frac{2}{5+1} = \frac{2}{6} = \frac{1}{3} \neq \frac{7}{19}$. Wait, let's re-verify $\frac{c_1}{c_2}$: $\frac{-7}{-(4k-1)} = \frac{3}{2k-1} \implies 7(2k-1) = 3(4k-1)$ $14k - 7 = 12k - 3 \implies 2k = 4 \implies k = 2$

Let's re-check carefully:

• For $k = 5$: $\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}$. $\frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}$. But $\frac{c_1}{c_2} = \frac{-7}{-(4(5)-1)} = \frac{7}{19} \neq \frac{1}{3}$.
• Hence $k=5$ makes $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ (Parallel/No solution).

Let's test $k=2$:

• $\frac{a_1}{a_2} = \frac{2}{3}$
• $\frac{b_1}{b_2} = \frac{3}{3} = 1 \neq \frac{2}{3}$.

What if there is no single value of $k$? Let's cross check:

1. $\frac{2}{k+1} = \frac{7}{4k-1} \implies 2(4k-1) = 7(k+1) \implies 8k - 2 = 7k + 7 \implies k = 9$.
2. $\frac{3}{2k-1} = \frac{7}{4k-1} \implies 3(4k-1) = 7(2k-1) \implies 12k - 3 = 14k - 7 \implies 2k = 4 \implies k = 2$.
3. $\frac{2}{k+1} = \frac{3}{2k-1} \implies 4k - 2 = 3k + 3 \implies k = 5$.

Since no single value of $k$ simultaneously satisfies $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, there is no real value of $k$ for which this system has infinitely many solutions.

Solution to Practice Question 2

Given system:

1. $\sqrt{2}x + \sqrt{3}y = 0 \quad \text{--- (Eq 1)}$
2. $\sqrt{3}x - \sqrt{8}y = 0 \quad \text{--- (Eq 2)}$

From Eq 1, express $x$ in terms of $y$: $\sqrt{2}x = -\sqrt{3}y \implies x = -\frac{\sqrt{3}}{\sqrt{2}}y \quad \text{--- (Eq 3)}$

Substitute $x$ into Eq 2: $\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - \sqrt{8}y = 0$ $-\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0$ Factoring out $y$: $y \left( -\frac{3}{\sqrt{2}} - 2\sqrt{2} \right) = 0$

Since the term inside the parenthesis is a non-zero real number: $y = 0$

Substitute $y = 0$ into Eq 3: $x = -\frac{\sqrt{3}}{\sqrt{2}}(0) = 0$

Final Answer: $x = 0, \quad y = 0$

Solution to Practice Question 3

Let the numerator of the fraction be $x$ and the denominator be $y$. The fraction is $\frac{x}{y}$.

Condition 1: $\frac{x + 2}{y + 2} = \frac{9}{11}$ $11(x + 2) = 9(y + 2)$ $11x + 22 = 9y + 18$ $11x - 9y = -4 \quad \text{--- (Eq 1)}$

Condition 2: $\frac{x + 3}{y + 3} = \frac{5}{6}$ $6(x + 3) = 5(y + 3)$ $6x + 18 = 5y + 15$ $6x - 5y = -3 \quad \text{--- (Eq 2)}$

Elimination Method: Multiply Eq 1 by $6$ and Eq 2 by $11$: $\text{Eq 1} \times 6 \implies 66x - 54y = -24 \quad \text{--- (Eq 3)}$ $\text{Eq 2} \times 11 \implies 66x - 55y = -33 \quad \text{--- (Eq 4)}$

Subtract Eq 4 from Eq 3: $(66x - 54y) - (66x - 55y) = -24 - (-33)$ $-54y + 55y = -24 + 33$ $y = 9$

Substitute $y = 9$ into Eq 1: $11x - 9(9) = -4$ $11x - 81 = -4$ $11x = 77 \implies x = 7$

Final Answer: The original fraction is $\frac{7}{9}$.

Exam Revision & Frequently Asked Questions (FAQs)

FAQ 1: What happens during substitution or elimination if the variables cancel out entirely and leave an identity like $0 = 0$?

Answer: When variables drop out leaving a true statement like $0 = 0$ or $12 = 12$, it algebraically indicates that the two given equations are mathematically identical (dependent). Graphically, the lines are coincident, and the system has infinitely many solutions.

FAQ 2: What happens if the variables cancel out and leave a false statement like $0 = 8$?

Answer: If all variables cancel out and leave a false identity, it means no set of values for $x$ and $y$ can ever satisfy both equations simultaneously. Graphically, the lines are parallel, and the system has no solution (inconsistent system).

FAQ 3: How should I choose between Substitution and Elimination during an exam if the question does not specify a method?

Answer:

• Use Substitution if one equation already presents $x$ or $y$ with a coefficient of $1$ or $-1$ (e.g., $x + 3y = 7$).
• Use Elimination if both equations have coefficients greater than $1$ (e.g., $3x + 4y = 10$ and $2x - 2y = 2$), as it avoids working with fractions in intermediate steps.

FAQ 4: How do we solve equations where variables appear in the denominator, such as $\frac{2}{x} + \frac{3}{y} = 13$?

Answer: Use substitution of variables to reduce them to standard linear equations. Let $\frac{1}{x} = u$ and $\frac{1}{y} = v$. The equation transforms into $2u + 3v = 13$. Solve for $u$ and $v$ using elimination or substitution, and finally convert back using $x = \frac{1}{u}$ and $y = \frac{1}{v}$.