NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2: Linear Equations in Two Variables

Algebraic Interpretation: Checking Consistency via Ratios

Before solving a pair of linear equations algebraically, we can predict the nature of their solutions by comparing the ratios of their coefficients. Consider the general form of a pair of linear equations in two variables:

1) a1x + b1y + c1 = 0
2) a2x + b2y + c2 = 0

By comparing the ratios of the coefficients (a1/a2, b1/b2, and c1/c2), we can determine whether the equations have a unique solution, infinitely many solutions, or no solution at all. This classification determines if the system is 'consistent' (having at least one solution) or 'inconsistent' (having no solution).

Let's analyze the three fundamental cases:

• Case 1: Intersecting Lines (Unique Solution)

If a1/a2 is not equal to b1/b2, the lines intersect at exactly one point. This means the system has a unique solution and is consistent.

• Case 2: Coincident Lines (Infinitely Many Solutions)

If a1/a2 = b1/b2 = c1/c2, the lines lie on top of each other. The system has infinitely many solutions and is consistent (dependent).

• Case 3: Parallel Lines (No Solution)

If a1/a2 = b1/b2 but is not equal to c1/c2, the lines never meet. The system has no solution and is inconsistent.

The Step-by-Step Substitution Method

1. Step 1: Express one variable in terms of the other — Choose either of the two equations and express one variable (say, x) in terms of the other variable (y). It is usually easiest to choose an equation where one of the coefficients is 1 or -1.
2. Step 2: Substitute this expression into the second equation — Substitute the value of x obtained in Step 1 into the OTHER equation. This converts the second equation into a linear equation in a single variable (y).
3. Step 3: Solve the single-variable equation — Solve the newly formed equation to find the numerical value of the variable (y).
4. Step 4: Back-substitute to find the second variable — Substitute the numerical value of y back into the expression obtained in Step 1 to calculate the value of x. Write the final answer as (x, y).

Worked NCERT-Pattern Examples

• Example 1: Solve the following pair of linear equations using the substitution method: x + y = 14 x - y = 4 Step 1: From the second equation (x - y = 4), express x in terms of y: x = y + 4 Step 2: Substitute x = y + 4 into the first equation (x + y = 14): (y + 4) + y = 14 2y + 4 = 14 2y = 10 y = 5 Step 3: Substitute y = 5 back into the expression x = y + 4: x = 5 + 4 = 9 Final Answer: x = 9, y = 5. (Verify: 9 + 5 = 14 and 9 - 5 = 4, which is correct!)
• Example 2: Check if the following pair of equations is consistent or inconsistent by comparing ratios: 2x + 3y - 9 = 0 4x + 6y - 18 = 0 Step 1: Identify the coefficients: a1 = 2, b1 = 3, c1 = -9 a2 = 4, b2 = 6, c2 = -18 Step 2: Calculate the ratios: a1/a2 = 2/4 = 1/2 b1/b2 = 3/6 = 1/2 c1/c2 = -9/-18 = 1/2 Step 3: Compare the ratios: Since a1/a2 = b1/b2 = c1/c2 = 1/2, the lines are coincident. Final Answer: The system has infinitely many solutions and is consistent.

Common Exam Mistakes & Board Tips

1. Sign Mistakes during Substitution: The most common error occurs when substituting an expression with a negative sign. Always use parentheses. For example, when substituting x = (3 - 2y) into 5x, write it as 5*(3 - 2y) = 15 - 10y instead of forgetting to distribute the multiplication.
2. Ignoring the standard form: Before comparing ratios (a1/a2, b1/b2, c1/c2), always rearrange both equations into the standard form: ax + by + c = 0. If one equation has the constant on the RHS and the other on the LHS, your sign for 'c' will be wrong, leading to an incorrect consistency check.
3. Double-Check by Substitution: Once you find the values of x and y, spend 10 seconds substituting them back into BOTH original equations. If they don't satisfy both, re-check your calculations immediately!

Practice Questions with Solutions

• Q: Solve the pair of linear equations using substitution: s - t = 3 s/3 + t/2 = 6 A: Step 1: Express s in terms of t from the first equation: s = t + 3 Step 2: Substitute s = t + 3 into the second equation: (t + 3)/3 + t/2 = 6 Step 3: Find a common denominator (6) to solve the fractional equation: 2(t + 3)/6 + 3t/6 = 6 (2t + 6 + 3t)/6 = 6 5t + 6 = 36 5t = 30 t = 6 Step 4: Substitute t = 6 back into the expression s = t + 3: s = 6 + 3 = 9 Final answer: s = 9, t = 6.
• Q: On comparing the ratios, find out if the lines representing the following pair of equations intersect at a point, are parallel, or coincide: 9x + 3y + 12 = 0 18x + 6y + 24 = 0 A: Step 1: Write down the coefficients: a1 = 9, b1 = 3, c1 = 12 a2 = 18, b2 = 6, c2 = 24 Step 2: Compute the ratios: a1/a2 = 9/18 = 1/2 b1/b2 = 3/6 = 1/2 c1/c2 = 12/24 = 1/2 Step 3: Analyze the relationship: Since a1/a2 = b1/b2 = c1/c2 = 1/2, the lines coincide. Final answer: The lines coincide (infinitely many solutions).
• Q: For what value of k will the following system of linear equations have no solution? 3x + y = 1 (2k - 1)x + (k - 1)y = 2k + 1 A: Step 1: For a system to have no solution, the condition is: a1/a2 = b1/b2 != c1/c2 Step 2: Write down the coefficients: a1 = 3, b1 = 1, c1 = -1 a2 = 2k-1, b2 = k-1, c2 = -(2k+1) Step 3: Equate the first two ratios: 3 / (2k - 1) = 1 / (k - 1) Cross-multiplying gives: 3(k - 1) = 1(2k - 1) 3k - 3 = 2k - 1 3k - 2k = 3 - 1 k = 2 Step 4: Verify the inequality with c1/c2: For k = 2, c1/c2 = -1 / -(22 + 1) = 1/5. Since a1/a2 = 3/(22-1) = 1, which is not equal to 1/5, the condition holds. Final answer: k = 2.
• Q: Solve 2x + 3y = 11 and 2x - 4y = -24. Hence, find the value of 'm' for which y = mx + 3. A: Step 1: Express 2x from the first equation: 2x = 11 - 3y Step 2: Substitute 2x = 11 - 3y into the second equation: (11 - 3y) - 4y = -24 11 - 7y = -24 -7y = -35 y = 5 Step 3: Find x using 2x = 11 - 3y: 2x = 11 - 3(5) 2x = 11 - 15 = -4 x = -2 Step 4: Substitute x = -2 and y = 5 into the equation y = mx + 3 to find m: 5 = m(-2) + 3 5 - 3 = -2m 2 = -2m m = -1 Final answer: x = -2, y = 5, and m = -1.

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