Master Pair of Linear Equations in Two Variables Class 10 NCERT

Understanding Consistency and Graphical Representation

A pair of linear equations in two variables $x$ and $y$ is represented as:

1) $a_1x + b_1y + c_1 = 0$
2) $a_2x + b_2y + c_2 = 0$

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

When we plot these equations on a graph, they represent two straight lines. The relationship between their coefficients determines their behavior:

• Intersecting Lines: If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a unique point. The system is consistent and has a unique solution.
• Coincident Lines: If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines overlap completely. The system is consistent (dependent) and has infinitely many solutions.
• Parallel Lines: If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines never meet. The system is inconsistent and has no solution.

Step-by-Step Algebraic Methods for Solving Equations

1. Substitution Method — Express one variable in terms of the other from one equation (e.g., $x = \frac{-b_1y - c_1}{a_1}$). Substitute this expression into the second equation to get a single-variable equation. Solve for that variable, then substitute its value back to find the first variable.
2. Elimination Method — Multiply one or both equations by suitable non-zero constants so that the coefficients of one variable (either $x$ or $y$) become numerically equal. Add or subtract the equations to eliminate that variable, solve for the remaining variable, and substitute back.

Pro-Tips for Board Exams

1. Sign Mistakes: While writing the standard form $ax + by + c = 0$, keep an eye on negative signs. A common mistake is writing $c$ with the wrong sign if it starts on the right-hand side.
2. Check for Consistency First: Before solving algebraically, check $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$. If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, do not waste time trying to find $x$ and $y$ as there is no solution!
3. Graph Scaling: In graphical questions, clearly mark the intersection coordinates $(x, y)$ and write the equations on their respective plotted lines.

Practice Questions with Solutions

• Q: Solve the following pair of linear equations using the elimination method: $9x - 4y = 2000$ $7x - 3y = 2000$ A: Step 1: Write down the equations: (i) $9x - 4y = 2000$ (ii) $7x - 3y = 2000$ Step 2: Multiply equation (i) by 3 and equation (ii) by 4 to make the coefficients of $y$ equal: (iii) $27x - 12y = 6000$ (iv) $28x - 12y = 8000$ Step 3: Subtract equation (iii) from equation (iv): $(28x - 27x) - (12y - 12y) = 8000 - 6000$ $x = 2000$ Step 4: Substitute $x = 2000$ in equation (i): $9(2000) - 4y = 2000$ $18000 - 4y = 2000$ $4y = 16000 \implies y = 4000$ Final answer: $x = 2000$, $y = 4000$.
• Q: Find the value of $k$ for which the pair of linear equations $kx + 3y - (k-3) = 0$ and $12x + ky - k = 0$ has infinitely many solutions. A: Step 1: Identify coefficients: $a_1 = k, b_1 = 3, c_1 = -(k-3)$ $a_2 = 12, b_2 = k, c_2 = -k$ Step 2: Apply the condition for infinitely many solutions: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies \frac{k}{12} = \frac{3}{k} = \frac{k-3}{k}$ Step 3: Solve the first equality $\frac{k}{12} = \frac{3}{k}$: $k^2 = 36 \implies k = 6$ or $k = -6$ Step 4: Solve the second equality $\frac{3}{k} = \frac{k-3}{k}$: $3 = k-3 \implies k = 6$ (for $k \neq 0$) Step 5: Verify $k = 6$ in the ratio: $\frac{6}{12} = \frac{3}{6} = \frac{6-3}{6} = \frac{1}{2}$. Final answer: $k = 6$.
• Q: Solve the pair of linear equations using the substitution method: $7x - 15y = 2$ $x + 2y = 3$ A: Step 1: Let the equations be: (i) $7x - 15y = 2$ (ii) $x + 2y = 3$ Step 2: Express $x$ in terms of $y$ from equation (ii): $x = 3 - 2y$ Step 3: Substitute $x = 3 - 2y$ in equation (i): $7(3 - 2y) - 15y = 2$ $21 - 14y - 15y = 2$ $21 - 29y = 2$ $29y = 19 \implies y = \frac{19}{29}$ Step 4: Substitute $y = \frac{19}{29}$ back into $x = 3 - 2y$: $x = 3 - 2\left(\frac{19}{29}\right) = 3 - \frac{38}{29} = \frac{87 - 38}{29} = \frac{49}{29}$ Final answer: $x = \frac{49}{29}$, $y = \frac{19}{29}$.
• Q: Check whether the equations $x + 3y = 6$ and $2x - 3y = 12$ are consistent. If so, find their graphical intersection point. A: Step 1: Identify coefficients: $a_1 = 1, b_1 = 3, c_1 = -6$ $a_2 = 2, b_2 = -3, c_2 = -12$ Step 2: Check consistency ratio: $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{-3} = -1$ Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the equations are consistent with a unique solution. Step 3: Solve the equations to find the intersection point. Add both equations directly: $(x + 3y) + (2x - 3y) = 6 + 12$ $3x = 18 \implies x = 6$ Step 4: Substitute $x = 6$ into the first equation: $6 + 3y = 6 \implies 3y = 0 \implies y = 0$ Final answer: The system is consistent. The unique intersection point is $(6, 0)$.

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