Mastering algebra is a crucial milestone in your high school journey, and nothing builds that foundation quite like the right study resources. If you are preparing for your upcoming board exams, finding reliable and precise ncert solutions class 10 maths chapter 3 is essential for securing top marks. Chapter 3, "Pair of Linear Equations in Two Variables," forms the backbone of algebraic problem-solving in the CBSE curriculum. Questions from this chapter frequently appear in board papers carrying significant weightage, making conceptual clarity non-negotiable.
Whether you are struggling to understand graphical representations or want to master the algebraic methods of elimination and substitution, having access to structured, step-by-step answers changes everything. Let us dive right into the core concepts and detailed solutions for this vital chapter to help you ace your NCERT solutions practice sessions.
Understanding the Basics of Linear Equations
Before jumping straight into the textbook problems, let's refresh what a linear equation in two variables looks like. A general equation of this type is written in the form:
$ax + by + c = 0$
Here, $a$, $b$, and $c$ are real numbers, and both $a$ and $b$ are non-zero. When we deal with a pair of such equations simultaneously, we call it a system of linear equations.
General Form of a Pair of Linear Equations
The standard algebraic representation for a pair of linear equations in two variables is:
- $a_1x + b_1y + c_1 = 0$
- $a_2x + b_2y + c_2 = 0$
Where $a_1, b_1, c_1, a_2, b_2,$ and $c_2$ are all real numbers.
Consistency and Graphical Meaning
When you plot these two equations on a Cartesian plane, they form two straight lines. Depending on the coefficients, these lines can intersect, be parallel, or coincide.
- Intersecting Lines: Exactly one solution (Consistent system). Condition: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
- Coincident Lines: Infinitely many solutions (Consistent and dependent system). Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
- Parallel Lines: No solution (Inconsistent system). Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
To review these foundational definitions alongside other topics, feel free to browse all chapter notes on our platform.
Graphical Method of Solution
The graphical method requires you to find at least two solutions for each linear equation, plot the corresponding points on graph paper, and draw the lines.
Example Problem: Solving Graphically
Solve the following pair of equations graphically: $x - y = 1$ $2x + 3y = 12$
Step-by-step Working:
For the first equation ($x - y = 1$ or $y = x - 1$):
- If $x = 0$, then $y = -1$. (Point: $(0, -1)$)
- If $x = 3$, then $y = 2$. (Point: $(3, 2)$)
For the second equation ($2x + 3y = 12$ or $y = \frac{12 - 2x}{3}$):
- If $x = 0$, then $y = 4$. (Point: $(0, 4)$)
- If $x = 3$, then $y = 2$. (Point: $(3, 2)$)
Final Answer:
Notice that the point $(3, 2)$ is common to both equations. Therefore, the unique solution to the pair of linear equations is $x = 3, y = 2$.
For more practice problems of this kind, check out our interactive practice tests to test your speed and accuracy.
Algebraic Methods: Substitution and Elimination
While graphs are great for visualization, algebraic methods are much faster and more precise during exams. As per the official NCERT guidelines, students must master both substitution and elimination methods.
1. The Substitution Method
In this method, you express one variable in terms of the other from one equation and substitute it into the second equation.
Sample Problem:
$7x - 15y = 2$ $x + 2y = 3$
- Step 1: From the second equation, express $x$ in terms of $y$: $x = 3 - 2y$
- Step 2: Substitute this value of $x$ into the first equation: $7(3 - 2y) - 15y = 2$ $21 - 14y - 15y = 2$ $-29y = 2 - 21$ $-29y = -19 \implies y = \frac{19}{29}$
- Step 3: Substitute $y$ back into the expression for $x$: $x = 3 - 2\left(\frac{19}{29}\right) = 3 - \frac{38}{29} = \frac{87 - 38}{29} = \frac{49}{29}$
Final Answer:
$x = \frac{49}{29}, y = \frac{19}{29}$
2. The Elimination Method
Here, you multiply one or both equations by a suitable non-zero constant to make the coefficients of one variable numerically equal. Then, you add or subtract the equations to eliminate that variable.
Sample Problem:
$3x + 4y = 10$ $2x - 2y = 2$
- Step 1: Multiply the second equation by 2 so the coefficients of $y$ match: $4x - 4y = 4$
- Step 2: Add this new equation to the first equation: $(3x + 4y) + (4x - 4y) = 10 + 4$ $7x = 14 \implies x = 2$
- Step 3: Substitute $x = 2$ into the second original equation: $2(2) - 2y = 2$ $4 - 2y = 2 \implies 2y = 2 \implies y = 1$
Final Answer:
$x = 2, y = 1$
Equations Reducible to a Pair of Linear Equations
Sometimes, equations are not linear in form initially, but they can be reduced to linear equations by applying simple substitutions.
Sample Problem:
$\frac{2}{x} + \frac{3}{y} = 13$ $\frac{5}{x} - \frac{4}{y} = -2$
Step-by-step Working:
Substitution: Let $\frac{1}{x} = p$ and $\frac{1}{y} = q$. The equations become: $2p + 3q = 13$ — (Equation 1) $5p - 4q = -2$ — (Equation 2)
Solve for $p$ and $q$: Multiply Equation 1 by 4 and Equation 2 by 3: $8p + 12q = 52$ $15p - 12q = -6$
Add both equations: $23p = 46 \implies p = 2$
Substitute $p = 2$ into Equation 1: $2(2) + 3q = 13 \implies 4 + 3q = 13 \implies 3q = 9 \implies q = 3$
Find $x$ and $y$: Since $p = \frac{1}{x} = 2$, we get $x = \frac{1}{2}$. Since $q = \frac{1}{y} = 3$, we get $y = \frac{1}{3}$.
Final Answer:
$x = \frac{1}{2}, y = \frac{1}{3}$
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Tips to Score Full Marks in Board Exams
Scoring a century in Class 10 Mathematics requires more than just knowing formulas; it demands immaculate presentation. Follow these expert tips:
- Always Write Conditions: Before solving, check the ratio of coefficients ($\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, $\frac{c_1}{c_2}$) to confirm whether a solution exists, especially for word problems.
- Label Your Graphs Clearly: If a question asks for a graphical solution, label the axes ($X$, $X'$, $Y$, $Y'$), origin ($0$), scale, and the intersection point clearly on the graph sheet.
- Double-Check Units in Word Problems: Age, speed, and distance problems require careful unit handling. Always write the final statement with appropriate units (e.g., "Speed = 40 km/h").
- Show Working Steps: Never jump straight from the equation to the final answer. Examiners award step marks, which can save your score even if a minor calculation error happens at the end.
- Practice Word Problems Daily: Many students lose marks in statement-based questions. Practice translating sentences into algebraic equations regularly.
Frequently Asked Questions
What are the main methods to solve a pair of linear equations in two variables?
There are two primary categories of methods: Graphical and Algebraic. The algebraic methods are further divided into substitution, elimination, and cross-multiplication methods (though cross-multiplication is omitted in some recent curriculum revisions, substitution and elimination remain essential).
How do I know if a pair of linear equations has no solution?
A pair of linear equations has no solution when the lines represented by them are parallel. Mathematically, this occurs when the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ is satisfied.
Can a pair of linear equations have more than one solution?
No. A pair of linear equations in two variables can either have a unique solution, no solution, or infinitely many solutions. It can never have a finite number of solutions greater than one (like two or three solutions).
Why are graphical solutions important if algebraic methods are faster?
Graphical solutions help you visually comprehend the geometric meaning of linear equations. They build your spatial reasoning and intuition, which is especially helpful when dealing with inequalities and linear programming in higher classes.
How many questions from Chapter 3 typically appear in the board exam?
On average, board question papers feature 1 to 3 questions from this chapter, ranging from 1-mark objective questions to 3 or 5-mark long-answer word problems.
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