NCERT Class 10 Maths: Pair of Linear Equations in Two Variables

The Concept of Linear Equations in Two Variables

A linear equation in two variables $x$ and $y$ is an equation that can be written in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero. Geometrically, every solution $(x, y)$ of a linear equation corresponds to a point on the straight line representing it.

When we have two such equations, we call them a pair of linear equations in two variables. Their general representation is:
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$

Graphical Behavior and Consistency Conditions

Before solving, we can predict whether the lines will intersect, be parallel, or coincide by comparing the coefficients:

1. Intersecting Lines (Unique Solution): If $a_1/a_2 \neq b_1/b_2$, the lines intersect at a single point. The system is consistent.
2. Coincident Lines (Infinitely Many Solutions): If $a_1/a_2 = b_1/b_2 = c_1/c_2$, the lines overlap completely. The system is dependent and consistent.
3. Parallel Lines (No Solution): If $a_1/a_2 = b_1/b_2 \neq c_1/c_2$, the lines never meet. The system is inconsistent.

Algebraic Methods to Solve Pair of Linear Equations

1. Substitution Method — Step 1: Pick one equation and express one variable (say $y$) in terms of the other variable ($x$). Step 2: Substitute this expression for $y$ in the other equation to get an equation in only one variable ($x$). Step 3: Solve this equation to find the value of $x$. Step 4: Substitute the value of $x$ back into the first equation to find $y$.
2. Elimination Method — Step 1: Multiply one or both equations by suitable non-zero constants so that the coefficients of one variable (either $x$ or $y$) become numerically equal. Step 2: Add or subtract the equations to eliminate that variable. Step 3: Solve the resulting single-variable equation to get its value. Step 4: Substitute this value back into any original equation to solve for the second variable.

Avoid Common Board Exam Mistakes

Here are some crucial tips from CBSE examiners to keep you ahead:

• Watch out for sign changes: When shifting terms from the LHS to the RHS, students frequently forget to reverse the positive/negative sign. Double-check your arithmetic steps.
• Convert equations to standard form first: Always arrange equations in the form $ax + by + c = 0$ before finding ratios $a_1/a_2$, $b_1/b_2$, and $c_1/c_2$.
• State your variables clearly: In word problems, begin by explicitly defining what $x$ and $y$ represent (e.g., Let the speed of the train be $x$ km/h).

Practice Questions with Solutions

• Q: Solve the pair of linear equations by the Substitution Method: $x + y = 14$ $x - y = 4$ A: Step 1: Let the given equations be: (i) $x + y = 14$ (ii) $x - y = 4$ Step 2: From equation (ii), express $x$ in terms of $y$: $x = 4 + y$ Step 3: Substitute the value of $x$ in equation (i): $(4 + y) + y = 14$ $4 + 2y = 14$ $2y = 10$ $y = 5$ Step 4: Substitute $y = 5$ in $x = 4 + y$: $x = 4 + 5 = 9$ Final answer: The unique solution is $x = 9$ and $y = 5$.
• Q: Solve the following pair of equations using the Elimination Method: $2x + 3y = 11$ $2x - 4y = -24$ A: Step 1: Write down the equations: (i) $2x + 3y = 11$ (ii) $2x - 4y = -24$ Step 2: Since the coefficients of $x$ are already equal, subtract equation (ii) from equation (i) to eliminate $x$: $(2x + 3y) - (2x - 4y) = 11 - (-24)$ $2x + 3y - 2x + 4y = 11 + 24$ $7y = 35$ $y = 5$ Step 3: Substitute $y = 5$ in equation (i): $2x + 3(5) = 11$ $2x + 15 = 11$ $2x = 11 - 15$ $2x = -4$ $x = -2$ Final answer: The solution is $x = -2$ and $y = 5$.
• Q: For which value of $k$ will the following pair of linear equations have infinitely many solutions? $kx + 3y - (k - 3) = 0$ $12x + ky - k = 0$ 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: Write the condition for infinitely many solutions (coincident lines): $a_1/a_2 = b_1/b_2 = c_1/c_2$ So, $k/12 = 3/k = -(k - 3)/(-k)$ Step 3: Solve the first ratio equality: $k/12 = 3/k$ $k^2 = 36$ $k = \pm 6$ Step 4: Solve the second ratio equality with the third to confirm: $3/k = (k - 3)/k$ If $k = 6$, then $3/6 = (6 - 3)/6 = 3/6$, which is true. If $k = -6$, then $3/(-6) = (-6 - 3)/(-6) = -9/-6 = 3/2$, which is false. Final answer: The value of $k$ for which the system has infinitely many solutions is $k = 6$.
• Q: Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Formulate equations and find their present ages. A: Step 1: Let Nuri's present age be $x$ years and Sonu's present age be $y$ years. Step 2: Formulate the first equation (5 years ago): Nuri's age = $x - 5$ Sonu's age = $y - 5$ Equation: $x - 5 = 3(y - 5)$ $x - 5 = 3y - 15$ $x - 3y = -10$ ... (i) Step 3: Formulate the second equation (10 years later): Nuri's age = $x + 10$ Sonu's age = $y + 10$ Equation: $x + 10 = 2(y + 10)$ $x + 10 = 2y + 20$ $x - 2y = 10$ ... (ii) Step 4: Subtract equation (i) from equation (ii): $(x - 2y) - (x - 3y) = 10 - (-10)$ $y = 20$ Step 5: Substitute $y = 20$ in equation (ii): $x - 2(20) = 10$ $x - 40 = 10$ $x = 50$ Final answer: Nuri is 50 years old and Sonu is 20 years old.

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