Elimination Method for Solving Linear Equations for CBSE Class 10

Among the algebraic techniques available for solving a system of two linear equations, the Elimination Method is universally considered the fastest, most direct, and least error-prone. While substitution often introduces awkward fractions early in the solution process, the elimination method works directly with whole numbers by equalizing coefficients.

In CBSE Class 10 Mathematics, mastering the elimination method is crucial for solving both direct algebraic systems and complex real-world word problems efficiently under exam time constraints.

What You Will Learn

• Conceptual principle of eliminating a variable by addition or subtraction
• Step-by-step algorithm for equalizing coefficients using the LCM
• When to add equations vs. when to subtract equations
• The famous interchanged coefficient shortcut for large numbers (e.g., $ax + by = c$ and $bx + ay = d$)
• Recognizing inconsistent systems (no solution) and dependent systems (infinitely many solutions)
• Solved CBSE board examination problems and step-by-step solutions

1. The Principle of the Elimination Method

The Elimination Method eliminates one of the two variables by creating identical (or opposite) coefficients for that variable in both equations.

$\text{If } a_1 x + b_1 y = c_1 \quad \text{and} \quad a_2 x + b_2 y = c_2$

By multiplying Equation (1) by a constant $k_1$ and Equation (2) by a constant $k_2$, we ensure that: $\text{Coefficient of one variable in (1)} = \pm \text{ Coefficient of that variable in (2)}$

Once equalized:

• Add the equations if the coefficients have opposite signs ($+k$ and $-k$).
• Subtract the equations if the coefficients have the same sign ($+k$ and $+k$, or $-k$ and $-k$).

Important: <u>When multiplying an equation by a constant, you must multiply EVERY term on BOTH the left-hand side and the right-hand side. Forgetting to multiply the constant term on the RHS is the most frequent student mistake!</u>

2. Systematic Step-by-Step Procedure

1. Step 1 (Arrange in Standard Form): Write both equations in the form $ax + by = c$.
2. Step 2 (Choose Variable to Eliminate): Inspect coefficients and pick the variable whose coefficients are easier to equalize (find the LCM of the coefficients).
3. Step 3 (Multiply): Multiply each equation by an appropriate non-zero factor so that the chosen variable has equal absolute coefficients in both equations.
4. Step 4 (Add or Subtract):
   • Opposite signs $\implies$ Add the two equations.
   • Same signs $\implies$ Subtract one equation from the other.
5. Step 5 (Solve Single-Variable Equation): Solve the resulting equation to obtain the value of the remaining variable.
6. Step 6 (Substitute to Find Second Variable): Substitute this value into either of the original equations to solve for the other variable.

3. Solved Step-by-Step Examples

Solved Example 1: Standard System

Problem: Solve the following system using the elimination method: $2x + 3y = 8 \quad \text{--- (1)}$ $4x + 5y = 14 \quad \text{--- (2)}$

Solution:

• Step 1: Choose to eliminate $x$. The coefficients of $x$ are $2$ and $4$. The LCM of $2$ and $4$ is $4$.
• Step 2: Multiply Equation (1) by $2$: $2 \times (2x + 3y = 8) \implies 4x + 6y = 16 \quad \text{--- (3)}$
• Step 3: Subtract Equation (2) from Equation (3): $(4x + 6y) - (4x + 5y) = 16 - 14$ $(4x - 4x) + (6y - 5y) = 2$ $y = 2$
• Step 4: Substitute $y = 2$ into Equation (1): $2x + 3(2) = 8$ $2x + 6 = 8 \implies 2x = 2 \implies x = 1$
• Final Solution: <u>$x = 1, \quad y = 2$</u>.

Solved Example 2: The Interchanged Coefficients Shortcut (CBSE Board Classic)

Problem: Solve the following pair of equations: $152x - 378y = -74 \quad \text{--- (1)}$ $-378x + 152y = -604 \quad \text{--- (2)}$

Notice: The coefficients are large numbers, but the coefficient of $x$ in (1) equals the coefficient of $y$ in (2), and vice versa. Using standard cross-multiplication or elimination would require huge products ($152 \times 378 = 57456$). Instead, use the Add-Subtract Technique:

Step 1: Add Equation (1) and Equation (2)

$(152 - 378)x + (-378 + 152)y = -74 + (-604)$ $-226x - 226y = -678$ Divide the entire equation by $-226$: $x + y = 3 \quad \text{--- (3)}$

Step 2: Subtract Equation (2) from Equation (1)

$[152 - (-378)]x + [-378 - 152]y = -74 - (-604)$ $(152 + 378)x - (378 + 152)y = -74 + 604$ $530x - 530y = 530$ Divide the entire equation by $530$: $x - y = 1 \quad \text{--- (4)}$

Step 3: Solve the Simplified Equations (3) and (4)

Adding (3) and (4): $(x + y) + (x - y) = 3 + 1 \implies 2x = 4 \implies x = 2$ Substitute $x = 2$ into (3): $2 + y = 3 \implies y = 1$

Therefore, the solution is $x = 2, y = 1$.

Exam Tip: Whenever you spot equations of the form $ax + by = c$ and $bx + ay = d$ with large coefficients, NEVER multiply by the large numbers directly! Always add the two equations to get $x + y = p$, and subtract them to get $x - y = q$.

4. Elimination with Fractional Coefficients

Problem: Solve $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$.

Solution:

1. Clear fractions first by multiplying by the LCM of denominators:
   • For Equation (1), multiply by $6$: $6\left(\frac{x}{2}\right) + 6\left(\frac{2y}{3}\right) = 6(-1) \implies 3x + 4y = -6 \quad \text{--- (A)}$
   • For Equation (2), multiply by $3$: $3(x) - 3\left(\frac{y}{3}\right) = 3(3) \implies 3x - y = 9 \quad \text{--- (B)}$
2. Notice that the coefficient of $x$ is already $3$ in both equations!
3. Subtract Equation (B) from Equation (A): $(3x + 4y) - (3x - y) = -6 - 9$ $5y = -15 \implies y = -3$
4. Substitute $y = -3$ into Equation (B): $3x - (-3) = 9 \implies 3x + 3 = 9 \implies 3x = 6 \implies x = 2$
5. Thus, $x = 2, y = -3$.

5. Summary and Comparison Guide

Remember: Check your final answer by plugging $x$ and $y$ back into BOTH original equations. If one equation balances but the other does not, re-check your subtraction step!

Common Mistake: Forgetting to distribute the negative sign to ALL terms when subtracting equations. In $(3x + 4y) - (3x - y)$, the term becomes $+4y - (-y) = +5y$, not $+3y$!