Substitution Method for Solving Linear Equations for CBSE Class 10

While the graphical method provides visual clarity, it suffers from practical limitations: reading fractional coordinates (such as $x = \frac{7}{13}$ or $y = \sqrt{2}$) from graph paper is imprecise and prone to reading errors. To obtain exact, error-free solutions, algebraic methods are required.

The Substitution Method is the first and most intuitive algebraic technique taught in CBSE Class 10 Mathematics. It works by using one equation to express a variable in terms of the other, effectively reducing a two-variable system into a simple single-variable linear equation.

What You Will Learn

• Underlying philosophy of algebraic elimination through substitution
• Step-by-step systematic algorithm for the Substitution Method
• Strategic selection of variables to avoid cumbersome fractions
• Solving systems with fractional and radical (square root) coefficients
• Detecting systems with infinitely many solutions and no solution algebraically
• Solved CBSE board examination questions and error-prevention tips

1. The Core Idea of Substitution

Consider a system of two linear equations in variables $x$ and $y$: $a_1 x + b_1 y + c_1 = 0 \quad \text{--- (1)}$ $a_2 x + b_2 y + c_2 = 0 \quad \text{--- (2)}$

The central obstacle in solving a system with two variables is the simultaneous presence of $x$ and $y$. The Substitution Method eliminates this obstacle by:

1. Re-arranging Equation (1) to write $x$ solely in terms of $y$: $x = f(y)$.
2. Substituting this expression $f(y)$ into Equation (2) in place of $x$.
3. The resulting equation contains only variable $y$, which can be solved easily using basic algebra.

Important: <u>Always substitute the expression into the OTHER equation, not the one from which it was derived. Substituting back into the same equation simply yields a trivial identity like $0 = 0$!</u>

2. Systematic Step-by-Step Algorithm

1. Step 1 (Select Variable): Look at both equations and select the variable with coefficient $1$ or $-1$ if available. This avoids introducing fractions.
2. Step 2 (Express Variable): Express that variable in terms of the other variable from Equation (1).
3. Step 3 (Substitute): Substitute this expression into Equation (2).
4. Step 4 (Solve for First Variable): Solve the resulting single-variable linear equation to find the numeric value of the first variable.
5. Step 5 (Back-Substitute): Substitute the value found in Step 4 back into the expression from Step 2 to obtain the value of the second variable.
6. Step 6 (Verification): Substitute both $(x, y)$ values into both original equations to confirm they satisfy both statements.

3. Solved Step-by-Step Examples

Solved Example 1: Standard Integer System

Problem: Solve the following pair of linear equations by the substitution method: $7x - 15y = 2 \quad \text{--- (1)}$ $x + 2y = 3 \quad \text{--- (2)}$

Solution:

• Step 1: In Equation (2), the coefficient of $x$ is $1$. This is the easiest term to isolate. $x = 3 - 2y \quad \text{--- (3)}$
• Step 2: Substitute $x = 3 - 2y$ into Equation (1): $7(3 - 2y) - 15y = 2$
• Step 3: Expand and solve for $y$: $21 - 14y - 15y = 2$ $21 - 29y = 2$ $-29y = 2 - 21 = -19$ $y = \frac{19}{29}$
• Step 4: Substitute $y = \frac{19}{29}$ into Equation (3): $x = 3 - 2\left(\frac{19}{29}\right) = 3 - \frac{38}{29} = \frac{87 - 38}{29} = \frac{49}{29}$
• Conclusion: <u>$x = \frac{49}{29}, \quad y = \frac{19}{29}$</u>.

Solved Example 2: Equations with Square Roots (CBSE Classic PYQ)

Problem: Solve the following system by substitution: $\sqrt{2}x + \sqrt{3}y = 0 \quad \text{--- (1)}$ $\sqrt{3}x - \sqrt{8}y = 0 \quad \text{--- (2)}$

Solution:

1. From Equation (1), express $x$ in terms of $y$: $\sqrt{2}x = -\sqrt{3}y \implies x = -\frac{\sqrt{3}}{\sqrt{2}}y \quad \text{--- (3)}$
2. Substitute $x$ into Equation (2): $\sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - \sqrt{8}y = 0$
3. Simplify the terms: $-\frac{3}{\sqrt{2}}y - \sqrt{8}y = 0$ $-\left(\frac{3}{\sqrt{2}} + \sqrt{8}\right)y = 0$
4. Since the coefficient $-\left(\frac{3}{\sqrt{2}} + \sqrt{8}\right) \ne 0$, we have: $y = 0$
5. Substitute $y = 0$ into Equation (3): $x = -\frac{\sqrt{3}}{\sqrt{2}}(0) = 0$
6. Therefore, the unique solution is $x = 0, y = 0$.

4. Special Cases: Infinitely Many Solutions and No Solution

When applying the substitution method, the variable terms may completely cancel out. Pay close attention to the resulting numerical statement:

Case A: Infinitely Many Solutions (True Statement)

Consider $2x + 3y = 9$ and $4x + 6y = 18$.

• From Equation (1): $x = \frac{9 - 3y}{2}$.
• Substitute into Equation (2): $4\left(\frac{9 - 3y}{2}\right) + 6y = 18 \implies 2(9 - 3y) + 6y = 18 \implies 18 - 6y + 6y = 18 \implies 18 = 18$
• The statement $18 = 18$ is a true statement independent of $y$.
• This indicates that both equations represent the same line. The system has infinitely many solutions.

Case B: No Solution (False Statement)

Consider $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$.

• From Equation (1): $x = 4 - 2y$.
• Substitute into Equation (2): $2(4 - 2y) + 4y - 12 = 0 \implies 8 - 4y + 4y - 12 = 0 \implies -4 = 0$
• The statement $-4 = 0$ is a false statement.
• This indicates that the lines are parallel. The system has no solution.

5. Summary and Examination Tips

Remember: Always look for a variable with coefficient $1$ or $-1$ before starting substitution. If no such coefficient exists, choose the variable with the smallest coefficient to minimize fraction size.

Common Mistake: Distributing negative signs incorrectly during substitution. In $7(3 - 2y)$, remember that $-7 \times (-2y) = +14y$, not $-14y$!