Cross-Multiplication Method and Reducible Equations for CBSE Class 10

While substitution and elimination are based on iterative algebraic manipulations, the Cross-Multiplication Method provides a direct determinant-based formula to compute solutions instantly from the coefficients. In addition, many advanced problems in CBSE Class 10 Mathematics involve systems that are not initially linear (such as variables in the denominator or reciprocal terms), but can be transformed into linear pairs using auxiliary substitutions.

Mastering both the cross-multiplication formula and the technique of reducing non-linear equations to linear pairs is essential for tackling the highest-weightage 4-mark and 5-mark questions in board examinations.

What You Will Learn

• Algebraic derivation and statement of the Cross-Multiplication Method
• The famous 2312 mnemonic diagram for error-free formula recall
• Conditions for unique, infinitely many, and no solutions using cross-multiplication
• Transforming reciprocal equations ($\frac{1}{x} = u, \frac{1}{y} = v$) into standard linear form
• Solving advanced reducible systems involving binomial denominators ($\frac{1}{x+y}, \frac{1}{x-y}$)
• Step-by-step solved board exam problems and common pitfalls

1. The Cross-Multiplication Method

Consider the general pair of linear equations in standard form: $a_1 x + b_1 y + c_1 = 0 \quad \text{--- (1)}$ $a_2 x + b_2 y + c_2 = 0 \quad \text{--- (2)}$

Important: <u>To apply cross-multiplication correctly, both equations MUST have all terms on the left-hand side so that the right-hand side is strictly zero ($= 0$).</u>

The 2312 Mnemonic Diagram

Write the coefficients in columns following the index order 2 - 3 - 1 - 2 (representing coefficients of $y$, constant terms, coefficients of $x$, and coefficients of $y$ again):

       x                   y                   1
   b1     c1           c1     a1           a1     b1
      \ /                 \ /                 \ /
      / \                 / \                 /    b2     c2           c2     a2           a2     b2

Cross-multiplying downwards (with positive sign) and upwards (with negative sign) gives the relationship:

$\frac{x}{b_1 c_2 - b_2 c_1} = \frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1}$

Explicit Formulas for $x$ and $y$

Provided $a_1 b_2 - a_2 b_1 \ne 0$ (which is equivalent to $\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$):

$x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$

2. Solved Example: Direct Cross-Multiplication

Problem: Solve the following pair of equations using the cross-multiplication method: $2x + 3y = 46$ $3x + 5y = 74$

Solution:

1. Write in Standard Form ($= 0$): $2x + 3y - 46 = 0 \implies a_1 = 2, b_1 = 3, c_1 = -46$ $3x + 5y - 74 = 0 \implies a_2 = 3, b_2 = 5, c_2 = -74$
2. Apply the 2312 Arrangement: $\frac{x}{(3)(-74) - (5)(-46)} = \frac{y}{(-46)(3) - (-74)(2)} = \frac{1}{(2)(5) - (3)(3)}$
3. Compute Denominators:
   • Under $x$: $(3)(-74) - (5)(-46) = -222 - (-230) = -222 + 230 = 8$
   • Under $y$: $(-46)(3) - (-74)(2) = -138 - (-148) = -138 + 148 = 10$
   • Under $1$: $(2)(5) - (3)(3) = 10 - 9 = 1$
4. Equate Ratios: $\frac{x}{8} = \frac{y}{10} = \frac{1}{1}$ $x = \frac{8}{1} = 8, \quad y = \frac{10}{1} = 10$
5. Therefore, <u>$x = 8, y = 10$</u>.

3. Equations Reducible to a Pair of Linear Equations

Certain systems of equations are not linear because the variables appear in the denominator, violating the condition that variable exponents must be $1$. However, these equations can be converted into standard linear equations by introducing auxiliary variables.

Type 1: Single Variable in Denominators

Consider: $\frac{2}{x} + \frac{3}{y} = 13 \quad \text{and} \quad \frac{5}{x} - \frac{4}{y} = -2$

Method of Solution:

1. Substitute $\frac{1}{x} = u$ and $\frac{1}{y} = v$ (where $x, y \ne 0$).
2. The equations transform into a standard linear system: $2u + 3v = 13 \quad \text{--- (1)}$ $5u - 4v = -2 \quad \text{--- (2)}$
3. Solve by elimination:
   • Multiply (1) by $4$: $8u + 12v = 52$
   • Multiply (2) by $3$: $15u - 12v = -6$
   • Adding both: $23u = 46 \implies u = 2$
4. Substitute $u = 2$ into (1): $2(2) + 3v = 13 \implies 4 + 3v = 13 \implies 3v = 9 \implies v = 3$
5. Convert back to $x$ and $y$: $u = \frac{1}{x} = 2 \implies x = \frac{1}{2}$ $v = \frac{1}{y} = 3 \implies y = \frac{1}{3}$
6. Final answer: $x = \frac{1}{2}, y = \frac{1}{3}$.

Type 2: Compound Denominators (CBSE Board 4-Mark Classic)

Problem: Solve for $x$ and $y$: $\frac{10}{x+y} + \frac{2}{x-y} = 4 \quad \text{--- (1)}$ $\frac{15}{x+y} - \frac{5}{x-y} = -2 \quad \text{--- (2)}$

Solution:

1. Let $\frac{1}{x+y} = u$ and $\frac{1}{x-y} = v$: $10u + 2v = 4 \implies 5u + v = 2 \quad \text{--- (3)}$ $15u - 5v = -2 \quad \text{--- (4)}$
2. From (3), $v = 2 - 5u$. Substitute into (4): $15u - 5(2 - 5u) = -2$ $15u - 10 + 25u = -2$ $40u = 8 \implies u = \frac{8}{40} = \frac{1}{5}$
3. Find $v$: $v = 2 - 5\left(\frac{1}{5}\right) = 2 - 1 = 1$
4. Reconstruct Equations in $x$ and $y$:
   • $\frac{1}{x+y} = u = \frac{1}{5} \implies x + y = 5 \quad \text{--- (5)}$
   • $\frac{1}{x-y} = v = 1 \implies x - y = 1 \quad \text{--- (6)}$
5. Solve the Resulting Linear Pair:
   • Adding (5) and (6): $2x = 6 \implies x = 3$
   • Subtracting (6) from (5): $2y = 4 \implies y = 2$
6. Therefore, <u>$x = 3$ and $y = 2$</u>.

4. Summary and Key Takeaways

Remember: In cross-multiplication, the denominator under $1$ is $a_1 b_2 - a_2 b_1$. If this equals $0$, you cannot divide by zero—this reflects either parallel lines (no solution) or coincident lines (infinite solutions)!

Common Mistake: Forgetting to invert $u$ and $v$ at the end of reducible equation problems. Students often stop after finding $u = 2$ and $v = 3$, forgetting that the original question asked for $x$ and $y$!