Solve the System of Equations by the Method of Cross-Multiplication

Video Explanation

Question

Solve the following system of equations by the method of cross-multiplication:

\[ (a-b)x + (a+b)y = 2a^2 – 2b^2, \]

\[ (a+b)(x+y) = 4ab \]

Solution

Step 1: Write the Equations in Standard Form

First equation:

\[ (a-b)x + (a+b)y = 2(a^2 – b^2) \quad \text{(1)} \]

Second equation:

\[ (a+b)x + (a+b)y = 4ab \quad \text{(2)} \]

Step 2: Compare with Standard Form

\[ a_1x + b_1y = c_1,\quad a_2x + b_2y = c_2 \]

From (1) and (2), we get:

\[ a_1 = a-b,\quad b_1 = a+b,\quad c_1 = 2(a^2-b^2) \]

\[ a_2 = a+b,\quad b_2 = a+b,\quad c_2 = 4ab \]

Step 3: Apply Cross-Multiplication Formula

\[ \frac{x}{(b_1c_2 – b_2c_1)} = \frac{y}{(a_2c_1 – a_1c_2)} = \frac{1}{(a_1b_2 – a_2b_1)} \]

Step 4: Substitute the Values

\[ \frac{x}{(a+b)4ab – (a+b)2(a^2-b^2)} = \frac{y}{(a+b)2(a^2-b^2) – (a-b)4ab} = \frac{1}{(a-b)(a+b) – (a+b)(a+b)} \]

\[ \frac{x}{-2(a+b)(a-b)^2} = \frac{y}{2(a-b)(a^2+b^2)} = \frac{1}{-2b(a+b)} \]

Step 5: Find the Values of x and y

\[ x = \frac{(a-b)^2}{b} \]

\[ y = -\frac{(a-b)(a^2+b^2)}{b(a+b)} \]

Conclusion

The solution of the given system of equations is:

\[ x = \frac{(a-b)^2}{b},\quad y = -\frac{(a-b)(a^2+b^2)}{b(a+b)} \]

\[ \therefore \quad \text{The solution is } \left( \frac{(a-b)^2}{b}, \; -\frac{(a-b)(a^2+b^2)}{b(a+b)} \right). \]