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:

\[ ax + by = a^2 \\ , bx + ay = b^2 \]

Solution

Step 1: Compare with the Standard Form

The standard form is:

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

From the given equations, we have:

\[ a_1 = a,\quad b_1 = b,\quad c_1 = a^2 \]

\[ a_2 = b,\quad b_2 = a,\quad c_2 = b^2 \]

Step 2: 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 3: Substitute the Values

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

\[ \frac{x}{(b^3 – a^3)} = \frac{y}{ab(a-b)} = \frac{1}{(a^2 – b^2)} \]

Step 4: Find the Values of x and y

\[ x = \frac{b^3 – a^3}{a^2 – b^2} = -\frac{a^2 + ab + b^2}{a + b} \]

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

Conclusion

The solution of the given system of equations is:

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

(Provided \(a \neq \pm b\))

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