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:

\[ \frac{x}{a} = \frac{y}{b}, \\ ax + by = a^2 + b^2 \]

Solution

Step 1: Convert the First Equation into Linear Form

From \[ \frac{x}{a} = \frac{y}{b} \] cross-multiplying, we get:

\[ bx = ay \quad \Rightarrow \quad bx – ay = 0 \quad \text{(1)} \]

The second equation is:

\[ ax + by = a^2 + b^2 \quad \text{(2)} \]

Step 2: Compare with the Standard Form

The standard form is:

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

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

\[ a_1 = b,\quad b_1 = -a,\quad c_1 = 0 \]

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

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

\[ \frac{x}{\big[(-a)(a^2 + b^2) – b\cdot 0\big]} = \frac{y}{\big[a\cdot 0 – b(a^2 + b^2)\big]} = \frac{1}{\big[b\cdot b – a(-a)\big]} \]

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

Step 5: Find the Values of x and y

\[ x = a \]

\[ y = b \]

Conclusion

The solution of the given system of equations is:

\[ x = a,\quad y = b \]

\[ \therefore \quad \text{The solution is } (a,\; b). \]