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:
\[ ax + by = a – b \\ , bx – ay = a + b \]
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:
\[ a_1 = a,\quad b_1 = b,\quad c_1 = a – b \]
\[ a_2 = b,\quad b_2 = -a,\quad c_2 = a + b \]
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
\[ \frac{x}{\big[b(a+b) – (-a)(a-b)\big]} = \frac{y}{\big[b(a-b) – a(a+b)\big]} = \frac{1}{\big[a(-a) – b(b)\big]} \]
\[ \frac{x}{(ab + b^2 + a^2 – ab)} = \frac{y}{(ab – b^2 – a^2 – ab)} = \frac{1}{(-a^2 – b^2)} \]
\[ \frac{x}{(a^2 + b^2)} = \frac{y}{-(a^2 + b^2)} = \frac{1}{-(a^2 + b^2)} \]
Step 4: Find the Values of x and y
\[ \frac{x}{(a^2 + b^2)} = \frac{1}{-(a^2 + b^2)} \Rightarrow x = -1 \]
\[ \frac{y}{-(a^2 + b^2)} = \frac{1}{-(a^2 + b^2)} \Rightarrow y = 1 \]
Conclusion
The solution of the given system of equations is:
\[ x = -1,\quad y = 1 \]
\[ \therefore \quad \text{The solution is } ( -1,\; 1 ). \]