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:
\[ (a+2b)x + (2a-b)y = 2 \\ , (a-2b)x + (2a+b)y = 3 \]
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:
\[ a_1 = a+2b,\quad b_1 = 2a-b,\quad c_1 = 2 \]
\[ a_2 = a-2b,\quad b_2 = 2a+b,\quad c_2 = 3 \]
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
\[ \frac{x}{\big[(2a-b)3 – (2a+b)2\big]} = \frac{y}{\big[(a-2b)2 – (a+2b)3\big]} = \frac{1}{\big[(a+2b)(2a+b) – (a-2b)(2a-b)\big]} \]
\[ \frac{x}{(6a-3b-4a-2b)} = \frac{y}{(2a-4b-3a-6b)} = \frac{1}{(4ab+4ab)} \]
\[ \frac{x}{(2a-5b)} = \frac{y}{(-a-10b)} = \frac{1}{8ab} \]
Step 4: Find the Values of x and y
\[ x = \frac{2a-5b}{8ab} \]
\[ y = -\frac{a+10b}{8ab} \]
Conclusion
The solution of the given system of equations is:
\[ x = \frac{2a-5b}{8ab},\quad y = -\frac{a+10b}{8ab} \]
\[ \therefore \quad \text{The solution is } \left( \frac{2a-5b}{8ab}, \; -\frac{a+10b}{8ab} \right). \]