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:

\[ 2(ax – by) + a + 4b = 0 \\ , 2(bx + ay) + b – 4a = 0 \]

Solution

Step 1: Simplify the Given Equations

First equation:

\[ 2ax – 2by = -a – 4b \quad \text{(1)} \]

Second equation:

\[ 2bx + 2ay = 4a – b \quad \text{(2)} \]

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:

\[ a_1 = 2a,\quad b_1 = -2b,\quad c_1 = -a – 4b \]

\[ a_2 = 2b,\quad b_2 = 2a,\quad c_2 = 4a – b \]

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

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

\[ \frac{x}{2(a^2+b^2)} = \frac{y}{-8(a^2+b^2)} = \frac{1}{4(a^2+b^2)} \]

Step 5: Find the Values of x and y

\[ x = \frac{1}{2} \]

\[ y = -2 \]

Conclusion

The solution of the given system of equations is:

\[ x = \frac{1}{2},\quad y = -2 \]

\[ \therefore \quad \text{The solution is } \left(\frac{1}{2},\; -2\right). \]