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:

\[ 6(ax + by) = 3a + 2b \\ , 6(bx – ay) = 3b – 2a \]

Solution

Step 1: Simplify the Given Equations

\[ ax + by = \frac{3a + 2b}{6} \quad \text{(1)} \]

\[ bx – ay = \frac{3b – 2a}{6} \quad \text{(2)} \]

Step 2: Compare with the Standard Form

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

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

\[ a_1 = a,\quad b_1 = b,\quad c_1 = \frac{3a+2b}{6} \]

\[ a_2 = b,\quad b_2 = -a,\quad c_2 = \frac{3b-2a}{6} \]

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

\[ \frac{x}{\left[b\left(\frac{3b-2a}{6}\right) – (-a)\left(\frac{3a+2b}{6}\right)\right]} = \frac{y}{\left[b\left(\frac{3a+2b}{6}\right) – a\left(\frac{3b-2a}{6}\right)\right]} = \frac{1}{(a(-a) – b^2)} \]

\[ \frac{x}{\frac{3(a^2+b^2)}{6}} = \frac{y}{\frac{2(a^2+b^2)}{6}} = \frac{1}{-(a^2+b^2)} \]

Step 5: Find the Values of x and y

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

\[ y = \frac{1}{3} \]

Conclusion

The solution of the given system of equations is:

\[ x = \frac{1}{2},\quad y = \frac{1}{3} \]

\[ \therefore \quad \text{The solution is } \left(\frac{1}{2},\; \frac{1}{3}\right). \]