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:

\[ bx + cy = a + b, \]

\[ ax\left(\frac{1}{a-b}-\frac{1}{a+b}\right) + cy\left(\frac{1}{b-a}-\frac{1}{b+a}\right) = \frac{2a}{a+b} \]

Solution

Step 1: Simplify the Second Equation

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

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

Substituting:

\[ \frac{2ab}{a^2-b^2}x – \frac{2ac}{a^2-b^2}y = \frac{2a}{a+b} \]

Dividing by \(2a\) and multiplying by \((a^2-b^2)\):

\[ bx – cy = a – b \quad \text{(2)} \]

The first equation is:

\[ bx + cy = a + b \quad \text{(1)} \]

Step 2: Compare with Standard Form

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

From (1) and (2):

\[ a_1 = b,\ b_1 = c,\ c_1 = a + b \]

\[ a_2 = b,\ b_2 = -c,\ c_2 = a – b \]

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

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

\[ \frac{x}{2ac} = \frac{y}{2b^2} = \frac{1}{-2bc} \]

Step 5: Find the Values of x and y

\[ x = -\frac{a}{b} \]

\[ y = -\frac{b}{c} \]

Conclusion

The solution of the given system of equations is:

\[ x = -\frac{a}{b},\quad y = -\frac{b}{c} \]

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