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 (where \(x \neq 0\) and \(y \neq 0\)):

\[ \frac{a^2}{x} – \frac{b^2}{y} = 0, \\ \frac{a^2 b}{x} + \frac{b^2 a}{y} = a + b \]

Solution

Step 1: Convert into Linear Equations

Let

\[ \frac{1}{x} = u,\quad \frac{1}{y} = v \]

Then the given equations become:

\[ a^2u – b^2v = 0 \quad \text{(1)} \]

\[ a^2b\,u + b^2a\,v = a + b \quad \text{(2)} \]

Step 2: Compare with Standard Form

\[ a_1u + b_1v = c_1,\quad a_2u + b_2v = c_2 \]

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

\[ a_1 = a^2,\quad b_1 = -b^2,\quad c_1 = 0 \]

\[ a_2 = a^2b,\quad b_2 = ab^2,\quad c_2 = a + b \]

Step 3: Apply Cross-Multiplication Formula

\[ \frac{u}{(b_1c_2 – b_2c_1)} = \frac{v}{(a_2c_1 – a_1c_2)} = \frac{1}{(a_1b_2 – a_2b_1)} \]

Step 4: Substitute the Values

\[ \frac{u}{\big[(-b^2)(a+b) – (ab^2)(0)\big]} = \frac{v}{\big[(a^2b)(0) – (a^2)(a+b)\big]} = \frac{1}{\big[a^2(ab^2) – (a^2b)(-b^2)\big]} \]

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

Step 5: Find u and v

\[ u = \frac{1}{a^2},\quad v = \frac{1}{b^2} \]

Step 6: Find x and y

\[ \frac{1}{x} = \frac{1}{a^2} \Rightarrow x = a^2 \]

\[ \frac{1}{y} = \frac{1}{b^2} \Rightarrow y = b^2 \]

Conclusion

The solution of the given system of equations is:

\[ x = a^2,\quad y = b^2 \]

\[ \therefore \quad \text{The solution is } (a^2,\; b^2). \]