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:

\[ x\left(a-b+\frac{ab}{a-b}\right) = y\left(a+b-\frac{ab}{a-b}\right), \\ x + y = 2a^2 \]

Solution

Step 1: Simplify the Coefficients

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

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

So the first equation becomes:

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

Multiplying both sides by \((a-b)\), we get:

\[ (a^2-ab+b^2)x – (a^2-ab-b^2)y = 0 \quad \text{(1)} \]

The second equation is:

\[ x + y = 2a^2 \quad \text{(2)} \]

Step 2: Compare with Standard Form

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

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

\[ a_1 = a^2-ab+b^2,\quad b_1 = -(a^2-ab-b^2),\quad c_1 = 0 \]

\[ a_2 = 1,\quad b_2 = 1,\quad c_2 = 2a^2 \]

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

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

Step 5: Find the Values of x and y

\[ x = a^2 + b^2 \]

\[ y = a^2 – b^2 \]

Conclusion

The solution of the given system of equations is:

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

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