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:

\[ \frac{x+y}{xy} = 2,\quad \frac{x-y}{xy} = 6 \]

Solution

Step 1: Simplify the Equations

Multiply both equations by \(xy\):

\[ x + y = 2xy \quad \text{(1)} \]

\[ x – y = 6xy \quad \text{(2)} \]

Step 2: Convert into Linear Equations

Divide equations (1) and (2) by \(xy\):

\[ \frac{1}{y} + \frac{1}{x} = 2 \quad \text{(3)} \]

\[ \frac{1}{y} – \frac{1}{x} = 6 \quad \text{(4)} \]

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

Then equations (3) and (4) become:

\[ u + v = 2 \quad \text{(5)} \]

\[ v – u = 6 \quad \text{(6)} \]

Step 3: Apply Cross-Multiplication Method

Comparing with standard form:

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

We get:

\[ a_1 = 1,\ b_1 = 1,\ c_1 = 2 \]

\[ a_2 = -1,\ b_2 = 1,\ c_2 = 6 \]

Using cross-multiplication:

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

\[ \frac{u}{(1\cdot6 – 1\cdot2)} = \frac{v}{((-1)\cdot2 – 1\cdot6)} = \frac{1}{(1\cdot1 – (-1)\cdot1)} \]

\[ \frac{u}{4} = \frac{v}{-8} = \frac{1}{2} \]

Step 4: Find the Values of x and y

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

\[ v = -4 \Rightarrow \frac{1}{y} = -4 \Rightarrow y = -\frac{1}{4} \]

Conclusion

The solution of the given system of equations is:

\[ x = \frac{1}{2},\quad y = -\frac{1}{4} \]

\[ \therefore \quad \text{The solution is } \left(\frac{1}{2},\; -\frac{1}{4}\right). \]