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 \(y \neq 0\)):

\[ \frac{5}{x+y} – \frac{2}{x-y} = -1, \quad \frac{15}{x+y} + \frac{7}{x-y} = 10 \]

Solution

Step 1: Substitute Variables

Let

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

Then the given equations become:

\[ 5u – 2v = -1 \quad \text{(1)} \]

\[ 15u + 7v = 10 \quad \text{(2)} \]

Step 2: Apply Cross-Multiplication Method

Comparing with standard form:

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

We get:

\[ a_1 = 5,\ b_1 = -2,\ c_1 = -1 \]

\[ a_2 = 15,\ b_2 = 7,\ c_2 = 10 \]

Using cross-multiplication:

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

\[ \frac{u}{((-2)\cdot10 – 7\cdot(-1))} = \frac{v}{(15\cdot(-1) – 5\cdot10)} = \frac{1}{(5\cdot7 – 15\cdot(-2))} \]

\[ \frac{u}{(-20 + 7)} = \frac{v}{(-15 – 50)} = \frac{1}{(35 + 30)} \]

\[ \frac{u}{-13} = \frac{v}{-65} = \frac{1}{65} \]

Step 3: Find u and v

\[ u = \frac{1}{5},\quad v = 1 \]

Step 4: Find x and y

\[ \frac{1}{x+y} = \frac{1}{5} \Rightarrow x + y = 5 \]

\[ \frac{1}{x-y} = 1 \Rightarrow x – y = 1 \]

Solving:

\[ x = 3,\quad y = 2 \]

Conclusion

The solution of the given system of equations is:

\[ x = 3,\quad y = 2 \]

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