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 \(y \neq 0\)):

\[ \frac{2}{x} + \frac{3}{y} = 13,\quad \frac{5}{x} – \frac{4}{y} = -2 \]

Solution

Step 1: Substitute Variables

Let

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

Then the given equations become:

\[ 2u + 3v = 13 \quad \text{(1)} \]

\[ 5u – 4v = -2 \quad \text{(2)} \]

Step 2: Apply Cross-Multiplication Method

Comparing with the standard form:

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

We get:

\[ a_1 = 2,\ b_1 = 3,\ c_1 = 13 \]

\[ a_2 = 5,\ b_2 = -4,\ c_2 = -2 \]

Using cross-multiplication:

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

\[ \frac{u}{(3\cdot(-2) – (-4)\cdot13)} = \frac{v}{(5\cdot13 – 2\cdot(-2))} = \frac{1}{(5\cdot3 – 2\cdot(-4))} \]

\[ \frac{u}{46} = \frac{v}{69} = \frac{1}{23} \]

Step 3: Find u and v

\[ u = 2,\quad v = 3 \]

Step 4: Find x and y

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

\[ \frac{1}{y} = 3 \Rightarrow y = \frac{1}{3} \]

Conclusion

The solution of the given system of equations is:

\[ x = \frac{1}{2},\quad y = \frac{1}{3} \]

\[ \therefore \quad \text{The solution is } \left(\frac{1}{2},\; \frac{1}{3}\right). \]