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:

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

Solution

Step 1: Substitute Variables

Let

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

Then the given equations become:

\[ u + v = a + b \quad \text{(1)} \]

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

Step 2: Write in Standard Linear Form

\[ 1\cdot u + 1\cdot v = a + b \]

\[ \frac{1}{a}u + \frac{1}{b}v = 2 \]

Step 3: Apply Cross-Multiplication Method

Comparing with:

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

We get:

\[ a_1 = 1,\ b_1 = 1,\ c_1 = a + b \]

\[ a_2 = \frac{1}{a},\ b_2 = \frac{1}{b},\ c_2 = 2 \]

Using cross-multiplication:

\[ \frac{u}{(c_1b_2 – c_2b_1)} = \frac{v}{(a_1c_2 – a_2c_1)} = \frac{1}{(a_1b_2 – a_2b_1)} \]

\[ \frac{u}{\left(\frac{a+b}{b} – 2\right)} = \frac{v}{\left(2 – \frac{a+b}{a}\right)} = \frac{1}{\left(\frac{1}{b} – \frac{1}{a}\right)} \]

\[ \frac{u}{\frac{a-b}{b}} = \frac{v}{\frac{a-b}{a}} = \frac{1}{\frac{a-b}{ab}} \]

Step 4: Find u and v

\[ u = a,\quad v = b \]

Step 5: Find x and y

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

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

Conclusion

The solution of the given system of equations is:

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

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