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:

\[ 5ax + 6by = 28 \\ , 3ax + 4by = 18 \]

Solution

Step 1: Reduce the Equations

Let \[ X = ax,\quad Y = by \]

Then the given equations become:

\[ 5X + 6Y = 28 \quad \text{(1)} \]

\[ 3X + 4Y = 18 \quad \text{(2)} \]

Step 2: Compare with Standard Form

\[ a_1X + b_1Y = c_1 \\ , a_2X + b_2Y = c_2 \]

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

\[ a_1 = 5,\quad b_1 = 6,\quad c_1 = 28 \]

\[ a_2 = 3,\quad b_2 = 4,\quad c_2 = 18 \]

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

\[ \frac{X}{(6\cdot18 – 4\cdot28)} = \frac{Y}{(3\cdot28 – 5\cdot18)} = \frac{1}{(5\cdot4 – 3\cdot6)} \]

\[ \frac{X}{(108 – 112)} = \frac{Y}{(84 – 90)} = \frac{1}{(20 – 18)} \]

\[ \frac{X}{-4} = \frac{Y}{-6} = \frac{1}{2} \]

Step 5: Find the Values of X and Y

\[ X = 2,\quad Y = 3 \]

Step 6: Find the Values of x and y

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

\[ by = 3 \Rightarrow y = \frac{3}{b} \]

Conclusion

The solution of the given system of equations is:

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

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