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:

\[ x + 2y + 1 = 0 \\ , 2x – 3y – 12 = 0 \]

Solution

Step 1: Write the Equations in Standard Form

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

\[ 2x – 3y = 12 \quad \text{(2)} \]

Step 2: Compare with Standard Form

The standard form is:

\[ a_1x + b_1y = c_1 \\ , a_2x + b_2y = c_2 \]

From equations (1) and (2), we get:

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

\[ a_2 = 2,\quad b_2 = -3,\quad c_2 = 12 \]

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

\[ \frac{x}{(2 \cdot 12 – (-3)(-1))} = \frac{y}{(2 \cdot (-1) – 1 \cdot 12)} = \frac{1}{(1 \cdot (-3) – 2 \cdot 2)} \]

\[ \frac{x}{(24 – 3)} = \frac{y}{(-2 – 12)} = \frac{1}{(-3 – 4)} \]

\[ \frac{x}{21} = \frac{y}{-14} = \frac{1}{-7} \]

Step 5: Find the Values of x and y

\[ \frac{x}{21} = \frac{1}{-7} \Rightarrow x = -3 \]

\[ \frac{y}{-14} = \frac{1}{-7} \Rightarrow y = 2 \]

Conclusion

The solution of the given system of equations is:

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

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