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:

\[ 3x + 2y + 25 = 0 \\ , 2x + y + 10 = 0 \]

Solution

Step 1: Write the Equations in Standard Form

\[ 3x + 2y = -25 \quad \text{(1)} \]

\[ 2x + y = -10 \quad \text{(2)} \]

Step 2: Compare with the Standard Form

The standard form is:

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

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

\[ a_1 = 3,\quad b_1 = 2,\quad c_1 = -25 \]

\[ a_2 = 2,\quad b_2 = 1,\quad c_2 = -10 \]

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

\[ \frac{x}{(2 \cdot (-10) – 1 \cdot (-25))} = \frac{y}{(2 \cdot (-25) – 3 \cdot (-10))} = \frac{1}{(3 \cdot 1 – 2 \cdot 2)} \]

\[ \frac{x}{(-20 + 25)} = \frac{y}{(-50 + 30)} = \frac{1}{(3 – 4)} \]

\[ \frac{x}{5} = \frac{y}{-20} = \frac{1}{-1} \]

Step 5: Find the Values of x and y

\[ \frac{x}{5} = \frac{1}{-1} \Rightarrow x = -5 \]

\[ \frac{y}{-20} = \frac{1}{-1} \Rightarrow y = 20 \]

Conclusion

The solution of the given system of equations is:

\[ x = -5,\quad y = 20 \]

\[ \therefore \quad \text{The solution is } ( -5,\; 20 ). \]