The system of equations $4x + 14y = 18$ and $6x + 21y = 33$ has

no solution

The correct answer is Option (4) → no solution

Step-by-Step Analysis:

To determine the number of solutions for a system of linear equations in the form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$, we compare the ratios of their coefficients:

1. Identify the coefficients:

• Equation 1: $4x + 14y = 18 ⇒a_1 = 4, b_1 = 14, c_1 = 18$
• Equation 2: $6x + 21y = 33 ⇒a_2 = 6, b_2 = 21, c_2 = 33$

2. Calculate the ratios:

• Ratio of x-coefficients: $\frac{a_1}{a_2} = \frac{4}{6} = \frac{\mathbf{2}}{\mathbf{3}}$
• Ratio of y-coefficients: $\frac{b_1}{b_2} = \frac{14}{21} = \frac{\mathbf{2}}{\mathbf{3}}$
• Ratio of constants: $\frac{c_1}{c_2} = \frac{18}{33} = \frac{\mathbf{6}}{\mathbf{11}}$

3. Apply the consistency rules:

• If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, there is a unique solution (lines intersect).
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, there are infinite solutions (lines are coincident).
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, there is no solution (lines are parallel).

In this case:

$\frac{2}{3} = \frac{2}{3} \neq \frac{6}{11}$

Since the ratios of the coefficients are equal but not equal to the ratio of the constants, the two lines are parallel and will never intersect. Therefore, the system has no solution.