Linear Equations in Two Variables - Classify system consistency: unique, no solution, or infinitely many solutions

Classify the following system of equations: $2x + 3y - 9 = 0$ and $4x + 6y - 18 = 0$.

Given: $a_1 = 2, b_1 = 3, c_1 = -9$ and $a_2 = 4, b_2 = 6, c_2 = -18$. Comparing the ratios: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$ $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$ $\frac{c_1}{c_2} = \frac{-9}{-18} = \frac{1}{2}$ Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the system has infinitely many solutions.

When all three ratios are equal, the two equations represent the same line (coincident lines), meaning every point on the line is a solution.

Determine if the system $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$ is consistent or inconsistent.

Given: $a_1 = 1, b_1 = 2, c_1 = -4$ and $a_2 = 2, b_2 = 4, c_2 = -12$. Calculating ratios: $\frac{a_1}{a_2} = \frac{1}{2}$ $\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}$ $\frac{c_1}{c_2} = \frac{-4}{-12} = \frac{1}{3}$ Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and the system has no solution.

Because the ratios of $x$ and $y$ coefficients are equal but not equal to the constant ratio, the lines are parallel. Parallel lines never intersect, making the system inconsistent.

Check the consistency of the system: $x - 2y = 0$ and $3x + 4y - 20 = 0$.

Given: $a_1 = 1, b_1 = -2, c_1 = 0$ and $a_2 = 3, b_2 = 4, c_2 = -20$. Comparing ratios: $\frac{a_1}{a_2} = \frac{1}{3}$ $\frac{b_1}{b_2} = \frac{-2}{4} = -\frac{1}{2}$ Since $\frac{1}{3} \neq -\frac{1}{2}$, we have $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

Since the ratio of the coefficients of $x$ is not equal to the ratio of the coefficients of $y$, the system has a unique solution and is consistent.