Concept: For a pair of linear equations in two variables, \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), the nature of their solutions can be determined by comparing the ratios of their coefficients:
(A) If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines are intersecting and there is exactly one unique solution.
(B) If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines are coincident (they are the same line), and there are infinitely many solutions.
(C) If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel and distinct, and there is no solution.
Step 1: Write the given equations in the standard form \(ax+by+c=0\)
Equation 1: \(x + 2y - 8 = 0\)
Here, \(a_1 = 1\), \(b_1 = 2\), \(c_1 = -8\).
Equation 2: \(2x + 4y = 16\)
Rewrite in standard form: \(2x + 4y - 16 = 0\)
Here, \(a_2 = 2\), \(b_2 = 4\), \(c_2 = -16\).
Step 2: Calculate the ratios of the coefficients
\(\frac{a_1}{a_2} = \frac{1}{2}\)
\(\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}\)
\(\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{8}{16} = \frac{1}{2}\)
Step 3: Compare the ratios
We observe that:
\[ \frac{a_1}{a_2} = \frac{1}{2} \]
\[ \frac{b_1}{b_2} = \frac{1}{2} \]
\[ \frac{c_1}{c_2} = \frac{1}{2} \]
So, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\).
Step 4: Determine the number of solutions
Since all three ratios are equal (\(\frac{1}{2}\)), the two linear equations represent coincident lines. This means they are essentially the same line, and every point on that line is a solution.
Therefore, there are infinitely many solutions.
This matches option (2).
(Note: If you multiply the first equation \(x + 2y - 8 = 0\) by 2, you get \(2(x + 2y - 8) = 2(0)\), which is \(2x + 4y - 16 = 0\). This is identical to the second equation \(2x + 4y = 16\), confirming they are the same line.)