Question:

The number of solutions of the pair of linear equations \(x + 2y - 8 = 0\) and \(2x + 4y = 16\) is :

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For two linear equations \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\): 1. Calculate ratios: \(\frac{a_1}{a_2}\), \(\frac{b_1}{b_2}\), \(\frac{c_1}{c_2}\). 2. Compare:
If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) \(\rightarrow\) One unique solution (intersecting lines).
If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) \(\rightarrow\) Infinitely many solutions (coincident lines).
If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) \(\rightarrow\) No solution (parallel distinct lines). Equation 1: \(x + 2y - 8 = 0\) (\(a_1=1, b_1=2, c_1=-8\)) Equation 2: \(2x + 4y - 16 = 0\) (\(a_2=2, b_2=4, c_2=-16\)) Ratios: \(\frac{1}{2}\), \(\frac{2}{4}=\frac{1}{2}\), \(\frac{-8}{-16}=\frac{1}{2}\). All are equal. \(\rightarrow\) Infinitely many solutions.
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  • No solution
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The Correct Option is B

Solution and Explanation

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.)
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