Question

How many solutions will \(x + 2y + 3 = 0\), \(3x + 6y + 9 = 0\) have?

• A. One solution
• B. No solution
• C. Infinitely many solutions
• D. None of these

Hint:

Notice that the second equation, \(3x + 6y + 9 = 0\), is just the first equation, \(x + 2y + 3 = 0\), multiplied by 3. When one equation is a multiple of the other, the lines are coincident.

Answer 1

The Correct Option is C

Step 1: Understanding the 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 number of solutions depends on the ratio of their coefficients.
1. If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines are intersecting and there is one unique solution.
2. If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel and there is no solution.
3. If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines are coincident and there are infinitely many solutions.

Step 2: Detailed Explanation:
For the given equations:
Equation 1: \(x + 2y + 3 = 0\), so \(a_1 = 1, b_1 = 2, c_1 = 3\).
Equation 2: \(3x + 6y + 9 = 0\), so \(a_2 = 3, b_2 = 6, c_2 = 9\).
Now, let's find the ratios:
\[ \frac{a_1}{a_2} = \frac{1}{3} \] \[ \frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3} \] \[ \frac{c_1}{c_2} = \frac{3}{9} = \frac{1}{3} \] Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the two lines are coincident (they are the same line).

Step 3: Final Answer:
The system of equations will have infinitely many solutions.