Question

If 5x+py+8=0 and 10x+15y+12=0 has no solution, then p =

• A. \(\frac{15}{2}\)
• B. \(\frac{13}{2}\)
• C. \(\frac{7}{2}\)
• D. \(\frac{5}{2}\)

Answer 1

The Correct Option is A

Step 1: Understand the condition for no solution.

Two linear equations of the form: \[ a_1x + b_1y + c_1 = 0 \\ a_2x + b_2y + c_2 = 0 \] have no solution if: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \] This is the condition for inconsistency — the lines are parallel but not coincident.

Step 2: Match coefficients.

Given equations:

• Equation 1: \(5x + py + 8 = 0\)
• Equation 2: \(10x + 15y + 12 = 0\)

Here, \[ a_1 = 5,\quad b_1 = p,\quad c_1 = 8 \\ a_2 = 10,\quad b_2 = 15,\quad c_2 = 12 \]

Now, apply the condition: \[ \frac{5}{10} = \frac{p}{15} \ne \frac{8}{12} \]

Step 3: Solve for \(p\)

From \( \frac{5}{10} = \frac{p}{15} \), cross-multiply: \[ 5 \cdot 15 = 10 \cdot p \Rightarrow 75 = 10p \Rightarrow p = \frac{75}{10} = \frac{15}{2} \]

Step 4: Check that \( \frac{c_1}{c_2} \ne \frac{8}{12} = \frac{2}{3} \)

And: \[ \frac{8}{12} = \frac{2}{3},\quad \frac{c_1}{c_2} = \frac{8}{12} \ne \frac{15}{2} \div 15 = \frac{1}{2} \Rightarrow \text{Condition for no solution is satisfied.} \]

The correct option is (A): \(\frac{15}{2}\)