Find the value(s) of p for the pair of equati | vedclass

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(B) The given pair of linear equations is:$2x + 3y - 5 = 0$ ... $(i)$$px - 6y - 8 = 0$ ... $(ii)$Comparing these with the standard form $a_1x + b_1y +

Vedclass · Accepted Answer

$p 
eq -4$

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(B) The given pair of linear equations is:$2x + 3y - 5 = 0$ ... $(i)$$px - 6y - 8 = 0$ ... $(ii)$Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 2, b_1 = 3, c_1 = -5$$a_2 = p, b_2 = -6, c_2 = -8$For a pair of linear equations to have a unique solution,the condition is:$\frac{a_1}{a_2} 
eq \frac{b_1}{b_2}$Substituting the values:$\frac{2}{p} 
eq \frac{3}{-6}$$\frac{2}{p} 
eq -\frac{1}{2}$By cross-multiplying:$p 
eq -4$Thus,the system has a unique solution for all real values of $p$ except $-4$,i.e.,$p \in \mathbb{R} - \{-4\}$.

Find the value$(s)$ of $p$ for the pair of equations $2x + 3y - 5 = 0$ and $px - 6y - 8 = 0$ if the system has a unique solution.

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