Question

12. For what value of p the pair oí lirear equations (p + 2)x - (2p + 1)y=3(2p - 1)
and 2x – 3y = 7 has a unique solution.​

Answer 1

☯︎ Aɴsᴡᴇʀ

we have, two equations which have a unique solution

• (p + 2)x - (2p + 1)y - 3(2p - 1) = 0 ......(i)

• 2x - 3y - 7 = 0 ........(ii)

first compare the eq.(i) with a1x + b1y + C1 = 0 and eq.(ii) with a2x + b2y + c2 = 0

so, here

• a1 = p + 2
• b1 = 2p - 1
• c1 = -6p + 3
• a2 = 2
• b2 = -3
• c2 = -7

we know that, if linear equations having a unique solution then,

$\frac{a1}{a2} ≠ \frac{b1}{b2}$

so here,

$\bold{: \implies \frac{p + 2}{2} ≠ \frac{2p - 1}{ - 3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies - 3(p + 2) ≠ 2(2p - 1)} \\ \\ \bold{ : \implies - 3p - 6 ≠ 4p - 2} \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies - 6 + 2 ≠4p + 3p} \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies - 4 ≠ 7p} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ : \implies \boxed{\: \bold{p ≠ \frac{ - 4}{ \: 7} }}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so, the given system of equation will have unique solution for all real values of p other than -4/7