Question

for what value of "p" the following pair of equation has a unique solution 2x+py =-5 and 3x+3y=-6​

Answer 1

☯︎ Aɴsᴡᴇʀ

we have, two equations which have a unique solution

• 2x + py + 5 = 0 ......(i)
• 3x + 3y + 6 = 0 ........(ii)

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

so, here

• a1 = 2
• b1 = p
• c1 = 5
• a2 = 3
• b2 = 3
• c2 = 6

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

$\frac{a1}{a2} ≠ \frac{b1}{b2}$

now, here

$\bold{ : \implies \frac{2}{3} ≠ \frac{p}{3} } \: \: \: \: \: \: \: \\ \\ \bold { : \implies p≠ \frac{ 2 \times \cancel{ 3}}{ \cancel{3}} } \\ \\ \bold{ : \implies \boxed{ \bold{p ≠ 2}}} \: \: \: \: \: \:$

so, the given system of equation will have unique solution for all real values of p other than 2

Answer 2

Aɴsᴡᴇʀ

we have, two equations which have a unique solution

2x + py + 5 = 0 ......(i)

3x + 3y + 6 = 0 ........(ii)

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

so, here

a1 = 2

b1 = p

c1 = 5

a2 = 3

b2 = 3

c2 = 6

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

$\frac{a1}{a2} ≠ \frac{b1}{b2}$

now, here

$\bold{ : \implies \frac{2}{3} ≠ \frac{p}{3} } \: \: \: \: \: \: \: \\ \\ \bold { : \implies p≠ \frac{ 2 \times \cancel{ 3}}{ \cancel{3}} } \\ \\ \bold{ : \implies \boxed{ \bold{p ≠ 2}}} \: \: \: \: \: \:$

so, the given system of equation will have unique solution for all real values of p other than 2