Question

What should be the value of p, for the given equations to have infinitely many
solutions?
5x + py = 4 and 15 x + 3y = 12​

Answer 1

Answer:

The equations (p−3)x+3y−p=0

px+py−12=0

For infinite many solution

a

2

a

1

=

b

2

b

1

=

c

2

c

1

Here a

1

=p−3,b

1

=3,c

1

−p

a

2

=p,b

2

=p,c

2

=−12

p

p−3

=

p

3

=

−12

−1

Solving

p

3

=

−12

−1

⟹p

2

=36⟹p=±6

Now solving

p

p−3

=

−12

−1

⟹p−3=3⟹p=6

Hence the value of p=6.

Answer 2

$\large\underline{\sf{Given- }}$

Two lines,

• 5x + py = 4 and 15x + 3y = 12 have infinitely many solutions.

$\large\underline{\sf{To\:Find - }}$

• Value of 'p' for which lines have infinitely many solutions.

Understanding the concept :-

$\sf \: Let \: consider \: lines \: a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 = 0$

then,

• two lines have infinitely many solutions iff

$\rm :\longmapsto\:\bf \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

$\large\underline{\sf{Solution-}}$

Given that,

Two lines,

• 5x + py = 4 and 15x + 3y = 12 have infinitely many solutions.

Now,

We know that,

• Two lines have infinitely many solutions iff

$\rm :\longmapsto\: \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

Here,

• • a₁ = 5

• • a₂ = 15

• • b₁ = p

• • b₂ = 3

• • c₁ = 4

• • c₂ = 12

Now,

On substituting the values, we get

$\rm :\longmapsto\:\dfrac{5}{15} = \dfrac{p}{3} = \dfrac{4}{12}$

$\rm :\longmapsto\:\dfrac{1}{3} = \dfrac{p}{3} = \dfrac{1}{3}$

$\bf\implies \:p \: = \: 1$

Additional Information :-

$\sf \: Let \: consider \: lines \: a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 = 0$

then

(1). System of equations have unique solution iff

$\rm :\longmapsto\:\bf \:\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}$

(2). System of equations have no solutions iff

$\rm :\longmapsto\:\bf \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$