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

2 thoughts on “What should be the value of p, for the given equations to have infinitely many<br />solutions?<br />5x + py = 4 and 15 x + 3y = 12”

1. [tex]\large\underline{\sf{Given- }}[/tex]

   Two lines,

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

   [tex]\large\underline{\sf{To\:Find – }}[/tex]

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

   Understanding the concept :-

   [tex]\sf \: Let \: consider \: lines \: a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 = 0[/tex]

   then,

   • two lines have infinitely many solutions iff

   [tex]\rm :\longmapsto\:\bf \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}[/tex]

   [tex]\large\underline{\sf{Solution-}}[/tex]

   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

   [tex]\rm :\longmapsto\: \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}[/tex]

   Here,

   • • a₁ = 5

   • • a₂ = 15

   • • b₁ = p

   • • b₂ = 3

   • • c₁ = 4

   • • c₂ = 12

   Now,

   On substituting the values, we get

   [tex]\rm :\longmapsto\:\dfrac{5}{15} = \dfrac{p}{3} = \dfrac{4}{12} [/tex]

   [tex]\rm :\longmapsto\:\dfrac{1}{3} = \dfrac{p}{3} = \dfrac{1}{3} [/tex]

   [tex]\bf\implies \:p \: = \: 1[/tex]

   Additional Information :-

   [tex]\sf \: Let \: consider \: lines \: a_1x + b_1y + c_1 = 0 \: and \: a_2x + b_2y + c_2 = 0[/tex]

   then

   (1). System of equations have unique solution iff

   [tex]\rm :\longmapsto\:\bf \:\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2}[/tex]

   (2). System of equations have no solutions iff

   [tex]\rm :\longmapsto\:\bf \:\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}[/tex]

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2. 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.

   Reply