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

[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] Reply

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

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

Two lines,

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

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,

[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,

Now,

We know that,

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

Here,

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]

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.