Question

The pair of linear equations 5x +ky= 4 and 15x + 3y = 12 have infinite many solutions if K is equal to *

Answer 1

EXPLANATION

• EQUATION GIVEN

5x + ky = 4 and 15x + 3y = 12

have infinite many solution.

TO FIND VALUE OF K.

Equation will be written as:-

5x + ky - 4 = 0

15x + 3y - 12 = 0

CONDITIONS FOR INFINITE MANY SOLUTION.

$\frac{</p><p>A1}{A2} = \frac{B1}{B2} = \frac{C1}{C2}$

a1 = 5, a2 = 15, b1 = k, b2 = 3, c1 = -4, c2 = -12

5/15 = k/3 = -4/-12

CASE 1

IF,

5/15 = k/3

1/3 = k/3

k = 3

CASE 2

IF,

k/3 = -4/-12

k/3 = 1/3

k = 3

Therefore,

value of k = 3 = ANSWER

Answer 2

Step-by-step explanation:

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}:⟼

a

2

a

1

=

b

2

b

1

=

c

2

c

1

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}:⟼

15

5

=

3

p

=

12

4

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

3

1

=

3

p

=

3

1

\bf\implies \:p \: = \: 1⟹p=1