Question

2x+3y+9=0, here x=2, y=4. solve the equation
​

Answer 1

Answer:

The required value of k is 6.

Step-by-step explanation:

Given :

The pair of equations 2x + 3y - 9 = 0 and 4x + ky - 18 = 0 has infinitely many solutions

To find :

the value of k

Solution :

Comparing the given two equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get

a₁ = 2 , b₁ = 3 , c₁ = -9

a₂ = 4 , b₂ = k , c₂ = -18

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 has infinitely many solutions when

\boxed{\tt \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

Substitute the values,

\begin{gathered}\sf \dfrac{2}{4}=\dfrac{3}{k}=\dfrac{-9}{-18} \\\\ \sf \dfrac{1}{2}=\dfrac{3}{k}=\dfrac{1}{2} \\\\ \implies \sf \dfrac{3}{k}=\dfrac{1}{2} \\\\ \implies \sf k=3 \times 2 \\\\ \implies \sf k=6\end{gathered}

4

2

=

k

3

=

−18

−9

2

1

=

k

3

=

2

1

⟹

k

3

=

2

1

⟹k=3×2

⟹k=6

The value of k is 6

_______________________

Know more :

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 has

1) no solution when \tt \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}

a

2

a

1

=

b

2

b

1



=

c

2

c

1

2) infinite solutions when \tt \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

3) unique solution when \tt \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}

a

2

a

1



=

b

2

b

1



=

c

2

c

1