Question

If the pair of equations 2x +3y=5 and 10x+15y=2k represent 2 coincident lines what is the value of k?​

Answer 1

Answer:

k = 12.5

Step-by-step explanation:

Given:

2 pairs of linear equation in 2 variables.

• 2x + 3y - 5 = 0
• 10x + 15y - 2k = 0

Solution:

Let

• a₁ = 2
• a₂ = 10
• b₁ = 3
• b₂ = 15
• c₁ = -5
• c₂ = -2k

For lines to be coincident,

$\tt{ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} }$

So,

$\dfrac{2}{10} = \dfrac{3}{15} = \dfrac{ - 5}{ - 2k}$

$\implies \sf{ \dfrac{1}{5} = \dfrac{1}{5} = \dfrac{ - 5}{ - 2k} }$

$\implies \sf{ \dfrac{1}{5} = \dfrac{ - 5}{ - 2k} }$

On cross multiplying and cancelling minus sign,

$\implies \sf{2k = 25}$

$\implies \sf{k = \dfrac{25}{2} }$

$\therefore \boxed{ \bf{ k = 12.5}}$

Answer 2

Answer:

k = 12.5

\begin{gathered}\\\end{gathered}

Step-by-step explanation:

Given:

2 pairs of linear equation in 2 variables.

2x + 3y - 5 = 0

10x + 15y - 2k = 0

\begin{gathered}\\\end{gathered}

Solution:

Let

a₁ = 2

a₂ = 10

b₁ = 3

b₂ = 15

c₁ = -5

c₂ = -2k

\begin{gathered}\\\end{gathered}

For lines to be coincident,

\begin{gathered} \tt{ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} } \\ \\ \end{gathered}

a

2

a

1

=

b

2

b

1

=

c

2

c

1

So,

\begin{gathered}\dfrac{2}{10} = \dfrac{3}{15} = \dfrac{ - 5}{ - 2k} \\ \\ \end{gathered}

10

2

=

15

3

=

−2k

−5

\begin{gathered} \implies \sf{ \dfrac{1}{5} = \dfrac{1}{5} = \dfrac{ - 5}{ - 2k} } \\ \\ \end{gathered}

⟹

5

1

=

5

1

=

−2k

−5

\begin{gathered} \implies \sf{ \dfrac{1}{5} = \dfrac{ - 5}{ - 2k} } \\ \\ \end{gathered}

⟹

5

1

=

−2k

−5

On cross multiplying and cancelling minus sign,

\begin{gathered} \implies \sf{2k = 25} \\ \\ \end{gathered}

⟹2k=25

\begin{gathered} \implies \sf{k = \dfrac{25}{2} } \\ \\ \end{gathered}

⟹k=

2

25

\begin{gathered} \therefore \boxed{ \bf{ k = 12.5}} \\ \\ \end{gathered}

∴

k=12.5