The value of k for which the lines (k+1)x + 3ky+15=0 and 5x+ky+5=0 are coincident if

Solution:

Given, the linear pair of equations are

3x + 2ky = 2

2x + 5y + 1 = 0

We have to find the value of k.

We know that,

For a pair of linear equations in two variables be a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\), then the graph will be a pair of parallel lines.

Here, a₁ = 3, b₁ = 2k, c₁ = -2

a₂ = 2, b₂ = 5, c₂ = 1

So, a₁/a₂ = 3/2

b₁/b₂ = 2k/5

c₁/c₂ = -2/1 = -2

By using the above result,

\(\frac{3}{2}=\frac{2k}{5}\)

On cross multiplication,

3(5) = 2(2k)

15 = 4k

So, k = 15/4

Therefore, the value of k is 15/4.

✦ Try This: If the lines given by 2x + 3ky = 2 and 3x + 5y + 1 = 0 are parallel, then the value of k is

Given, the linear pair of equations are

2x + 3ky = 2

3x + 5y + 1 = 0

We are required to find the value of k.

Here, a₁ = 2, b₁ = 3k, c₁ = -2

a₂ = 3, b₂ = 5, c₂ = 1

So, a₁/a₂ = 2/3

b₁/b₂ = 3k/5

c₁/c₂ = -2/1 = -2

By using the above result,

\(\frac{2}{3}=\frac{3k}{5}\)

On cross multiplication,

2(5) = 3(3k)

10 = 9k

So, k = 10/9

Therefore, the value of k is 10/9

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 3

NCERT Exemplar Class 10 Maths Exercise 3.1 Problem 7

If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is, a. -5/4, b. ⅖, c. 15/4, d. 3/2

Summary:

If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is 15/4

☛ Related Questions:

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• One equation of a pair of dependent linear equations is -5x + 7y = 2. The second equation can be, a. . . . .
• A pair of linear equations which has a unique solution x = 2, y = -3 is, a. x + y = -1; 2x - 3y = -5 . . . .

The given equations are:       k + 1x + 3ky  + 15 = 0and 5x + ky + 5 = 0Here, a1 = k  + 1,   b1 = 3k,  c1 = 15            a2 = 5,  b2 = k,  c2 = 5The given equations represent coincident  lines if they have infinitely many solution.the condition for infinitely many solutions is:       a1a2 = b1b2 = c1c2⇒ k  + 15 = 3kk = 155⇒ k + 15 = 31 =  31⇒ k + 15 = 31⇒ k + 1 = 15⇒  k = 15 - 1⇒ k = 14

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Find the value of k for which the line (K +1)x + 3ky + 15 =0 and 5x + ky + 5 coincident

Posted by Shreya Chatterjee 1 year, 11 months ago

• 1 answers

The given equations are:

(k+1)x+3ky+15=0, and

5x+ky+5=0

Since, the given equations are coincident, therefore

{tex}\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \frac{k+1}{5}=\frac{3k}{k}=\frac{15}{5}{/tex}

Taking the first two terms, we have

k=14

Or

The given equations are:

, and

Since, the given equations are coincident, therefore

Taking the first two terms, we have

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