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:
- The value of c for which the pair of equations cx - y = 2 and 6x - 2y = 3 will have infinitely many . . . .
- 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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