Q3. Find the value of k for which the lines of pair of equations 3x+2ky=2 and 2x+5y= -1 are parallel. Question Q3. Find the value of k for which the lines of pair of equations 3x+2ky=2 and 2x+5y= -1 are parallel. in progress 0 Math Alice 7 months 2021-10-13T18:01:06+00:00 2021-10-13T18:01:06+00:00 2 Answers 0 views 0

## Answers ( )

## Answer :

k = 15/4

## Note:

★ A linear equation is two variables represent a straight line .

★ The word consistent is used for the system of equations which consists any solution .

★ The word inconsistent is used for the system of equations which doesn’t consists any solution .

★ Solution of a system of equations : It refers to the possibile values of the variable which satisfy all the equations in the given system .

★ A pair of linear equations are said to be consistent if their graph ( Straight line ) either intersect or coincide each other .

★ A pair of linear equations are said to be inconsistent if their graph ( Straight line ) are parallel .

★ If we consider equations of two straight line

ax + by + c = 0 and a’x + b’y + c’ = 0 , then ;

• The lines are intersecting if a/a’ ≠ b/b’ .

→ In this case , unique solution is found .

• The lines are coincident if a/a’ = b/b’ = c/c’ .

→ In this case , infinitely many solutions are found .

• The lines are parallel if a/a’ = b/b’ ≠ c/c’ .

→ In this case , no solution is found .

## Solution :

Here ,

The given linear equations are ;

3x + 2ky = 2

2x + 5y = -1

These equations can be rewritten as ;

3x + 2ky – 2 = 0

2x + 5y + 1 = 0

Clearly , we have ;

a = 3 , b = 2k , c = -2

a’ = 2 , b’ = 5 , c’ = 1

For the given lines to be parallel ,

a/a’ = b/b’ ≠ c/c’

3/2 = 2k/5 ≠ -2/1

Clearly , 3/2 ≠ -2/1

Thus ,

=> 3/2 = 2k/5

=> 2k/5 = 3/2

=> k = (3/2)×(5/2)

=> k = 15/4

## Hence , k = 15/4 .

3/2=2k/5

3*5=4k

15=4k

15/4=k