For what value of ‘k’ the pair of equations x-ky=2 and 3x+2y=-50 has a unique solution? Question For what value of ‘k’ the pair of equations x-ky=2 and 3x+2y=-50 has a unique solution? in progress 0 Math Serenity 5 months 2021-12-13T02:53:56+00:00 2021-12-13T02:53:56+00:00 2 Answers 0 views 0

## Answers ( )

Answer:Find the value of k for which the system of linear equation

x +2y = 3 and 5x + k y = -7 has unique solution

## Answer:

k ≠ -2/3

## 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 equations are ;

x – ky = 2

3x + 2y = -50

The given equations can be rewritten as ;

x – ky – 2 = 0

3x + 2y + 50 = 0

Clearly ,

a = 1 , b = -k , c = -2

a’ = 3 , b = 2 , c = 50

Now,

a/a’ = 1/3

b/b’ = -k/2

c/c’ = -2/50 = -1/25

Now,

We know that , for unique solution ;

=> a/a’ ≠ b/b’

=> 1/3 ≠ -k/2

=> -k/2 ≠ 1/3

=> k ≠ -2/3

## Hence ,

## k ≠ -2/3 ( k can be any real number other than -2/3 ) .