Question

find the value of k so that the following system of equation no solutions 3x-y-5=0 abd 6x-2y+k=0​

Answer 1

Answer :

k ≠ 10 (ie. k can be any real number other than -10)

Note:

★ A linear equation in 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 system of linear equations is ;

3x - y - 5 = 0 -----(1)

6x - 2y + k = 0 ------(2)

Now ,

Comparing the given linear equations (1) and (2) with the general linear equations ax + by + c = 0 and a'x + b'y + c' = 0 , we get ;

a = 3

a' = 6

b = -1

b' = -2

c = -5

c' = k

Now ,

a/a' = 3/6 = 1/2

b/b' = -1/-2 = 1/2

c/c' = -5/k

The given system will have no solution if ;

=> a/a' = b/b' ≠ c/c'

=> ½ = ½ ≠ -5/k

=> ½ ≠ -5/k

=> k ≠ -5 × 2

=> k ≠ -10

Hence ,

k ≠ 10 (ie. k can be any real number other than -10)