Question

Write the value of k for which the system of equations x + y – 4 = 0 and 2x + ky – 3 = 0 has no solution.

Answer 1

Value of k = 2

If two equations such as :-

$a_{1}x \: + \: b_{1}y \: + c_{1} \: = 0 \\ a_{2}x \: + \: b_{2}y \: + c_{2} \: = 0$

has no solutions.

Then :-

$\\ \frac{ a_{1} }{ a_{2}} = \frac{ b_{1} }{ b_{2} } \: ≠ \: \frac{ c_{1} }{ c_{2} }$

Now, we will compare the given equations with the general equations.

Given :-

1st Equation :-

$x + y - 4 = 0$

Here,

$a_{1} = 1 \\ b_{1} = 1 \\ c_{1} = - 4$

2nd Equation :-

$2x + ky - 3 = 0$

Here,

$a_{2} = 2 \\ b_{2} = k \\ c_{2} = - 3$

Now, substituting these values.

$\\ \frac{1}{2} = \frac{1}{k} ≠ \frac{ - 4}{ - 3}$

By looking at the above equation,

we can see that 1/2 cannot be equal to 4/3

Therefore, 1/k cannot be equal to 4/3 as well as it is equal to 1/2.

So,

$\\ \frac{1}{2} = \frac{1}{k} \\ \\ { \: k = 2}$

Answer 2

Given:

x + y – 4 = 0

2x + ky – 3 = 0

To find:

The value of k for which the system of equations have no solution.

Answer:

• For system of equations not having any solution, condition is
• a1/a2 = b1/b2 ≠ c1/c2
• Here a1 =1, a2 =2,  b1 =1,  b2 =k,  c1 =-4, c2 =-3
• By putting values in this equation we get,

k=2