For which value S of K Will the pair of equations KX 3y K 3 12x Ky K have B many solutions?

Find the value of k for which the system of equations has a unique solution:
kx + 3y = (k – 3),
12x + ky = k

Nội dung chính Show

For which value(s) of k will the pair of equations kx + 3y = k - 3; 12x + ky = k have no solution
For what value of k is the system of equations KX 3y K 2 12x Ky K inconsistent?
For what value of k has no solution?
What value of k the pair of equations 4x 3y 9 2x KY 11 has no solution?
For which value S of K Will the pair of equations KX 3y K 3 12x Ky?
For what value of k is the system of equations KX 3y K 2 12x Ky K inconsistent?
For which values of k will the pair of equations have no solution?
What value of k the pair of equations 4x 3y 9 2x KY 11 has no solution?

Nội dung chính

For which value(s) of k will the pair of equations kx + 3y = k - 3; 12x + ky = k have no solution
For what value of k is the system of equations KX 3y K 2 12x Ky K inconsistent?
For what value of k has no solution?
What value of k the pair of equations 4x 3y 9 2x KY 11 has no solution?

The given system of equations:
kx + 3y = (k – 3)
⇒ kx + 3y – (k - 3) = 0 ….(i)
And, 12x + ky = k
⇒12x + ky - k = 0 …(ii)
These equations are of the following form:
`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`
Here, `a_1 = k, b_1= 3, c_1= -(k – 3) and a_2 = 12, b_2 = k, c_2= -k`
For a unique solution, we must have:
`(a_1)/(a_2) ≠ (b_1)/(b_2)`
i.e., `k /12 ≠ 3/k`
⇒ `k^2 ≠ 36 ⇒ k ≠ ±6`
Thus, for all real values of k, other than ±6, the given system of equations will have a unique solution.

Solution : Given pair of linear equations is `kx + 3y = k - 3 " " ...(i) ` and `" " 12x + ky = k " " ...(ii)` On comparing with `ax + by + c = 0`, we get `a_(1) = k, b_(1) = 3` and `c_(1) = -(k - 3) " " ` [from Eq. (i)] `a_(2) = 12, b_(2) = k` and ` c_(2) = -k " " `[from Eq. (ii)] For no solution of the pair of linear equations, `(a_(1))/(a_(2)) = (b_(2))/(b_(2)) != (c_(1))/(c_(2))` `rArr " " (k)/(12) = (3)/(k) != (-(k - 3))/(-k)` Taking first two parts, we get `rArr " " (k)/(12) = (3)/(k)` `rArr " " k^(2) = 36` `rArr " " k +- 6` Taking last two parts, we get `(3)/(k) != (k - 3)/(k)` `rArr " " 3k != k (k - 3)` `rArr " " 3k - k (k - 3) != 0` `rArr " "k( 3-k + 3) != 0` `rArr " " k(6 - k) != 0` `rArr " " k !=0` and `!= 6` Hence, required value of k for which the given pair of linear equations has no solution is - 6.

Solution:

Given, the pair of linear equations are

kx + 3y = k - 3

12x + ky = k

We have to determine the value of k for which the pair of linear equations will have no solution.

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 and so the pair of equations will have no solution.

Here, a₁ = k, b₁ = 3, c₁ = k - 3

a₂ = 12, b₂ = k, c₂ = k

So, a₁/a₂ = k/12

b₁/b₂ = 3/k

c₁/c₂ = (k - 3)/k

For no solution,

\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\)

So, k/12 = 3/k ≠ (k - 3)/k

Case 1) k/12 = 3/k

k(k) = 3(12)

k2 = 36

k = ±6

Case 2) 3/k ≠ (k - 3)/k

3(k) ≠ k(k - 3)

3k ≠ k2 - 3k

k2 - 3k - 3k ≠ 0

k2 - 6k ≠ 0

k(k - 6) ≠ 0

So, k = 6, 0

Therefore, for the value of k = -6, the pair of linear equations have no solution.

✦ Try This: For which value(s) of λ, do the pair of linear equations λx + y = 2λ/3 and x/2 + λy = 10 have no solution

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 3

NCERT Exemplar Class 10 Maths Exercise 3.3 Problem 2

For which value(s) of k will the pair of equations kx + 3y = k - 3; 12x + ky = k have no solution

Summary:

For the value of k = -6, the pair of linear equations kx + 3y = k - 3; 12x + ky = k has no solution

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For what value of k is the system of equations KX 3y K 2 12x Ky K inconsistent?

For k = ±6, the system of equations kx + 3y = k - 2, 12x + ky = k is inconsistent.

For what value of k has no solution?

Hence, the given system of equations has no solution when k = −1.

What value of k the pair of equations 4x 3y 9 2x KY 11 has no solution?

k=−32.

For which value S of K Will the pair of equations KX 3y K 3 12x Ky?

For which value(s) of k will the pair of equations kx + 3y = k – 3 ; 12x + ky = k have no solution? Therefore, value of k for which the given pair of linear equations has no solution is k = – 6.