For what value of k will the pair of equations KX 3y K 3?

Find the value of k for which the system of equations
kx + 3y + 3 - k = 0, 12x + ky - k = 0
has no solution.

Solution

The given system of equations can be written as
kx + 3y + 3 - k = 0                       ….(i)
12x + ky - k = 0                           ….(ii)
This system of the form:
`a_1x+b_1y+c_1 = 0`
`a_2x+b_2y+c_2 = 0`
where, `a_1 = k, b_1= 3, c_1 = 3 - k and a_2 = 12, b_2 = k, c_2= –k`

For the given system of linear equations to have no solution, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`
`⇒ k/12 = 3/k ≠ (3−k)/(−k)`
`⇒k/12 = 3/k and 3/k ≠ (3−k)/(−k)`
`⇒ k^2 = 36 and -3 ≠ 3 - k`
⇒ k = ±6 and k ≠ 6
⇒k = -6
Hence, k = -6.

Concept: Pair of Linear Equations in Two Variables

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

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.

The given pair of linear equations is

kx + 3y = k – 3 …(i)

12x + ky = k …(ii)

On comparing the equations (i) and (ii) with ax + by = c = 0,

We get,

a1 = k, b1 = 3, c1 = -(k – 3)

a2 = 12, b2 = k, c2 = – k

Then,

a1 /a2 = k/12

b1 /b2 = 3/k

c1 /c2 = (k-3)/k

For no solution of the pair of linear equations,

a1/a2 = b1/b2≠ c1/c2

k/12 = 3/k ≠ (k-3)/k

Taking first two parts, we get

k/12 = 3/k

k2 = 36

k = + 6

Taking last two parts, we get

3/k ≠ (k-3)/k

3k ≠ k(k – 3)

k2 – 6k ≠ 0

so, k ≠ 0,6

Therefore, value of k for which the given pair of linear equations has no solution is k = – 6.

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Solution

kx+3y=k-3 12x+ky=k compare with a1x+b1y=c1 a2x+b2y=c2 a1=k,a2=12,b1=3,b2=k c1=k−3,c2=k For infinite solutions, a1a2=b1b2=c1c2 ∴k12=3k=k−3k k12=3k⇒k2=36 ∴k=6.

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