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 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.
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.
Question
For which value[s] of k will the pair of equations kx+3y=k-3 and 12x+ky=k have no solution?
Solution
Step 1: Convert the equations to standard form
Given pair of linear equations,
kx+3y=k-3 and 12x+ky=k
Converting the given equations to standard form,
kx+3y=k-3⇒kx+3y+[3-k]=0
12x+ky=k⇒12x+ky-k=0
Step 2: Compare the equations with the standard equation
Comparing the two equations with the standard form, we define
a1=k, b1=3 and c1=3-k
a2=12, b2=k and c2 =-k
Step 3: Define the condition for no solutions
For no solution, the given set of equations must satisfy the condition,
a1a2=b1b2≠c1c2
Thus,
k12=3k≠3-k-k
Step 4: Solve for k
Considering the first two parts,
k12=3k⇒k2-36=0⇒k=±6
Considering the last two parts,
3k≠3-k-k⇒k-3≠3⇒ k≠6
Considering the last two results, we can say that for k=-6, the set of equations will have no solutions
Therefore, for k=-6, the set of equations has no solutions.
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.