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
Is there an error in this question or solution?
APPEARS IN
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.
Question
Open in App
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.
Suggest Corrections
19
Similar questions
Q. For which value[s] of k will the pair of equations have no
solution?
kx+3y=k−3;
12x+ky=k;
k≠0
Q. If the system of equations
kx+3y−[k−3]=0,12x+ky−k= has
infinitely many solutions, then k=
Q. For which value of the given system of equations
have infinitely many solution,
[k−3]x+3y=k and
kx+ky=12
Q. Find the value of k for which the given system of equations has no
solution.kx+3y=k−3;12x+ky=k
Q. The value of k for which
kx+3y−k+3=0 and 12x+ky=k, have infinite solutions, is?
View More
Solve
Textbooks
Question Papers