For what value of k will the pair of linear equations KX 3y K 3 12x Ky K 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 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 (adsbygoogle = window.adsbygoogle || []).push({}); Given pair of linear equations,kx+3y=k-3 and 12x+ky=kConverting the given equations to standard form,kx+3y=k-3⇒kx+3y+(3-k)=012x+ky=k⇒12x+ky-k=0Step 2: Compare the equations with the standard equationComparing the two equations with the standard form, we definea1=k, b1=3 and c1=3-ka2=12, b2=k and c2 =-kStep 3: Define the condition for no solutionsFor no solution, the given set of equations must satisfy the condition,a1a2=b1b2≠c1c2 Thus,k12=3k≠3-k-kStep 4: Solve for kConsidering the first two parts,k12=3k⇒k2-36=0⇒k=±6 (adsbygoogle = window.adsbygoogle || []).push({}); Considering the last two parts,3k≠3-k-k⇒k-3≠3⇒ k≠6Considering the last two results, we can say that for k=-6, the set of equations will have no solutionsTherefore, 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.

For what value of k will the pair of equations have unique solution?

Thus for all real values of k other than 10, the given system of equations will have a unique solution. Hence, the required value of k is 10. There is no value of k for which the given system of equations has an infinite number of solutions.

For which value of k will the pair of linear equation have unique solution KX 2y 5 and 3x y 1?

Concept used: If (a1/a2) ≠ (b1/b2), then the system of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has a unique solution. ∴ The system has a unique solution if k ≠ 6.

How do you find K in a unique solution?

If a1a2≠b1b2 then the equations have unique solution, if a1a2 = b1b2 = c1c2 then the equations have infinitely many solutions and if a1a2 = b1b2 ≠ c1c2 then the equations have no solutions. Doing this will solve your problem and will give you the value of k while solving the equations.

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.