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 Show
Solution : Given pair of linear equations is Find the value of k for which the system of equations has a unique solution: The given system of equations: 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.
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