InterviewSolution
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InterviewSolution
| 1. |
Kx+3y=k- - 3 12x+ky=k. Find the value of k. |
| Answer» The given equations arekx + 3y - (k - 3) = 0 ......... (i)12x + ky - k = 0 ........... (ii)The system of linear equations is in the form of\xa0a1x + b1y + c1\xa0= 0a2x + b2y + c2\xa0= 0Compare (i) and (ii), we geta1= k\xa0,b1= 3, c1\xa0= -(k - 3),a2=12\xa0,b2= k\xa0,c2\xa0= -kFor a unique solution, we must have{tex}\\frac { a _ { 1 } } { a _ { 2 } } \\neq \\frac { b _ { 1 } } { b _ { 2 } }{/tex}{tex}\\frac { k } {1 2 } \\neq \\frac { 3 } { k }{/tex}{tex}\\Rightarrow k ^ { 2 } \\neq 36 {/tex}{tex}\\Rightarrow k \\neq \\pm 6{/tex}Thus, for all real value of k other than {tex}\\pm 6{/tex}, the given system of equations will have a unique solution. | |