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For what value of k will the following equations have no solution.(3k+1)x+3y =2 ; (k2+1)x+(k-2)y=5 |
| Answer» Given linear equation is(3k + 1)x + 3 y - 2 = 0 .......... (i)(k2 + 1)x + (k - 2)y - 5 = 0 ............ (ii)Compare with a1x + b1y + c = 0 and a2x + b2y and c2 = 0a1 = 3k + 1 , b1 = 3 , c1 = -2and a2 = k2+ 1 , b2 = k - 2, c2 = -5The given system of equations will have no solution, if{tex} \\frac { a_1 } { a_2 } = \\frac { b_1 } { b_2} \\neq \\frac { c_1 } { c_2 }{/tex}{tex} \\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \\frac { 3 } { k - 2 } \\neq \\frac { - 2 } { - 5 }{/tex}{tex}\\Rightarrow \\quad \\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \\frac { 3 } { k - 2 } \\text { and } \\frac { 3 } { k - 2 } \\neq \\frac { 2 } { 5 }{/tex}Now,\xa0{tex}\\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \\frac { 3 } { k - 2 }{/tex}{tex}\\Rightarrow{/tex}\xa0(3k + 1)(k - 2)=3(k2 + 1){tex}\\Rightarrow{/tex}\xa03k2 - 5k - 2 =3k2 + 3{tex}\\Rightarrow{/tex}\xa0-5k - 2 =3{tex}\\Rightarrow{/tex}\xa0-5k = 5{tex}\\Rightarrow{/tex}\xa0k = -1Clearly,\xa0{tex}\\frac { 3 } { k - 2 } \\neq \\frac { 2 } { 5 }{/tex}\xa0for k = -1.Hence, the given system of equations will have no solution for k = -1. | |