Find the value of k in 2x + 3y=7 and (k-1)x + (k+2)y=3k

1.	Find the value of k in 2x + 3y=7 and (k-1)x + (k+2)y=3k
Answer» {tex}2x + 3y = 7{/tex}{tex}(k - 1) x + (k + 2)y = 3k{/tex}These are of the form{tex}a_1x + b_1y + c_1\xa0= 0 ,\\ a_2x + b_2y + c_2\xa0= 0{/tex}where,{tex}a_1= 2\xa0,\\ b_1= 3,\\ c_1\xa0= -7,{/tex}{tex}a_2=k - 1\xa0\\ ,b_2= k + 2 ,\\ c_2\xa0= -3k {/tex} for infinitely many solutions, we must have{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { c _ { 1 } } { c _ { 2 } }{/tex}This hold only when{tex}\\frac { 2 } { k - 1 } = \\frac { 3 } { k + 2 } = \\frac { - 7 } { - 3 k }{/tex}{tex}\\frac { 2 } { k - 1 } = \\frac { 3 } { k + 2 } = \\frac { 7 } { 3 k }{/tex}Now the following cases arises:Case I:{tex}\\frac { 2 } { k - 1 } = \\frac { 3 } { k + 2 }{/tex}{tex}\\Rightarrow{/tex}2(k + 2) = 3(k -1){tex}\\Rightarrow{/tex}2k + 4= 3k - 3{tex}\\Rightarrow{/tex}\xa0k = 7Case II:{tex}\\frac { 3 } { k + 2 } = \\frac { 7 } { 3 k }{/tex}{tex}\\Rightarrow{/tex}7(k + 2) = 9k{tex}\\Rightarrow{/tex}7k + 14= 9k{tex}\\Rightarrow{/tex}\xa0k = 7Case III:{tex}\\frac { 2 } { k - 1 } = \\frac { 7 } { 3 k }{/tex}{tex}\\Rightarrow{/tex}\xa07k - 7 = 6k{tex}\\Rightarrow{/tex}\xa0k = 7For k = 7, there are infinitely many solutions of the given system\xa0of equations.

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