For what value of k does the pair of equations 5x 2y 2k and 2 K 1 * X Ky 3k 4 have an infinite number of solutions?

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5x+2y=2k 2(k+1)x+ky=(3k+4)

Posted by Sneha Agrawal 3 years, 2 months ago

{tex}5x + 2y = 2k{/tex}
{tex}2(k + 1)x + ky = (3k + 4){/tex} These are of the form

{tex}a_1x + b_1y + c_1 = 0 , a_2x + b_2y + c_2 = 0{/tex}

where,

{tex}a_1= 5 ,b_1= 2, c_1 = -2k,{/tex}

{tex}a_2= 2(k +1) ,b_2= k ,c_2 = -(3k + 4){/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 holds only when

{tex}\frac { 5 } { 2 ( k + 1 ) } = \frac { 2 } { k } = \frac { - 2 k } { - ( 3 k + 4 ) }{/tex}

{tex}\Rightarrow \frac { 5 } { 2 ( k + 1 ) } = \frac { 2 } { k } = \frac { 2 k } { ( 3 k + 4 ) }{/tex} Now, the following cases arises:

Case 1:

{tex}\frac { 5 } { 2 ( k + 1 ) } = \frac { 2 } { k }{/tex}[Taking I and II]
{tex}\Rightarrow 5 k = 4 ( k + 1 ) \Rightarrow 5 k = 4 k + 4{/tex} k = 4

Case 2:

{tex}\frac { 2 } { k } = \frac { 2 k } { ( 3 k + 4 ) }{/tex}[Taking II and III]
{tex}\Rightarrow{/tex}2(3k + 4) = 2k2 {tex}\Rightarrow{/tex}6k + 8 = 2k2
{tex}\Rightarrow{/tex} {tex}2k^2 - 6k + 8 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}2(k^2 - 3k + 4)= 0{/tex}
{tex}\Rightarrow{/tex}{tex}k^2 - 3k - 4 = 0{/tex}
{tex}\Rightarrow{/tex}{tex}k^2 - 4k + k - 4 = 0{/tex}
{tex}\Rightarrow{/tex}k(k - 4) + 1(k - 4) = 0
{tex}\Rightarrow{/tex}(k - 4)(k + 1) = 0 (k - 4) = 0 or k + 1 = 0

{tex}\Rightarrow{/tex} k = 4 or k = -1

Case 3:
{tex}\frac { 5 } { 2 ( k + 1 ) } = \frac { 2 k } { ( 3 k + 4 ) }{/tex}[Taking I and II]
{tex}\Rightarrow 15 k + 20 = 4 k ^ { 2 } = 4 k{/tex}
{tex}\Rightarrow{/tex}4k2 + 4k - 15k - 20 = 0
{tex}\Rightarrow{/tex} 4k2 - 11k - 20 = 0
{tex}\Rightarrow{/tex}4k2 - 16k + 5k - 20 =0
{tex}\Rightarrow{/tex}4k(k - 4) + 5(k - 4) = 0
{tex}\Rightarrow{/tex}(k - 4)(4k + 5) = 0
{tex}\Rightarrow k = 4 \text { or } k = \frac { - 5 } { 4 }{/tex}
Thus, k = 4, is the common value of which there are infinitely many solutions.