Find the value of $k$ for which the following system of equations having infinitely many solution:

$x\ +\ (k\ +\ 1)y\ =\ 4$
$(k\ +\ 1)x\ +\ 9y\ =\ 5k\ +\ 2$


 Given: The given equation are  $x\ +\ (k\ +\ 1)y\ =\ 4$ ;$(k\ +\ 1)x\ +\ 9y\ =\ 5k\ +\ 2$

To do:  Find the value of $k$ for which the following system of equations having infinitely many solutions.


Solution:

The given system of equation is:

$x\ +\ (k\ +\ 1)y\ =\ 4$


$(k\ +\ 1)x\ +\ 9y\ =\ 5k\ +\ 2$

The system of equation is of the form $a_{1} x+b_{1} y=c_{1}\ and\ a_{2} x+b_{2} y=c_{2}$


Here, $a_1 = 1, b_1=k+1, c_1=-4 \ and \ a_2=k+1, b_2=9, c_2=-(5k+2) $

For the infinitely many solutions there is a condition

$\frac{a_{1}}{a_{2}} \ =\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}} \ $

$\frac{1}{k+1}  =\frac{k+1}{9} =\frac{-4}{-(5k+2)} \ $

Now , $\frac{1}{k+1}  =\frac{k+1}{9} $ and $\frac{k+1}{9} =\frac{4}{5k+2} $

$(k+1)^2=9$ and $(5k+2)(k+1)=4\times9$

$k^2+2k+1=9$ and  $5k^2+5k+2k+2=36$

$k^2+2k+1-9=0$ and $5k^2+7k+2-36=0$

$k^2+2k-8=0$ and $5k^2+17k-10k-34=0$

$k(k+4)-2(k+4)=0$ and $(5k+17)-2(5k+17)=0$ 

$(k+4)(k-2)=0$ and  $(5k+17)(k-2)=0$

$k=-4 , k=2$ and $k=-\frac{17}{5}, k=2$

$k=2$ satisfies the conditions

Hence, the  system of equations having infinitely many solutions if $k = 2$

Updated on: 10-Oct-2022

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