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Find the value of $k$ for which the following system of equations having infinitely many solution:
$2x\ +\ 3y\ =\ 7$
$(k\ +\ 1)x\ +\ (2k\ -\ 1)y\ =\ 4k\ +\ 1$
Given: The given equation are $2x\ +\ 3y\ =\ 7$ ;$(k\ +\ 1)x\ +\ (2k\ -\ 1)y\ =\ 4k\ +\ 1$
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:
$2x\ +\ 3y\ =\ 7$
$(k\ +\ 1)x\ +\ (2k\ -\ 1)y\ =\ 4k\ +\ 1$
The system of equation is of the form $a_{1} x+b_{1} y=c_{1}\ and\ a_{2} x+b_{2} y=c_{2}$
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{2}{k+1} =\frac{3}{2k-1} =\frac{7}{4k+1} \ $
Now , $\frac{2}{k+1} =\frac{3}{2k-1}$
$\Rightarrow 3(k+1)=2(2k-1)$
$\Rightarrow 3k+3 = 4k-2$
$\Rightarrow 3k-4k=-5$
$Rightarrow k = 5$
Hence, the system of equations having infinitely many solutions if $k = 5$
Updated on: 10-Oct-2022
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