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

$2x\ +\ 3y\ =\ k$$(k\ -\ 1)x\ +\ (k\ +\ 2)y\ =\ 3k$

Given:

The given system of equations is:

$2x\ +\ 3y\ =\ k$

$(k\ -\ 1)x\ +\ (k\ +\ 2)y\ =\ 3k$

To do:

We have to find the value of $k$ for which the given system of equations has infinitely many solutions.

Solution:

The given system of equations can be written as:

$2x\ +\ 3y\ -\ k=0$

$(k\ -\ 1)x\ +\ (k\ +\ 2)y\ -\ 3k\ =\ 0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

The condition for which the above system of equations has infinitely many solutions is

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

Comparing the given system of equations with the standard form of equations, we have,

$a_1=2, b_1=3, c_1=-k$ and $a_2=k-1, b_2=k+2, c_2=-3k$

Therefore,

$\frac{2}{k-1}=\frac{3}{k+2}=\frac{-k}{-3k}$

$\frac{2}{k-1}=\frac{3}{k+2}=\frac{1}{3}$

$\frac{2}{k-1}=\frac{1}{3}$ and $\frac{3}{k+2}=\frac{1}{3}$

$3\times2=(k-1)\times1$ and $3\times3=(k+2)\times1$

$6=k-1$ and $9=k+2$

$k=6+1$ and $k=9-2$

$k=7$

**The value of $k$ for which the given system of equations has infinitely many solutions is $7$.**