System of linear equation Determine the value for k for which the system of linear equation has infinitely many solution.
\begin{cases}
2x - y  = 2\\
4x + ky = 4
\end{cases}
 A: $2x - y = 2$
$4x + ky = 4$
Multiply first equation by 2 and subtract it from the second one.
So, you have: 
$2x - y = 2$
$(k+2)y = 0$.
If $k \ne -2$ then $y = 0$ and $x = 1$. 
If $k = -2$ then any $y$ is solution (in this case $x = (y+2)/2$)
A: You have a linear system of two unknowns with two equations.
If these equations are linearly independent you have at most one solution.
If these two equations are linearly dependent you have an infinity of solutions.
You can see that by interpreting each equation as a straight line. So the straight lines can be different so they have one or zero intersection point or they are the same and all points on the straight line are solutions.
In this case, to have an infinity of solution you have to find a number $\lambda$ such as the first eaquation multiplied by $\lambda$ give the seconde one.
So $\lambda*2=4$, $\lambda*(-1)=k$ and $\lambda*2=4$
Thus $\lambda = 2$ and $k=-2$
A: Using the matrix approach: 
$ \left( \begin{array}{cc}
2 & -1 \\
4 & k \end{array} \right)$
This is the matrix given by the coefficients of $x$ and $y$
For infinitely many solutions, the rank of this matrix has to be equal to the rank of the augmented matrix and less than the order of the matrix.
(Rank of a matrix, in layman's terms, will be the number of non-zero rows left in the matrix after performing elementary row operations on the matrix, starting from the bottom row.)
Here, the matrix is a $2\times2$ matrix, and so its order is $2$. 
So, the rank of the matrix has to be less than 2, i.e. 1. (Rank is always non-negative, non-zero).
So, we need to have exactly 1 non-zero row, i.e. the first row and the second row needs to be zero.
What operation should we use to make it zero?
We have to make the bottom row elements zero. $4$ can be made zero by subtracting $2$(the corresponding element in the top row! Observe!) $\times2$
so, multiplying first row with 2 and subtracting that from the second row gives us: 
$\left( \begin{array}{cc}
2 & -1 \\
(4-2\times2) & (k-2(-1)) \end{array} \right)$
which is 
$\left( \begin{array}{cc}
2 & -1 \\
0 & (k+2) \end{array} \right)$
The second row here will be zero if $k+2=0$, i.e. $k=-2$. That is your answer.
I know this method seems complicated, but once you get the grasp of it, it can be used to find k for any number of equations.  
A: I am giving you the simpliest solution accept it
General form of pair of linear equations in two variables: 
The general form for a pair of linear equations in two variables x and y is
a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0,
Where a1, b1, c1, a2, b2, c2 are all real numbers and a12 + b12 ≠ 0, a22 + b22 ≠ 0.
Coincident lines
Infinite number of solutions. The pair of linear equations is consistent and dependent.4
which is a1/a2=b1/b2=c1/c2
so we can write 2/4=-1/k=2/4
SO WE GET K=-2
A: Hint:-
Follow the general rule.
For,infinite solution,of two system of linear equations,
$$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2=0$$
the coefficients must satisfy the condition-$$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$
