# Thread: Need help with this equation using reduction methods

1. ## Need help with this equation using reduction methods

For which values of
k are there no solutions, many solutions, or a unique solution to this
system?

x
y = 1

3
x 3y = k

This is what i did

1 -1 | 1
3 -3 | k

Line 2 - (3 x line 1)

1 -1 | 1
0 0 |k-3

So k = 3 there are inifinte solutions
And when k doesnt = 3 there are no solutions

But what about one solution? Is there one?

For which values of
k are there no solutions, many solutions, or a unique solution to this
system?

x
y = 1

3
x 3y = k

This is what i did

1 -1 | 1
3 -3 | k

Line 2 - (3 x line 1)

1 -1 | 1
0 0 |k-3

So k = 3 there are inifinte solutions
And when k doesnt = 3 there are no solutions

But what about one solution? Is there one?
i hope this is what you wrote :

$x-y=1$

$3x-3y=k$

there is no unique (one) solution because determinant is not different from zero

$D=\begin{vmatrix}
1 &-1 \\
3 &-3
\end{vmatrix} = 1\cdot (-3) - 3\cdot (-1) = 0$

$D_x= \begin{vmatrix}
1 &-1 \\
k &-3
\end{vmatrix} = -3+k = k-3$

$D_y= \begin{vmatrix}
1 &1 \\
3 &k
\end{vmatrix} = k-3$

so to conclude ... (this is theory that you should know)

$D\neq 0 \Rightarrow$ there is unique solution

$\displaystyle x = \frac {D_x}{D}$

$\displaystyle y = \frac {D_y}{D}$

$D= 0 \Rightarrow$ and $D_x\neq 0 \vee D_y\neq 0$ there is no solution

$D= D_x=D_y=0 \Rightarrow$ there is infinite many solutions