# 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?

2. Realized i've already asked this exact same question woops, mods delete please. Sorry.

3. Originally Posted by adam_leeds
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 :

$\displaystyle x-y=1$

$\displaystyle 3x-3y=k$

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

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

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

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

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

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

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

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

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

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