Thread: 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

3x − 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?

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

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

3x − 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