We have a linear system of equations if and only if:

• We have several equations that have to be fulfilled at the same time.

• Those equations are linear, that means every variable occurs at most with power 1

Now, let's check:

Is not a linear system, since $xx$ and $yy$ have power $-1-1$ in the first line.

Is not a linear system, since $xx$ has power $22$ in the second line.

All other systems are indeed linear, since $xx$ and $yy$ only appear with power $11$.

You may show that the equations $\mathrm{I}\backslash mathrm\{I\}$ and $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ cannot be satisfied at the same time. The best way to do this is by subtraction.

Subtract equation $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ from $\mathrm{I}\backslash mathrm\{I\}$.

The equation is obviously wrong. Thus, the given system of equations is also "wrong" and has no solution.

Thus the solution set is $\mathbb{L}=\{\}\backslash mathbb\{L\}=\backslash \{\backslash \}$.

You need to show that equations $\mathrm{I}\backslash mathrm\{I\}$ and $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ are equivalent. That is, any $xx$ and $yy$ that solves equation $\mathrm{I}\backslash mathrm\{I\}$ also solves equation $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ (and vice versa).

One way to show this is by the substitution method.

First, solve $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ for y .

Now substitute ${\mathrm{I}\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{II\}\text{'}$ into equation $\mathrm{I}\backslash mathrm\{I\}$. You get the new equation ${\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{I\text{'}\}$.

Summarize the left side of the equation.

The equation ${\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{I\text{'}\}$ is true and independent of $xx$. So there are infinitely many solutions.

$\mathbb{L}=\{(x\mathrm{\mid }y)\mathrm{\mid }y=3x-2\}\backslash mathbb\{L\}=\backslash \{(x|y)|\backslash ;y=3x-2\backslash \}$

A linear system of equations has exactly one solution, if it can be solved unambiguously for $xx$ and $yy$.

In order to find this out, the substitution method is useful, because for example equation $\mathrm{I}\backslash mathrm\{I\}$ can be solved for $yy$ very easily.

First solve equation $\mathrm{I}\backslash mathrm\{I\}$ for $yy$.

Now plug ${\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{I\text{'}\}$ into $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$. You get the new equation $\mathrm{I}{\mathrm{I}}^{\mathrm{\prime }}\backslash mathrm\{II\text{'}\}$.

Summarize the left side.

Solve for $xx$.

$x=\frac{11}{15}x\; =\; \backslash frac\{11\}\{15\}$

Plug $x=\frac{11}{15}x=\backslash frac\{11\}\{15\}$ into $\mathrm{I}\backslash mathrm\{I\}$ , in order to get $yy$.

Solve for $yy$.

$y=-\frac{19}{30}y\; =\; -\backslash frac\{19\}\{30\}$

So there is exactly one solution.