Exercises: systems with two variables

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:

$\begin{array}{rrllll}\mathrm{I}& \frac{7}{x}& -& \frac{12}{y}& =& \frac{5}{6}\\ \mathrm{II}& x& -& y& =& 2\end{array}$

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

$\begin{array}{rrlllll}\mathrm{I}& -3& =& x& +& 2y& \\ \mathrm{II}& 1& =& x^2& +& y& \end{array}$

Is not a linear system, since $x$ has power $2$ in the second line.

All other systems are indeed linear, since $x$ and $y$ only appear with power $1$.

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

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

$\begin{array}{rlrl}\mathrm{I}-\mathrm{II}\to & -4& =& 0\end{array}$

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

Thus the solution set is $𝕃=\{\}$.

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

One way to show this is by the substitution method.

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

$\begin{array}{rrlllll}\mathrm{II}& 3x& =& 2& +& y& |-2\end{array}$

$\begin{array}{lrllll}{\mathrm{II}}^{\prime }& 3x& -& 2& =& y\end{array}$

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

$\begin{array}{rrllll}{\mathrm{I}}^{\prime }& \frac{9}{2}x& -& \frac{3}{2}(3x-2)& =& 3\end{array}$

Summarize the left side of the equation.

$\begin{array}{rrll}{\mathrm{I}}^{\prime }& 3& =& 3\end{array}$

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

$𝕃=\{(x|y)|y=3x-2\}$

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

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

First solve equation $\mathrm{I}$ for $y$.

$\begin{array}{rrlllll}\mathrm{I}& 2x& +& y& =& \frac{5}{6}& |-2x\end{array}$

$\begin{array}{rrllll}{\mathrm{I}}^{\prime }& y& =& \frac{5}{6}& -& 2x\end{array}$

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

$\begin{array}{rrllll}\mathrm{I}{\mathrm{I}}^{\prime }& x& -& 2(\frac{5}{6}-2x)& =& 2\end{array}$

Summarize the left side.

$\begin{array}{rrllll}\mathrm{I}{\mathrm{I}}^{\prime }& 5x& -& \frac{5}{3}& =& 2\end{array}$

Solve for $x$.

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

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

$\begin{array}{rrllll}\mathrm{I}& 2\cdot (\frac{11}{15})& +& y& =& \frac{5}{6}\end{array}$

Solve for $y$.

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

So there is exactly one solution.