#### Chapter 4 System of Linear Equations

Section 4.4 More general Systems

# 4.4.3 Exercises

##### Exercise 4.4.3
Find the $y$-intercept $b$ and the slope $m$ of a line described by the equation $y=mx+b$ which is defined by two points. The first point at ${x}_{1}=\alpha$ lies on the line described by the equation ${y}_{1}\left(x\right)=-2\left(1+x\right)$. The second point at ${x}_{2}=\beta$ lies on the line described by the equation ${y}_{2}\left(x\right)=x-1$. The following figure illustrates the situation.
1. Find the system of equations for the parameters $b$ and $m$.
The first equation reads $m\alpha +b$$=$
;
the second equation reads $m\beta +b$$=$
.
The constants $\alpha$ and $\beta$ have to remain in the solution; for these alpha and beta can be entered.
2. Solve this system of equations for $b$ and $m$. For which values of $\alpha$ and $\beta$ does the system have a unique solution, no solution, or an infinite number of solutions?
For $\alpha =-2$ and $\beta =2$ one obtains, for example, the solution $m$$=$
and $b$$=$
, the case $\alpha =2$ and $\beta =-2$ results in the solution $m$$=$
and $b$$=$ .
The LS has an infinite number of solutions if $\alpha$$=$
and $\beta$$=$ .
The corresponding solutions can be parameterised by $m=r$ and $b$$=$
, $r\in ℝ$.
3. What is the graphical interpretation of the last two cases, i.e. no solution and an infinite number of solutions?

##### Exercise 4.4.4
Find the solution set of the following LS depending on the parameter $t\in ℝ$.

$\begin{array}{cccc}\hfill x-y+tz& \hfill =\hfill & t\hfill & \text{equation} \left(1\right) ,\hfill \\ \hfill tx+\left(1-t\right)y+\left(1+{t}^{2}\right)z& \hfill =\hfill & -1+t\hfill & \text{equation} \left(2\right) ,\hfill \\ \hfill \left(1-t\right)x+\left(-2+t\right)y+\left(-1+t-{t}^{2}\right)z& \hfill =\hfill & {t}^{2}\hfill & \text{equation} \left(3\right) .\hfill \end{array}$

The LS only has solutions for the following values of the parameter: $t\in \text{}$
.
Set can be entered in the form $\left\{$a;b;c;$\dots \right\}$. The empty set can be entered as $\left\{\right\}$.
For the smallest value of the parameter $t$ the solution is
$x$$=$
, $y$$=$
, $z=r$, $r\in ℝ$.
For the greatest value of the parameter $t$ the solution is
$x$$=$
, $y$$=$
, $z=r$, $r\in ℝ$.