4.3.5 Exercises

1. For example, the first equation ($2x-y+5z=1$) is solved for $y$ resulting in $y=2x+5z-1$, which is then substituted into the third equation ($-4x+y-3z=-1$):
   
   ${\displaystyle -4x+(2x+5z-1)-3z=-1\iff -2x+2z=0\iff z=x\hspace{0.5em}.}$
   
   The last result, which is already solved for the variable $z$, is substituted into the second equation ($11x+8z=2$) that does not depend on $y$:
   
   ${\displaystyle 11x+8x=2\iff 19x=2\iff x=\frac{2}{19}\hspace{0.5em}.}$
   
   Then also
   
   ${\displaystyle z=\frac{2}{19}\hspace{0.5em},}$
   
   and for $y$ one obtains from the first equation:
   
   ${\displaystyle y=2·\frac{2}{19}+5·\frac{2}{19}-1=\frac{4+10-19}{19}=\frac{-5}{19}\hspace{0.5em}.}$
   
   Hence, the system of linear equations has a unique solution and the solution set is $L=\{(x=\frac{2}{19};y=-\frac{5}{19};z=\frac{2}{19})\}$.
   However, other approaches are equally possible.
   
2. By adding the first and the third equation the variable $y$ is eliminated:
   
   ${\displaystyle (2x-y+5z)+(-4x+y-3z)=1+(-1)\iff -2x+2z=0\hspace{0.5em}.}$
   
   Multiplying the last equation by $(-4)$ results in
   
   ${\displaystyle 8x-8z=0\hspace{0.5em},}$
   
   and adding the second equation ($11x+8z=2$) to this result the variable $z$ is eliminated:
   
   ${\displaystyle (8x-8z)+(11x+8z)=0+2\iff 19x=2\iff x=\frac{2}{19}\hspace{0.5em}.}$
   
   As in the first part of this exercise, the variables $z$ and $y$ can be determined subsequently; of course, the solution set is again $L=\{(x=\frac{2}{19};y=-\frac{5}{19};z=\frac{2}{19})\}$.
   However, other approaches are equally possible.
   

For example, equation ($1$) is solved for $I_1$

${\displaystyle I_1=I_2+I_3\hspace{0.5em},}$

which is substituted into equation ($2$):

${\displaystyle R_1(I_2+I_3)+R_2{I}_2=U\iff (R_1+R_2)I_2+R_1{I}_3=U\mathrm{\hspace{0.5em}\hspace{0.5em}}:\hspace{0.5em}\text{equation}\hspace{0.5em}(2\text{'})\hspace{0.5em}.}$

This last equation ($2\text{'}$) and equation ($3$) only depend on the variables $I_2$ and $I_3$. They form a system of two linear equations in two variables, which is now solved: e.g. equation ($3$) can be solved for $I_3$:

${\displaystyle I_3=\frac{R_2}{R_3}{I}_2\mathrm{\hspace{0.5em}\hspace{0.5em}}:\hspace{0.5em}\text{equation}\hspace{0.5em}(3\text{'})\hspace{0.5em},}$

which is then substituted into equation $(2\text{'})$:

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{}& \hfill & (R_1+R_2)I_2+R_1·\frac{R_2}{R_3}{I}_2=U\hfill \\ \multicolumn{1}{c}{}& \iff \hfill & (R_1+R_2+R_1·\frac{R_2}{R_3})I_2=U\hfill \\ \multicolumn{1}{c}{}& \iff \hfill & (\frac{R_1{R}_3}{R_3}+\frac{R_2{R}_3}{R_3}+\frac{R_1{R}_2}{R_3})I_2=U\hfill \\ \multicolumn{1}{c}{}& \iff \hfill & \frac{R_1{R}_2+R_1{R}_3+R_2{R}_3}{R_3}{I}_2=U\hfill \\ \multicolumn{1}{c}{}& \iff \hfill & I_2=\frac{R_3}{R_1{R}_2+R_1{R}_3+R_2{R}_3}U\hspace{0.5em}.\hfill \end{array}}$
Hence, for $I_3$ it follows, using equation $(3\text{'})$,

${\displaystyle I_3=\frac{R_2}{R_3}·\frac{R_3}{R_1{R}_2+R_1{R}_3+R_2{R}_3}U=\frac{R_2}{R_1{R}_2+R_1{R}_3+R_2{R}_3}U}$

and for $I_1$

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{I_1=I_2+I_3}& =\hfill & \frac{R_3}{R_1{R}_2+R_1{R}_3+R_2{R}_3}·U+\frac{R_2}{R_1{R}_2+R_1{R}_3+R_2{R}_3}·U\hfill \\ \multicolumn{1}{c}{}& =\hfill & \frac{R_2+R_3}{R_1{R}_2+R_1{R}_3+R_2{R}_3}·U\hspace{0.5em}.\hfill \end{array}}$
Now, the values of the resistances of the resistors ($R_1=1\hspace{0.17em}\mathrm{}\mathrm{\Omega }$, $R_2=2\hspace{0.17em}\mathrm{}\mathrm{\Omega }$, and $R_3=3\hspace{0.17em}\mathrm{}\mathrm{\Omega }$) and the voltage ($U=5.5\hspace{0.17em}\mathrm{V}$) can be inserted. Using $1\hspace{0.17em}\mathrm{V}=1\hspace{0.17em}\mathrm{}\mathrm{\Omega }·1\hspace{0.17em}\mathrm{A}$ this results for $I_1$ in:

${\displaystyle I_1=\frac{(2+3)\hspace{0.17em}\mathrm{}\mathrm{\Omega }}{(1·2+1·3+2·3)\hspace{0.17em}\mathrm{}\mathrm{\Omega }{}^2}·5.5·\mathrm{\Omega }·\mathrm{A}=2.5\hspace{0.17em}\mathrm{A}\hspace{0.5em}.}$

Likewise, for $I_2$ and $I_3$ this results in: $I_2=1.5\hspace{0.17em}\mathrm{A}$ and $I_3=1\hspace{0.17em}\mathrm{A}.$
However, other approaches are equally possible.

Since the first equation ($x+2z=5$) does not depend on the variable $y$ at all, it is reasonable to eliminate $y$ from the second and third equation as well. For this, the second equation ($3x+y-2z=-1$) is multiplied by $2$ and added to the third equation ($-x-2y+4z=7$):

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{}& \hfill & (-x-2y+4z)+2·(3x+y-2z)=7+2·(-1)\hfill \\ \multicolumn{1}{c}{}& \iff \hfill & -x-2y+4z+6x+2y-4z=7-2\hfill \\ \multicolumn{1}{c}{}& \iff \hfill & 5x=5\hfill \\ \multicolumn{1}{c}{}& \iff \hfill & x=1\hspace{0.5em}.\hfill \end{array}}$
At the same time the variable $z$ is also eliminated. Inserting the value for $x$ in the first equations results in

${\displaystyle 1+2z=5\iff 2z=4\iff z=2\hspace{0.5em}.}$

Using the second equation the value of $y$ is

${\displaystyle 3·1+y-2·2=-1\iff 3+y-4=-1\iff y=0\hspace{0.5em}.}$

Hence, the solution set is $L=\{(x=1;y=0;z=2)\}$.