Addition method

Procedure with examples

System of equations with two variables and two equations

Notation and Explanation

$\def\arraystretch{1.25} \begin{array}{r|ccccc}\hphantom{II-I}\mathrm{I}&-4a&+&3b&=&6\\\mathrm{II}&3a&-&6b&=&3\end{array}$

A variable is selected and the least common multiple (lcm) of the coefficients is determined. Here $b$ is chosen and calculated:

$\mathrm{\operatorname{lcm}(3;6)}=6$

Now the equations must be multiplied so that the coefficients of the variable in both equations are equal to the lcm.

In this case: the first equation is multiplied by $2$.

$\def\arraystretch{1.25} \begin{array}{r|ccccc}\mathrm{2\cdot I\to I'}&-8a&+&6b&=&12\\\hphantom{II-I\to'}\mathrm{II} &3a&-&6b&=&3\end{array}$

Now the equations are added (or subtracted) from each other to eliminate the selected variable.

In this case: Equations $\mathrm{I}'$ and $\mathrm{II}$ are added.

$\def\arraystretch{1.25} \begin{array}{r|rrll}\mathrm{I'}&-8a&+6b&=12\\\mathrm{II+I'\to II'} &3a\color{blue}{-6b}-8a\color{blue}{+6b}&&=3+12\end{array}$

Summing up yields:

$\def\arraystretch{1.25} \begin{array}{r|ccccc}\mathrm{I'}&-8a&+&6b&=&12\\\hphantom{II-I\to}\mathrm{II'} &-5a&&&=&15\end{array}$

Now $\color{green}{a}$ is the only variable in the second equation. It is solved for $\color{green}{a}$.

$\def\arraystretch{1.25} \begin{array}{r|ccccc}\mathrm{I'}&-8a&+&6b&=&12\\\hphantom{II-I\to}\mathrm{II'} &\color{green}{a}&&&\color{green}{=}&\color{green}{-3}\end{array}$

The value for $\color{red}{b}$ is still missing. Since it is known that $a=-3$, the value is put into $\mathrm{I}'$ and the equation is solved for $\color{red}{b}$.

$\def\arraystretch{1.25} \begin{array}{r|ccccc}\;a \mathrm{\text{ in }I'\to I''}&-8\cdot \color{green}{(-3)}&+6\color{red}{b}&=12\end{array}$

$\def\arraystretch{1.25} \begin{array}{r|rcl}\hphantom{II-I'\to}\mathrm{I}''&24+6b&=&12&|-24\\\mathrm{I}''&6b&=&-12&|:6\\\mathrm{I}''&\color{red}{b}&\color{red}{=}&\color{red}{-2}\end{array}$

The values of $a$ and $b$ are $a=-3,\;b=-2$. The values are written into a tuple. The solution set is:

$L=\{(-3;-2)\}$

System of equations with three variables and three equations

Notation and Explanation

$\def\arraystretch{1.25} \begin{array}{r|ccccc}\mathrm{I}&2 a&+&3b&-&c&=&11\\\mathrm{II}&a&-&b&+&2c&=&3\\\hphantom{2\cdot IIII'\to}\mathrm{III}&3a&-&2b&+&3c&=&8\end{array}$

As above, one selects a variable and finds the least common multiple of the coefficients. Here $a$ is chosen and the lcm is calculated:

$\mathrm{lcm(1;2;3)}=6$

Now the equations are multiplied so that the coefficients of the variable in all equations are equal to [lcm]().

In this case: the first equation is multiplied by $3$, the second by $6$, the third by $2$.

$\def\arraystretch{1.25} \begin{array}{r|ccccc}\mathrm{3\cdot I\to I'}&6a&+&9b&-&3c&=&33\\\mathrm{6\cdot II\to II'}&6a&-&6b&+&12c&=&18\\\quad\mathrm{2\cdot III\to III'}&6a&-&4b&+&6c&=&16\end{array}$

Now the equations are added (or subtracted) from each other to eliminate the chosen variable.

In this case: Equation $\mathrm{I}'$ is subtracted from $\mathrm{II}'$ and from $\mathrm{III}'$.

$\def\arraystretch{1.25} \begin{array}{r|ccccc}\mathrm{I'}&6a&+&9b&-&3c&=&33\\\mathrm{II'-I'\to II''}&6a-6a&-&6b-9b&+&12c-(-3c)&=&18-33\\\mathrm{III'-I'\to III''}&6a-6a&-&4b-9b&+&6c-(-3c)&=&16-33\end{array}$

$\def\arraystretch{1.25} \begin{array}{r|ccccc}\mathrm{I'}&6a&+&9b&-&3c&=&33\\\mathrm{II'-I'\to II''}&6a-6a&-&6b-9b&+&12c-(-3c)&=&18-33\\\mathrm{III'-I'\to III''}&6a-6a&-&4b-9b&+&6c-(-3c)&=&16-33\end{array}$

Now $\mathrm{II}''$ and $\mathrm{III}''$ form a system of equations with two unknowns. This new system of equations is solved as described above and it yields:

$b=2;\;c=1$.

These two values are then substituted into $\mathrm{I}'$ and solved for $\color{green}{a}$.

$\def\arraystretch{1.25} \begin{array}{r|rcl}b\text{ and }c\text{ in }\mathrm{I'\to I''}&6\color{green}{a}+9\cdot2-3\cdot1&=&33\\&6a+15&=&33&|-15\\&6a&=&18&|:6\\&\color{green}{a}&\color{green}{=}&\color{green}{3}\end{array}$

The values of $a$, $b$ and $c$ are $a=3,\;b=2; c=1$. They are written into a tuple. The solution set is:

$L=\{(3;2;1)\}$