Addition method

Procedure with examples

System of equations with two variables and two equations

Notation and Explanation

$\begin{array}{rccccc}\phantom{II-I}\mathrm{I}& -4a& +& 3b& =& 6\\ \mathrm{II}& 3a& -& 6b& =& 3\end{array}$

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

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

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 $2$.

$\begin{array}{rccccc}2\cdot \mathrm{I}\to {\mathrm{I}}^{\prime }& -8a& +& 6b& =& 12\\ \phantom{II-I{\to }^{\prime }}\mathrm{II}& 3a& -& 6b& =& 3\end{array}$

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

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

$\begin{array}{rrrl}{\mathrm{I}}^{\prime }& -8a& +6b& =12\\ \mathrm{I}\mathrm{I}+{\mathrm{I}}^{\prime }\to \mathrm{I}{\mathrm{I}}^{\prime }& 3a-6b-8a+6b& & =3+12\end{array}$

Summing up yields:

$\begin{array}{rccccc}{\mathrm{I}}^{\prime }& -8a& +& 6b& =& 12\\ \phantom{II-I\to }\mathrm{I}{\mathrm{I}}^{\prime }& -5a& & & =& 15\end{array}$

Now $a$ is the only variable in the second equation. It is solved for $a$.

$\begin{array}{rccccc}{\mathrm{I}}^{\prime }& -8a& +& 6b& =& 12\\ \phantom{II-I\to }\mathrm{I}{\mathrm{I}}^{\prime }& a& & & =& -3\end{array}$

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

$\begin{array}{rccc}a\text{ in }{\mathrm{I}}^{\prime }\to {\mathrm{I}}^{\prime \prime }& -8\cdot (-3)& +6b& =12\end{array}$

$\begin{array}{rrcll}\phantom{II-I^{\prime }\to }{\mathrm{I}}^{\prime \prime }& 24+6b& =& 12& |-24\\ {\mathrm{I}}^{\prime \prime }& 6b& =& -12& |:6\\ {\mathrm{I}}^{\prime \prime }& b& =& -2& \end{array}$

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

$\begin{array}{rccccccc}\mathrm{I}& 2a& +& 3b& -& c& =& 11\\ \mathrm{II}& a& -& b& +& 2c& =& 3\\ \phantom{2\cdot III{I}^{\prime }\to }\mathrm{III}& 3a& -& 2b& +& 3c& =& 8\end{array}$

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

$\mathrm{l}\mathrm{c}\mathrm{m}(1;2;3)=6$

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 $2$.

$\begin{array}{rccccccc}3\cdot \mathrm{I}\to {\mathrm{I}}^{\prime }& 6a& +& 9b& -& 3c& =& 33\\ 6\cdot \mathrm{I}\mathrm{I}\to \mathrm{I}{\mathrm{I}}^{\prime }& 6a& -& 6b& +& 12c& =& 18\\ 2\cdot \mathrm{I}\mathrm{I}\mathrm{I}\to \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime }& 6a& -& 4b& +& 6c& =& 16\end{array}$

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

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

$\begin{array}{rccccccc}{\mathrm{I}}^{\prime }& 6a& +& 9b& -& 3c& =& 33\\ \mathrm{I}{\mathrm{I}}^{\prime }-{\mathrm{I}}^{\prime }\to \mathrm{I}{\mathrm{I}}^{\prime \prime }& 6a-6a& -& 6b-9b& +& 12c-(-3c)& =& 18-33\\ \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime }-{\mathrm{I}}^{\prime }\to \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime \prime }& 6a-6a& -& 4b-9b& +& 6c-(-3c)& =& 16-33\end{array}$

$\begin{array}{rccccccc}{\mathrm{I}}^{\prime }& 6a& +& 9b& -& 3c& =& 33\\ \mathrm{I}{\mathrm{I}}^{\prime }-{\mathrm{I}}^{\prime }\to \mathrm{I}{\mathrm{I}}^{\prime \prime }& 6a-6a& -& 6b-9b& +& 12c-(-3c)& =& 18-33\\ \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime }-{\mathrm{I}}^{\prime }\to \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime \prime }& 6a-6a& -& 4b-9b& +& 6c-(-3c)& =& 16-33\end{array}$

Now ${\mathrm{II}}^{\prime \prime }$ and ${\mathrm{III}}^{\prime \prime }$ 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 ${\mathrm{I}}^{\prime }$ and solved for $a$.

$\begin{array}{rrcl}b\text{ and }c\text{ in }{\mathrm{I}}^{\prime }\to {\mathrm{I}}^{\prime \prime }& 6a+9\cdot 2-3\cdot 1& =& 33\\ & 6a+15& =& 33& |-15\\ & 6a& =& 18& |:6\\ & a& =& 3\end{array}$

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

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