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Addition method

The addition method is a method for solving systems of equations.

To solve a system of equations using the addition method, the equations or their multiples are added or subtracted until there is only one variable in each equation.

Procedure with examples

System of equations with two variables and two equations

Notation and Explanation

II−II−4a+3b=6II3a−6b=3\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 bb is chosen and calculated:

lcm⁡(3;6)=6\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 22.

2⋅I→I′−8a+6b=12II−I→′II3a−6b=3\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 I′\mathrm{I}' and II\mathrm{II} are added.

I′−8a+6b=12II+I′→II′3a−6b−8a+6b=3+12\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:

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

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

I′−8a+6b=12II−I→II′a=−3\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 b\color{red}{b} is still missing. Since it is known that a=−3a=-3, the value is put into I′\mathrm{I}' and the equation is solved for b\color{red}{b}.

  a in I′→I′′−8⋅(−3)+6b=12\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}

II−I′→I′′24+6b=12∣−24I′′6b=−12∣:6I′′b=−2\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 aa and bb are a=−3,  b=−2a=-3,\;b=-2. The values are written into a tuple. The solution set is:

L={(−3;−2)}L=\{(-3;-2)\}

System of equations with three variables and three equations

Notation and Explanation

I2a+3b−c=11IIa−b+2c=32⋅IIII′→III3a−2b+3c=8\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 aa is chosen and the lcm is calculated:

lcm(1;2;3)=6\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 33, the second by 66, the third by 22.

3⋅I→I′6a+9b−3c=336⋅II→II′6a−6b+12c=182⋅III→III′6a−4b+6c=16\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 I′\mathrm{I}' is subtracted from II′\mathrm{II}' and from III′\mathrm{III}'.

I′6a+9b−3c=33II′−I′→II′′6a−6a−6b−9b+12c−(−3c)=18−33III′−I′→III′′6a−6a−4b−9b+6c−(−3c)=16−33\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}%%

I′6a+9b−3c=33II′−I′→II′′6a−6a−6b−9b+12c−(−3c)=18−33III′−I′→III′′6a−6a−4b−9b+6c−(−3c)=16−33\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 II′′\mathrm{II}'' and III′′\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=1b=2;\;c=1.

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

b and c in I′→I′′6a+9⋅2−3⋅1=336a+15=33∣−156a=18∣:6a=3\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 aa, bb and cc are a=3,  b=2;c=1a=3,\;b=2; c=1. They are written into a tuple. The solution set is:

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


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