Installation procedure

The substitution procedure is used to solve systems of equations . The idea behind this procedure is to solve one of the equations for a variable and then plug that variable into the other equations. This eliminates one variable.

This procedure can also be used for larger or non-linear systems of equations, but it then quickly becomes confusing. However, if you proceed according to the following algorithm, you can keep an overview even with large systems of equations:

There are n equations with n variables.

• Step 1: Solve the first equation (any equation) to the last variable.
• Step 2: Substituting this equation into all of the other equations.

A system of equations with n - 1 variables is created. Steps 1 and 2 are repeated until there is only one equation with one variable left.

Now insert all the variables from below.

Note: As confusing expressions arise when inserting them, it is advisable to make simplifications in between. If you can group constants together, this is what you should do. Fractions with constants should be combined into a new constant if necessary: ​​For example ( a + b + c ) / ( e + f ) = h , where a , b , c , e , f , h are all constant.

Example with two variables

For a system of equations with two equations and two variables , proceed as follows:

• Step 1: Solving an equation for a variable
• Step 2: Substituting these variables into the other equation
• Step 3: Solve the equation obtained in step 2 for the contained variable
• Step 4: Substituting the solution into the equation transformed after step 1

Numerical example

The following system of equations is given:

${\displaystyle {\begin{matrix}({\text{I}})&12x&-&5y&=&29\\({\text{II}})&18x&+&2y&=&34\end{matrix}}}$

Step 1 :

One of the two equations must be solved for or . In this example the 2nd equation is solved for . ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle y}$

${\displaystyle {\begin{matrix}({\text{II}})&18x&+&2y&=&34&|&-18x\\({\text{II}})&&&2y&=&34-18x&|&:2\\({\text{II}})&&&y&=&17-9x\end{matrix}}}$

Step 2 :

Then you can replace this in the first equation with the term : ${\displaystyle y}$ ${\displaystyle (17-9x)}$

${\displaystyle {\begin{matrix}({\text{II in I}})&12x&-&5\cdot (17-9x)&=&29\end{matrix}}}$

Step 3 :

This equation can now be solved for. ${\displaystyle x}$

${\displaystyle {\begin{matrix}12x-5\cdot (17-9x)&=&29&|&{\text{Klammer aufloesen}}\\12x-85+45x&=&29&|&{\text{zusammenfassen}}\\57x-85&=&29&|&+85\\57x&=&114&|&:57\\x&=&2\end{matrix}}}$

Step 4 :

The solution is inserted into the rearranged equation (II): ${\displaystyle x=2}$

${\displaystyle {\begin{matrix}x=2{\text{ in (II) einsetzen:}}&y&=&17-9\cdot 2\\&y&=&-1\end{matrix}}}$

And checked:

${\displaystyle {\begin{matrix}{\text{Probe:}}&({\text{I}})&12\cdot 2-5\cdot (-1)&=&29\\&&24-(-5)&=&29&{\text{wahr}}\\&({\text{II}})&18\cdot 2+2\cdot (-1)&=&34\\&&36+(-2)&=&34&{\text{wahr}}\end{matrix}}}$

The solution set is thus: . ${\displaystyle \mathbb {L} =\{(2|{-1})\}}$

See also

• Equation procedure
• Addition method