Substitution method

Procedure

In order to solve a system of equations with the substitution method, one should select the equation for which it is easiest to solve for an unknown. This is best possible if one unknown stands alone, i.e. with coefficient 1.

After this, the unknown is solved and then inserted or replaced in the remaining equations.

Example

$\begin{array}{cccccccc}\mathrm{I}& a& +& \frac{1}{2}b& -& 10c& =& 5\\ \mathrm{II}& 2a& -& b& & & =& 6\\ \mathrm{III}& -2a& +& 4b& -& 5c& =& -12\end{array}$

Here the $b$ in the equation $\mathrm{II}$ or also the $a$ in equation $\mathrm{II}$ are suitable. In this example, the variable b was selected:

$\begin{array}{rrcllc}\mathrm{II}& 2a-b& =& 6& & |+b\\ & 2a& =& 6+b& & |-6\\ & 2a-6& =& b& & & \end{array}$

Now you can substitute $b=2a-6$ for example in the equation $\mathrm{I}$ and then simplify as much as possible:

$\begin{array}{rrcccc}{\mathrm{I}}^{\prime }& a+\frac{1}{2}(2a-6)-10c& =& 5& & \\ & a+a-3-10c& =& 5& & \\ & 2a-3-10c& =& 5& & |+3\\ & 2a-10c& =& 8& & \end{array}$

Now you can solve this equation for $c$ and then substitute both $b$ and $c$ in the equation $\mathrm{III}$ to find out the unknown $a$.

$\begin{array}{rrcll}{\mathrm{I}}^{\prime }& 2a-10c& =& 8& |+10c\\ & 2a& =& 8+10c& |-8\\ & 2a-8& =& 10c& |:10\\ & \frac{1}{5}a-\frac{4}{5}& =& c& \end{array}$

So now we know:

• $b=2a-6$

• $c=\frac{1}{5}a-\frac{4}{5}$

Putting this into $\mathrm{III}$ gives:

$\begin{array}{cccc}\mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime }& -2a+4(2a-6)-5(\frac{1}{5}a-\frac{4}{5})& =& -12\\ & -2a+8a-24-a+4& =& -12\\ & 5a-20& =& -12& |+20\\ & 5a& =& 8& |:5\\ & a& =& \frac{8}{5}\end{array}$

Now that you have calculated a, you can calculate the other quantities:

• $\begin{array}{l}1a=\frac{8}{5}=\mathrm{1,6}\end{array}$

• $\begin{array}{l}b=2\cdot \left(\frac{8}{5}\right)-6=\frac{16}{5}-6=-2\frac{4}{5}=-\mathrm{2,8}\end{array}$

• $c=\frac{1}{5}\cdot \left(\frac{8}{5}\right)-\frac{4}{5}=\frac{8}{25}-\frac{4}{5}=-\frac{12}{25}=-\mathrm{0,48}$