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Substitution method

The substitution method is a procedure for solving systems of linear equations. If one of the equations is solved for a variable xx, the term on the other side is substituted for xx in all other equations. This reduces both the number of variables and the number of equations by one.  

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

Ia+12b10c=5II2ab    =6III2a+4b5c=12\def\arraystretch{1.25} \begin{array}{cccccccc}\mathrm{I}&a&+&\frac12b&-&10c&=&5\\\mathrm{II}&2a&-&b&\;&\;&=&6\\\mathrm{III}&-2a&+&4b&-&5c&=&-12\end{array}

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

II2ab=6+b2a=6+b62a6=b\def\arraystretch{1.25} \begin{array}{rrcllcccccc} \mathrm{II}&2a-b&=&6&&|+b\\ &2a&=&6+b&&\vert-6\\ &2a-6&=&b&&& \end{array}

Now you can substitute b=2a6b=2a-6 for example in the equation I\mathrm{I} and then simplify as much as possible:

Ia+12(2a6)10c=5a+a310c=52a310c=5+32a10c=8\def\arraystretch{1.25} \begin{array}{rrcccccccc}\mathrm{I'}&a+\frac12\left(2a-6\right)-10c&=&5&&\\&a+a-3-10c&=&5&&\\&2a-3-10c&=&5&&\vert+3\\&2a-10c&=&8&&\end{array}

Now you can solve this equation for cc and then substitute both bb and cc in the equation III\mathrm{III} to find out the unknown aa.

I2a10c=8+10c2a=8+10c82a8=10c:1015a45=c\def\arraystretch{1.25} \begin{array}{rrcllccccccc}\mathrm{I'}&2a-10c&=&8&\vert+10c\\&2a&=&8+10c&\vert-8\\&2a-8&=&10c&\vert:10\\&\frac15a-\frac45&=&c\end{array}

So now we know:

  • b=2a6b=2a-6

  • c=15a45c=\frac15a-\frac45

Putting this into III\mathrm{III} gives:

III2a+4(2a6)5(15a45)=122a+8a24a+4=125a20=12+205a=8:5a=85\def\arraystretch{1.25} \begin{array}{ccccccccc}\mathrm{III'}&-2a+4\left(2a-6\right)-5\left(\frac15a-\frac45\right)&=&-12\\&-2a+8a-24-a+4&=&-12\\&5a-20&=&-12&\vert+20\\&5a&=&8&\vert:5\\&a&=&\frac85\end{array}

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

  • 1a=85=1,6\def\arraystretch{1.25} \begin{array}{l}{1}a=\frac85=1{,}6\end{array}

  • b=2(85)6=1656=245=2,8\def\arraystretch{1.25} \begin{array}{l}b=2\cdot\left(\frac85\right)-6=\frac{16}5-6=-2\frac45=-2{,}8\end{array}

  • c=15(85)45=82545=1225=0,48c=\frac15\cdot\left(\frac85\right)-\frac45=\frac8{25}-\frac45=-\frac{12}{25}=-0{,}48


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