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Elimination by equating coefficients

The elimination by equating coefficients\textbf{elimination by equating coefficients} is a method to solve linear systems of equations.

This method is useful when the same term\textbf{the same term} appears in (at least) two different equations\textbf{(at least) two different equations} in a system of equations.

For example, in the following system of equations, the variable xx is multiplied by the factor 3 in each equation:

I3x+2y\displaystyle \displaystyle \sf {I} \quad \textcolor{red}{3x} + 2y ==8\displaystyle 8
II3xy\displaystyle \displaystyle \sf {II} \quad \textcolor{red}{3x} - y ==9\displaystyle 9

One can then solve for xx and equate the other side in each case. Thus, one eliminates one variable (here: xx) and can determine the other variable (here: yy) by inserting it into one of the equations.

Example

The procedure will now be demonstrated with the following system of equations with 22 equations and 22 variables:

Ia+12b\displaystyle \displaystyle \sf {I} \quad a + \frac12b \\ ==5\displaystyle 5

In school, the variables in equations or systems of equations are often denoted by x\sf x, y\sf y, z\sf z and so on. However, the variables can of course also be designated with the other letters of the alphabet (here: a,b\sf a,b).

II2ab\displaystyle \displaystyle \sf {II} \quad 2a - b \\ ==6\displaystyle 6
  • Solve both equations for one variable\textbf{Solve both equations for one variable}

First, both sides are solved for one variable. In this case, for example, it will be solved for aa .

Ia+12b\displaystyle \displaystyle \sf {I} \quad a+\frac12b ==5\displaystyle \displaystyle \sf 512b\displaystyle \displaystyle \sf -\frac12b
Ia\displaystyle \displaystyle \sf {I'} \quad \textcolor{green} a ==512b\displaystyle \displaystyle \sf \textcolor{green}{5-\frac12b}
II2ab\displaystyle \displaystyle \sf {II} \quad 2a-b ==6\displaystyle \displaystyle \sf 6+b\displaystyle \displaystyle \sf +b
2a\displaystyle \displaystyle \sf 2a ==6+b\displaystyle \displaystyle \sf 6+b:2\displaystyle \displaystyle \sf :2
IIa\displaystyle \displaystyle \sf {II'} \quad \textcolor{green}a \\==3+12b\displaystyle \displaystyle \sf \quad \textcolor{green}{3+\frac12b} \\

Since now at I\sf {I'} and II\sf {II'} the left-hand sides are both equal, the right-hand sides must also be equal, therefore 512b=3+12b\sf 5-\dfrac12b = 3+\dfrac12b.

This step is called "equating"\textbf{"equating"}.

  • Equating I’and II’\textbf{Equating \sf I'and \sf II'}

512b=3+12b\sf 5-\dfrac12b = 3+\dfrac12b

  • Solve equation\textbf{Solve equation}

This new equation has only one variable\textbf{only one variable} and can therefore be solved as usual.

512b\displaystyle \displaystyle \sf 5-\dfrac12b ==3+12b\displaystyle \displaystyle \sf 3+\dfrac12b+12b\displaystyle \sf +\dfrac12b
5\displaystyle \displaystyle \sf 5==3+b\displaystyle \sf 3+b3\displaystyle \sf -3
2\displaystyle \displaystyle \sf 2==b\displaystyle \sf b

This solution can now be substituted into one of the upper equations to calculate the value of the second variable.

  • Substitution into equation I’ or II’ \textbf{Substitution into equation \sf {I'} or \sf {II'} }

It does not matter which equation you use! To check the result, you can also insert it into both equations and check if the same value comes out.

Substitution of b into II\sf {II'}

a=3+122=3+1=4\sf a = 3+ \dfrac12\cdot2=3+1=4

This yields the solution set:

L={(4;2)}\sf L = \{(4;2)\}

  • Proof (can also be ommited) \textbf{Proof (can also be ommited) }

To check the solution, substitute it into the original equation and check that they are satisfied.

I4+122\displaystyle \displaystyle \sf {I} \quad 4+ \dfrac12 \cdot 2==5\displaystyle \sf 5 \quad \textcolor{green}{\checkmark}
II242\displaystyle \displaystyle \sf {II} \quad 2\cdot4-2==6\displaystyle \sf 6 \quad \textcolor{green}{\checkmark}


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