Elimination by equating coefficients

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

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

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

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

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

Example

The procedure will now be demonstrated with the following system of equations with $2$ equations and $2$ variables:

$\displaystyle \displaystyle \sf {I} \quad a + \frac12b \\$	$=$	$\displaystyle 5$
	↓	In school, the variables in equations or systems of equations are often denoted by $\sf x$ , $\sf y$ , $\sf z$ and so on. However, the variables can of course also be designated with the other letters of the alphabet (here: $\sf a,b$ ).
$\displaystyle \displaystyle \sf {II} \quad 2a - b \\$	$=$	$\displaystyle 6$

$\textbf{Solve both equations for one variable}$

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

$\displaystyle \displaystyle \sf {I} \quad a+\frac12b$	$=$	$\displaystyle \displaystyle \sf 5$	$\displaystyle \displaystyle \sf -\frac12b$
$\displaystyle \displaystyle \sf {I'} \quad \textcolor{green} a$	$=$	$\displaystyle \displaystyle \sf \textcolor{green}{5-\frac12b}$
$\displaystyle \displaystyle \sf {II} \quad 2a-b$	$=$	$\displaystyle \displaystyle \sf 6$	$\displaystyle \displaystyle \sf +b$
$\displaystyle \displaystyle \sf 2a$	$=$	$\displaystyle \displaystyle \sf 6+b$	$\displaystyle \displaystyle \sf :2$
$\displaystyle \displaystyle \sf {II'} \quad \textcolor{green}a \\$	$=$	$\displaystyle \displaystyle \sf \quad \textcolor{green}{3+\frac12b} \\$

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

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

$\textbf{Equating \sf I'and \sf II'}$

$\sf 5-\dfrac12b = 3+\dfrac12b$

$\textbf{Solve equation}$

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

$\displaystyle \displaystyle \sf 5-\dfrac12b$	$=$	$\displaystyle \displaystyle \sf 3+\dfrac12b$	$\displaystyle \sf +\dfrac12b$
$\displaystyle \displaystyle \sf 5$	$=$	$\displaystyle \sf 3+b$	$\displaystyle \sf -3$
$\displaystyle \displaystyle \sf 2$	$=$	$\displaystyle \sf b$

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

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

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 $\sf {II'}$

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

This yields the solution set:

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

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

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

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

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