Sometimes the elimination method is easier than the substitution method
for solving
systems of equations.
The elimination method is socalled because the original system is replaced (if needed) by an equivalent
system,
where ‘addition’ of the two equations eliminates one of the variables.
Here's the very simple key concept behind this method:
Here's the same concept, arranged vertically:
if  $A$  $=$  $B$  
and  $C$  $=$  $D$  
then  $A+C$  =  $B+D$ 
This observation allows you to generate a new equation
(the ‘$\,A+C=B+D\,$’)
by adding the lefthand and righthand sides of two existing equations.
This new equation will be true, whenever the
original two equations are true.
The method is illustrated with three examples below.
To use the elimination method, you always want each equation ‘arranged’ in the same way,
so that when the equations are stacked on top of each other,
like terms end up in the same column.
For example, if a system is presented to you as
$
\begin{gather}
\cssId{s21}{3x2y5=0}\cr
\cssId{s22}{7y=2 + 3x}
\end{gather}
$
then you'll need to rearrange it
Write each equation in the same form, with terms in the same order in both equations.
In the examples below, the choice was made to put each equation in the form $\,ax + by = c\,$:
$3x$  $$  $2y$  $=$  $5$ 
$3x$  $+$  $7y$  $=$  $2$ 

Notice that there is a $\,3x\,$ in the first equation
and a $\,3x\,$ in the second equation. ‘Adding’ the two equations eliminates the variable $\,x\,$. 

$\displaystyle y =\frac{7}{5}\,$  Solve for $\,y\,$.  
$ \begin{gather} \cssId{s40}{3x2(\frac{7}{5})=5}\cr \cssId{s41}{3x=5+\frac{14}{5}}\cr \cssId{s42}{3x=\frac{39}{5}}\cr \cssId{s43}{x=\frac{39}{15}} \end{gather} $ 
Substitute this value of $\,y\,$ back into either original equation (preferably, a simple one) and solve for $\,x\,$. 

$\displaystyle x=\frac{13}{5}$  Report all numbers as fractions in simplest form.  
The unique solution is the ordered pair $\displaystyle \bigl(\frac{13}{5},\frac{7}{5}\bigr)\,$.  Clearly report the unique solution. 
Of course, it's too much to hope that a system will always be ‘readymade’ for elimination, like the example above.
A nextbest scenario is when you can replace just one equation with an equivalent one.
This happens when a variable has a coefficient (in one equation) that is a
multiple of the coefficient in the other equation,
as in the next example:
$2x$  $$  $5y$  $=$  $1$ 
$6x$  $+$  $7y$  $=$  $8$ 
$ \begin{gather} \cssId{s60}{3(2x5y)=3(1)}\cr \cssId{s61}{6x+7y=8} \end{gather} $ 
There is a $\,2x\,$ in the first equation
and a $\,6x\,$ in the second equation.
Notice that $\,6\,$ is a multiple of $\,2\,$; this is the situation you want to look for. We really want a $\,6x\,$ in the first equation, so that $\,x\,$ can be eliminated. Therefore, multiply both sides of the first equation by $\,3\,$ and replace the first equation in the system with this new one. 
$ \begin{gather} \cssId{s68}{6x+15y=3}\cr \cssId{s69}{6x+7y=8} \end{gather} $ 
Multiply out in the first equation. In this equivalent system, it's now easy to eliminate $\,x\,$. From here, everything proceeds the same as the prior example. 
$ \begin{gather} \cssId{s73}{22y=5}\cr \cssId{s74}{y=\frac{5}{22}} \end{gather} $  ‘Add’ the two equations to eliminate $\,x\,$ and then solve for $\,y\,$. 
$ \begin{gather} \cssId{s76}{2x5(\frac{5}{22})=1}\cr\cr \cssId{s77}{2x =\frac{22}{22} + \frac{25}{22}}\cr\cr \cssId{s78}{x=\frac{47}{44}} \end{gather} $  Substitute this value of $\,y\,$ back into a simple earlier equation, and solve for $\,x\,$. 
The unique solution is the ordered pair $ \displaystyle \bigl(\frac{47}{44},\frac{5}{22}\bigr)\,. $ 
Clearly report the unique solution. 
In the next example, there are no multiples in sight.
In this case, both equations are replaced by equivalent ones,
to produce an equivalent system where one of the variables is easily eliminated.
This is as hard as it will get!
$2x$  $$  $3y$  $=$  $5$ 
$5x$  $+$  $4y$  $=$  $7$ 
$ \begin{gather} \cssId{s91}{2x3y=5}\cr \cssId{s92}{5x+4y=7} \end{gather} $ 
Looking at the $\,x\,$ terms, $\,5\,$ is not a multiple of $\,2\,$. Looking at the $\,y\,$ terms, $\,4\,$ is not a multiple of $\,3\,$. (You can ignore plus/minus signs at this analysis stage.) This is the situation where both equations need to be adjusted. In this example, we will choose to eliminate the $\,y\,$ terms; you could just as well choose to eliminate the $\,x\,$ terms. 
$ \begin{gather} \cssId{s99}{4(2x3y)=4(5)}\cr \cssId{s100}{3(5x+4y)=3(7)}\cr\cr\cr \cssId{s101}{8x12y=20}\cr \cssId{s102}{15x+12y=21} \end{gather} $ 
So, what are the sizes of the coefficients of the
$\,y\,$ terms in the two equations?
Answer: $\,3\,$ and $\,4\,$ The least common multiple of $\,3\,$ and $\,4\,$ is $\,12\,$. (Any multiple of both coefficients will work; the least common multiple is simplest.) So, we want to end up with one coefficient of $\,y\,$ equal to $\,12\,$, and the other equal to $\,12\,$. To do this, we can multiply both sides of the first equation by $\,4\,$, and multiply both sides of the second equation by $\,3\,$. (There's another way you can do it, too—you choose!) 
$ \begin{gather} \cssId{s111}{23x = 41}\cr \cssId{s112}{x=\frac{41}{23}} \end{gather} $ 
From here, everything is the same as in the earlier examples. ‘Add’ the two equations together to eliminate $\,y\,$. Then, solve for $\,x\,$ in the resulting equation. 
$ \begin{gather} \cssId{s116}{2(\frac{41}{23})  3y = 5}\cr\cr \cssId{s117}{3y = 5\cdot\frac{23}{23}  \frac{82}{23}}\cr\cr \cssId{s118}{y = \frac{11}{23}} \end{gather} $  Substitute this value of $\,x\,$ back into a simple earlier equation, and solve for $\,y\,$. 
The unique solution is the ordered pair $ \displaystyle \bigl(\frac{41}{23},\frac{11}{23}\bigr)\,. $ 
Clearly report the unique solution. 
If you want, go to wolframalpha.com and type this in
(you can just cutandpaste):
2x3y=5, 5x+4y=7
You'll see the two lines graphed, and the solution clearly labeled and reported.
How easy is that!?
You can also cutandpaste the equations from the online exercise below into WolframAlpha;
the variables must be $\,x\,$ and $\,y\,$, so adjust if needed;
and, you'll need to insert the comma between the two equations.
If, upon eliminating a variable, the resulting equation is always false (like $\,2 = 3\,$),
then the lines are parallel—there are no solutions.
If, upon eliminating a variable, the resulting equation is always true (like $\,2=2\,$),
then the lines are the same—there are infinitely many solutions.
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
