Systems of linear equations: Two equations with two unknowns
Solving systems of equations by elimination
We have seen how we can solve a system of equations by means of substitution. There is another method to solve systems of linear equations, namely the elimination method. This method can also be applied to systems with multiple linear equations with multiple unknowns, whereas using substitution can be quite complicated in the case of such systems.
Procedure 
Example  
With the elimination method for solving a system of two linear equations with two unknowns, we proceed as follows. 
Solve the following system: #\lineqs{\quad 2 \cdot x +4 \cdot y+5&= \quad 0 \cr \quad 3 \cdot x +2 \cdot y 4&= \quad 0 \cr}# 

Step 1 
Multiply the first and/or second equation with a number such that the coefficient of #x# or of #y# of both equations become equal. 
#\lineqs{2 \cdot x +4 \cdot y+5=0 \cr 6 \cdot x +4 \cdot y 8=0 \cr}# 
Step 2 
Subtract the second equation from the first. We are now left with an equation with only #x# or #y#. 
#\begin{array}{rcl}2 \cdot x +4 \cdot y+5&=&0 \\ 6 \cdot x +4 \cdot y 8&=&0\\ \hline 8\cdot x+13&=&0\end{array}# 
Step 3 
Solve the linear equation of step 2 using reduction. 
#x=\frac{13}{8}# 
Step 4 
Substitute the value obtained in step 3 in one of the original equations to determine the value of the remaining unknown. 
#\begin{array}{rcl}2 \cdot \frac{13}{8}+4 \cdot y +5&=&0\\ 4 \cdot y +\frac{7}{4}&=&0\\ 4 \cdot y&=&\frac{7}{4} \\ y&=&\frac{7}{16} \end{array}# 
Step 5 
Give the answer in the form \[\lineqs{ x & =\;\; \ldots \\ y &=\;\; \ldots }\] 
#\lineqs{ x &= \;\; \frac{13}{8} \\ y &= \;\; \frac{7}{16} }# 
#\lineqs{x&=&{{1}\over{5}}\cr y&=&{{31}\over{15}}\cr }#
In the system, the coefficient of #y# in the first equation is equal to the coefficient of #y# in the second equation. We can therefore eliminate the variable #y# from the first equation by replacing this equation with the difference of the given equations. Hence, we start with step 2 of the procedure.
Step 2  \[\begin{array}{rcl}x+3\cdot y&=&6 \\ 6\cdot x+3\cdot y&=&5\\ \hline 5\cdot x&=&1\end{array}\] 
Step 3  We can reduce this equation in order to find #x#. This goes as follows: \[\begin{array}{rcl} 5\cdot x&=&1 \\&&\phantom{x}\blue{\text{the equation we need to solve}} \\ x&=& {{1}\over{5}} \\&&\phantom{x}\blue{\text{both sides divided by }5} \end{array}\] 
Step 4  We substitute #x={{1}\over{5}}# in the second original equation. We subsequently solve this by reduction. This goes as follows: \[\begin{array}{rcl} 6\cdot {{{1}\over{5}}}+3\cdot y &=&5 \\&&\phantom{xyz}\blue{\text{substituted }x={{1}\over{5}} \text{ in the second equation}}\\ {{6}\over{5}}+3\cdot y&=&5\\&&\phantom{xyz}\blue{\text{calculated}}\\ 3\cdot y&=&{{31}\over{5}}\\&&\phantom{xyz}\blue{\text{added }{{6}\over{5}}\text{ on both sides}}\\ y&=&{{31}\over{15}}\\&&\phantom{xyz}\blue{\text{both sides divided by }3} \end{array}\] 
Step 5  Hence, the solution to the system is \[\lineqs{x&=&{{1}\over{5}}\cr y&=&{{31}\over{15}}\cr }\] 
Or visit omptest.org if jou are taking an OMPT exam.