Essentially, the method of elimination is similar to that of substitution – using the two equations, we try to form an equation in only unknown, which enables us to calculate that unknown. Then, we plug that value into any of the two equations to determine the other unknown. However, there is a slight difference in how the single-variable equation is formed. In elimination, we manipulate both equations, and then add or subtract them, so that one of the two variables gets eliminated. Let us see how, with an example.

Consider the following pair of linear equations:

\[\begin{array}{l} 2x + 3y - 11 = 0\\ 3x + 2y - 9 = 0 \end{array}\]

The coefficients of x in the two equations are 2 and 3 respectively. Let us multiply the first equation by 3 and the second equation by 2, so that the coefficients of x in the two equations become equal:

\[\begin{array}{l}\left\{ \begin{array}{l}3 \times \left( {2x + 3y - 11 = 0} \right)\\2 \times \left( {3x + 2y - 9 = 0} \right)\end{array} \right.\\ \Rightarrow \;\;\;6x + 9y - 33 = 0\\\qquad6x + 4y - 18 = 0\end{array}\]

Now, let us subtract the two equations, which means that we subtract the left hand sides of the two equations, and the right hand side of the two equations, and the equality will still be preserved (this should be obvious: if I = II and III = IV, then I – III will be equal to II – IV):

\[\begin{array}{l}\left\{ \begin{array}{l}\,\,\,\,6x + 9y - 33 = 0\\\,\,\,\,6x + 4y - 18 = 0\\\underline { - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,0 + 5y\,\,\, - 15 = 0\end{array} \right.\\ \Rightarrow \;\;\;5y = 15\\ \Rightarrow \;\;\;y = 3\end{array}\]

Note how x gets eliminated, and we are left with an equation in y alone. Once we have the value of y, we proceed as earlier – we plug this into any of the two equations. Let us put this into the first equation:

\[\begin{array}{l}2x + 3y - 11 = 0\\ \Rightarrow \;\;\;2x + 3\left( 3 \right) - 11 = 0\\ \Rightarrow \;\;\;2x + 9 - 11 = 0\\ \Rightarrow \;\;\;2x = 2\\ \Rightarrow \;\;\;x = 1\end{array}\]

Thus, the solution is:

\[x = 1,\;y = 3\]

Let’s take another example. Consider the following pair of linear equations:

\[\begin{array}{l}3x - 2y = 7\\ - 2x + 5y =  - 1\end{array}\]

This time, let’s eliminate y:

\[\begin{align}& \left\{ \begin{array}{l} 5 \times \left( {3x - 2y - 7 = 0} \right)\\ 2 \times \left( { - 2x + 5y + 1 = 0} \right) \end{array} \right.\\& ADD\,\,\,\,\, \Rightarrow \;\;\;\frac{\begin{align}15x - 10y - 35 = 0\\ - 4x + 10y + 2 = 0 \end{align}}{{11x + 0 - 33 = 0}}\\&\qquad\quad\;\Rightarrow \;\;\;11x = 33\,\,\, \Rightarrow \;\;\;x = 3 \end{align}\]

Note that to eliminate y, we added the two equations (in the previous example, we subtracted them). Now, we plug the value of x we have obtained into the first equation:

\[\begin{array}{l}3x - 2y = 7\\ \Rightarrow \;\;\;3\left( 3 \right) - 2y = 7\\ \Rightarrow \;\;\;2y = 2\\ \Rightarrow \;\;\;y = 1\end{array}\]

Thus, the solution is:

\[x = 3,\;y = 1\]

Example 1: Solve the following pair of linear equations by elimination:

\[\begin{array}{l}2x - 3y + 5 = 0\\5x + 4y - 22 = 0\end{array}\]

Solution: We have:

\[\begin{array}{l}\left\{ \begin{array}{l}4 \times \left( {2x - 3y + 5 = 0} \right)\\3 \times \left( {5x + 4y - 22 = 0} \right)\end{array} \right.\\ \Rightarrow \;\;\;\left\{ \begin{array}{l}\,\,\,\,\,8x - 12y + 20 = 0\\\,\,\,\,\,15x + 12y - 66 = 0\\\underline { + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,23x + 0\,\,\,\,\,\, - 46 = 0\end{array} \right.\\ \Rightarrow \;\;\;23x = 46\\ \Rightarrow \;\;\;x = 2\end{array}\]

Now, we plug this value of x into the first equation to get y:

\[\begin{array}{l}2x - 3y + 5 = 0\\ \Rightarrow \;\;\;2\left( 2 \right) - 3y + 5 = 0\\ \Rightarrow \;\;\;3y = 9\\ \Rightarrow \;\;\;y = 3\end{array}\]

Thus, the solution is:

\[x = 2,\;y = 3\]

Example 2: Solve the following pair of linear equations by elimination:

\[\begin{array}{l}x + 3y + 1 = 0\\7x - 2y + 3 = 0\end{array}\]

Solution: We eliminate x in this case:

\[\begin{array}{l}\left\{ \begin{array}{l}7 \times \left( {x + 3y + 1 = 0} \right)\\1 \times \left( {7x - 2y + 3 = 0} \right)\end{array} \right.\\ \Rightarrow \;\;\;\left\{ \begin{array}{l}\,\,\,\,\,7x + 21y + 7 = 0\\\,\,\,\,\,7x - 2y + 3 = 0\\\underline { - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,0 + 23y + 4 = 0\end{array} \right.\\ \Rightarrow \;\;\;y =  - \frac{4}{{23}}\end{array}\]

Now, we plug this value of y into the first equation to get x:

\[\begin{align}&x + 3y + 1 = 0\\ &\Rightarrow \;\;\;x + 3\left( {\frac{{ - 4}}{{23}}} \right) + 1 = 0\\ &\Rightarrow \;\;\;x - \frac{{12}}{{23}} + 1 = 0\\ &\Rightarrow \;\;\;x = 1 - \frac{{12}}{{23}} = \frac{{11}}{{23}}\end{align}\]

Thus, the solution is:

\[x = \frac{{11}}{{23}},\;y =  - \frac{4}{{23}}\]

Example 3: Solve the following pair of linear equations by elimination:

\[\begin{array}{l}x + 4y - 1 = 0\\ - 2x - 8y + 7 = 0\end{array}\]

Solution: We have:

\[\begin{array}{l}\left\{ \begin{array}{l}2 \times \left( {x + 4y - 1 = 0} \right)\\1 \times \left( { - 2x - 8y + 7 = 0} \right)\end{array} \right.\\ \Rightarrow \;\;\;\left\{ \begin{array}{l}\,\,\,2x + 8y - 2 = 0\\ - 2x - 8y + 7 = 0\\\underline { + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,0 + 0\,\,\,\,\, + 5 = 0\end{array} \right.\\ \Rightarrow \;\;\;5 = 0\end{array}\]

This is an impossible equality, which means that the given pair of equations is inconsistent.

Is there a difference between substitution and elimination method?

  • Substitution method: – solving an equation for a particular variable and then substituting or replacing that equation into the other one to arrive at the solution
  • Elimination method: – multiplying an equation by a number, then add the two equations together and arrive at the solution

Steps to solve a problem using elimination method

  • Here are 4 simple steps to solve linear pair equations using elimination method. Take the example of two equations (a) 2x+3y=8 and (b) 3x+2y=7
  • Step i- multiply each equation by a number so that the two equations have the same leading coefficient (multiply equation a with 3 and b with 2)
  • Step ii- Subtract equation b from a [(6x+9y=24)-(6x+4y=14)=5y=10]
  • Step iii- Solve new equation to find y [ y=10/5=2 ]
  • Step iv- Substitute y=2 in equation a or b and solve for x [(a) 2x+3(2)=8]
  • You should get x=1, y=2 or (1,2)

How to know when to use substitution or elimination in a problem?

  • Elimination method: – When the equation is in standard form of Ax+By=C or when all variables have a coefficient other than 1
  • Substitution method: – When one or both of the equations is already solved for one of the variables or when one of the vriables has a coefficient of 1

which method is better? Elimination or Substitution?

  • Elimination method – is easier as compared to substitution method because it eliminates one variable from the equation and let's us solve for the remaining unknown variable. This is applicable only when all variables have coefficients. Incase one variable doesn't a coefficient, then Substitution would be easier.
Download SOLVED Practice Questions of Elimination Method for FREE
Pair of Linear Equations in 2 Variables
grade 10 | Answers Set 1
Pair of Linear Equations in 2 Variables
grade 10 | Questions Set 2
Pair of Linear Equations in 2 Variables
grade 10 | Answers Set 2
Pair of Linear Equations in 2 Variables
grade 10 | Questions Set 1
Download SOLVED Practice Questions of Elimination Method for FREE
Pair of Linear Equations in 2 Variables
grade 10 | Answers Set 1
Pair of Linear Equations in 2 Variables
grade 10 | Questions Set 2
Pair of Linear Equations in 2 Variables
grade 10 | Answers Set 2
Pair of Linear Equations in 2 Variables
grade 10 | Questions Set 1
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