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.

Download practice questions along with solutions 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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Pair of Linear Equations in 2 Variables
grade 10 | Answers Set 1
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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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