Solutions of a Linear Equation

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Consider the following linear equation:

\[x + 2y = 4\]

Note that \(x = 2\) and \(y = 1\) (together) satisfy this equation. We state this fact succinctly by saying that \(\left( {2,1} \right)\) is a solution of the equation. In general, if \(\left( {p,q} \right)\) is a solution of the equation \(ax + by + c = 0\), this means that \(x = p\) and \(y = q\) satisfy the equation, that is, \(ap + bq + c = 0\).

Example 1: Which of the following are solutions of the equation \(3x - 4y + 1 = 0\)?

(A)  \(\left( {1,2} \right)\)

(B)  \(\left( {5,4} \right)\)

(C)  \(\left( { - 3, - 2} \right)\)

(D)   \(\left( {2,7} \right)\)

 (E)  \(\left( {1,1} \right)\)

Solution: Let’s consider option (A). If we use \(x = 1,\,\,y = 2\) in the given equation, we see that it is not satisfied:

\[3\left( 1 \right) - 4\left( 2 \right) + 1 \ne 0\]

Thus, \(\left( {1,2} \right)\) is not a solution for the given equation.

Now, consider option (B). We use \(x = 5,\,\,y = 4\) and find that

\[3\left( 5 \right) - 4\left( 4 \right) + 1 = 0\]

Thus, \(\left( {5,4} \right)\) is a solution for the given equation.

Similarly, you can verify that options (C) and (E) are valid solutions for the given equation, but option (D) is not.

Example 2: Write three solutions of the equation \(x - 2y + 3 = 0\).

Solution: If we substitute \(x = 0\), we obtain \(y = \frac{3}{2}\). Thus, \(\left( {0,\frac{3}{2}} \right)\) is a solution of this equation.

Substituting \(y = 0\) gives \(x =  - 3\), which means that \(\left( { - 3,0} \right)\) is another possible solution.

Substituting \(x = 1\) gives \(y = 2\), so that \(\left( {1,2} \right)\) is yet another solution.

By now, it should be obvious to you that a linear equation in two variables will have infinitely many solutions. Substituting a particular value for one variable gives us a particular value for the other variable, and hence a possible solution.

Example 3: Write two solutions of the equation \(2x - 3y + 1 = 0\) in which the value of x is a negative non-integer rational number.

Solution: We can use any two negative rational values for x. Substituting \(x =  - \frac{1}{2}\) gives:

\[\begin{align}&2\left( { - \frac{1}{2}} \right) - 3y + 1 = 0\\& \Rightarrow \,\,\, - 1 - 3y + 1 = 0\\& \Rightarrow \,\,\,y = 0\end{align}\]

 Thus, \(\left( { - \frac{1}{2},0} \right)\) is a solution. Substituting \(x =  - \frac{2}{5}\) gives:

\[\begin{align}&2\left( { - \frac{2}{5}} \right) - 3y + 1 = 0\\ &\Rightarrow \,\,\, - \frac{4}{5} - 3y + 1 = 0\\ &\Rightarrow \,\,\,y = \frac{1}{{15}}\end{align}\]

Thus, \(\left( { - \frac{2}{5},\frac{1}{{15}}} \right)\) is another possible solution.

Example 4: Write two solutions of the equation \( - \frac{1}{2}x + \sqrt 2 y = \sqrt 3 \) in which the value of y is an irrational number.

Solution: We can take any two irrational values for y. Substituting \(y = \sqrt 2 \) gives:

\[\begin{align}& - \frac{1}{2}x + \sqrt 2 \left( {\sqrt 2 } \right) = \sqrt 3 \\& \Rightarrow \,\,\, - \frac{1}{2}x + 2 = \sqrt 3 \\& \Rightarrow \,\,\,x = 2\left( {2 - \sqrt 3 } \right)\end{align}\]

Thus, \(\left( {2\left( {2 - \sqrt 3 } \right),\sqrt 2 } \right)\) is one such solution. Next, substituting \(y = \sqrt 3 \) gives:

\[\begin{align}& - \frac{1}{2}x + \sqrt 2 \left( {\sqrt 3 } \right) = \sqrt 3 \\ &\Rightarrow \,\,\, - \frac{1}{2}x + \sqrt 6  = \sqrt 3 \\& \Rightarrow \,\,\,x = 2\left( {\sqrt 6  - \sqrt 3 } \right)\end{align}\]

Thus, \(\left( {2\left( {\sqrt 6  - \sqrt 3 } \right),\sqrt 3 } \right)\) is another such solution.

Example 5. If \(\left( { - 2,3} \right)\) is a solution of the equation \(4x - ky = 1\), what is the value of k?

Solution: \(x =  - 2,\,\,y = 3\) must satisfy the given equation:

\[\begin{array}{l}4\left( { - 2} \right) - k\left( 3 \right) = 1\\ \Rightarrow \,\,\, - 8 - 3k = 1\\ \Rightarrow \,\,\,k =  - 3\end{array}\]

Example 6: If \(\left( {2, - t} \right)\) is a solution of \(ax + k = by\), what is the value of t?

Solution: \(x = 2,\,\,y =  - t\) must satisfy the given equation:

\[\begin{align}&a\left( 2 \right) + k = b\left( { - t} \right)\\& \Rightarrow \,\,\,t =  - \left( {\frac{{2a + k}}{b}} \right)\end{align}\]

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