# 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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