Question

Write four solutions to the equation $\pi x+y=9$.

Answer 1

Hint: In this question, we are given the relation between x and y in the form of a linear equation, therefore there would be an infinite number of solutions. As we are asked to find only four solutions, we can rewrite the given equation to express y in terms of x and then put 4 values of x to find the corresponding values of y. For each x, the corresponding pair (x,y) will be a solution to the given equation.

Complete step-by-step answer:

The given relation between x and y is $\pi x+y=9$. We can take the $\pi x$ term to the RHS by introducing a minus sign to rewrite the relation as
$y=9-\pi x................(1.1)$
Now, as we are asked to find four solutions, we can take four arbitrary values of x and then find the corresponding values of y satisfying equation (1.1).
If we take x=0, then from equation (1.1), we obtain
$y=9-\pi \times 0=9................(1.2)$
Thus, (0,9) will be a solution to the given equation.
If we take x=1, then from equation (1.1), we obtain
$y=9-\pi \times 1=9-\pi ................(1.3)$
Thus, $\left( 1,9-\pi \right)$ will be a solution to the given equation.
If we take $x=\dfrac{1}{\pi }$, then from equation (1.1), we obtain
$y=9-\pi \times \dfrac{1}{\pi }=9-1=8................(1.4)$
Thus, $\left( \dfrac{1}{\pi },8 \right)$ will be a solution to the given equation.
If we take x=2, then from equation (1.1), we obtain
$y=9-\pi \times 2=9-2\pi ................(1.5)$
Thus, $\left( 2,9-2\pi \right)$ will be a solution to the given equation.

Thus, from equations (1.2), (1,3), (1.4) and (1.5), we find that four solutions to the given equation should be (0,9), $\left( 1,9-\pi \right)$ , $\left( \dfrac{1}{\pi },8 \right)$ and $\left( 2,9-2\pi \right)$.

Note: We should note that we could have taken any values of x convenient to us to find the values of y satisfying the given equation. Also, we can take the value of x as fractions as taken in equation (1.4) because the linear equation will hold for any real value of x and thus x can be also taken as a fraction.