Write four solutions for each of the following equations:
(i) \( 2 x+y=7 \)
(ii) \( \pi x+y=9 \)
(iii) \( x=4 y \).

To do:

We have to write four solutions for each of the given equations.

Solution:

(i) Given,

$2x+y=7$

Let us substitute $x=0, 1$ and $y=0, 1$

$x=0$

This implies,

$2(0)+y=7$

$0+y=7$

$y=7$

Therefore, $x, y=0, 7$

$x=1$

This implies,

$2(1)+y=7$

$2+y=7$

$y=7-2$

$y=5$

Therefore, $x, y=1, 5$

$y=0$

This implies,

$2x+(0)=7$

$2x=7$

$x=\frac{7}{2}$

Therefore, $x, y=\frac{7}{2}, 0$

$y=1$

This implies,

$2x+(1)=7$

$2x=7-1$

$2x=6$

$x=\frac{6}{2}$

$x=3$
Therefore, $x, y=3, 1$

Hence, the solutions are $x, y=(0,7), (1, 5), (\frac{7}{2}, 0), (3, 1)$.

(ii) Given,

$\pi x+y=9$

Let us substitute $x= 0, 1, $ and $y=0, 1$ to write four solutions.

$x=0$

This implies,

$\pi x+y=9$

$\pi (0)+y=9$

$0+y=9$

$y=9$

Therefore, $x, y=0, 9$

$x=1$

This implies,

$\pi x+y=9$

$\pi (1)+y=9$

$\pi+y=9$

$y=9-\pi$

Therefore, $x, y=1, 9-\pi$

$y=0$

This implies,

$\pi x+0=9$

$\pi x=9$

$x=\frac{9}{\pi}$

Therefore, $x, y=\frac{9}{\pi}, 0$

$y=1$

This implies,

$\pi x+1=9$

$\pi x=9-1$

$\pi x=8$

$x=\frac{8}{\pi}$

Therefore, $x, y=\frac{8}{\pi}, 1$

Hence, the solutions are $x, y=(0,9), (1, 9-\pi ), (\frac{9}{\pi}, 0), (\frac{8}{\pi}, 1)$.

(iii) Given,

$x=4y$

Let us substitute $x=0, 1$ and $y=0, 1$ to write four solutions.

$x=0$

This implies,

$0=4y$

$y=\frac{0}{4}$

$y=0$

Therefore, $x, y=0, 0$

$x=1$

This implies,

$1=4y$

$\frac{1}{4}=y$

$y=\frac{1}{4}$

Therefore, $x, y=1,\frac{1}{4}$

$y=2$

This implies,

$x=4(2)$

$x=8$

Therefore, $x, y=8, 0$

Let $y=1$

This implies,

$x=4(1)$

$x=4$

Therefore, $x, y=4, 1$

Hence, the solutions are $x, y=(0,0), (1, \frac{1}{4}), (8, 0), (4, 1)$.

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Updated on: 10-Oct-2022

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