Question

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

Answer 1

$\left(i\right)2x+y=7$
$\Rightarrow y=7-2x$
$\to Putx=0,$
$y=7-2\times 0\Rightarrow y=7$
$(0,7)$ is a solution.
$\to Now,putx=1$
$y=7-2\times 1\Rightarrow y=5$
$(1,5)$ is a solution.
$\to Now,putx=2$
$y=7-2\times 2\Rightarrow y=3$
$(2,3)$ is a solution.
$\to Now,putx=-1$
$y=7-2\times -1\Rightarrow y=9$
$(-1,9)$ is a solution.

$\left(ii\right)\pi x+y=9$
$\Rightarrow y=9-\pi x$
$\to Putx=0,$
$y=9-\pi \times 0\Rightarrow y=9$
$(0,9)$ is a solution.
$\to Now,putx=1$
$y=9-\pi \times 1\Rightarrow y=9-\pi$
$(1,9-\pi )$ is a solution.
$\to Now,putx=2$
$y=9-\pi \times 2\Rightarrow y=9-2\pi$
$(2,9-2\pi )$ is a solution.
$\to Now,putx=-1$
$y=9-\pi \times -1\Rightarrow y=9+\pi$
$(-1,9+\pi )$ is a solution.

$\left(iii\right)x=4y$
$\to Putx=0,$
$0=4y$
$\Rightarrow y=0$
$(0,0)$ is a solution.
$\to Now,puty=1$
$x=4\left(1\right)\Rightarrow 4$
$(4,1)$ is a solution.
$\to Now,puty=-1$
$x=4(-1)$
$x=-4$
Therefore, $(-4,-1)$ is a solution of this equation.
For $x=2,$
$2=4y$
$y=\frac{2}{4}=\frac{1}{2}$
Therefore, $(2,\frac{1}{2})$ is a solution of this equation.