Question

let r = {(x-y):x,y E w , 2x+y=8} then, find the domin and range of r

Answer 1

$\large\underline{\sf{Solution-}}$

Given relation is

$\rm :\longmapsto\:R = \{(x,y) : 2x + y = 8 \: \forall \: x,y \: \in \: W \}$

So,

$\rm :\longmapsto\:2x + y = 8$

can be rewritten as

$\rm :\longmapsto\: y = 8 - 2x$

$\red{ \sf \: When \: x = 0 \: \implies \: y = 8 - 2 \times 0\bf\implies \:y = 8}$

$\green{ \sf \: When \: x = 1 \: \implies \: y = 8 - 2 \times 1\bf\implies \:y = 6}$

$\blue{ \sf \: When \: x = 2 \: \implies \: y = 8 - 2 \times 2\bf\implies \:y = 4}$

$\purple{ \sf \: When \: x = 3 \: \implies \: y = 8 - 2 \times 3\bf\implies \:y = 2}$

$\pink{ \sf \: When \: x = 4 \: \implies \: y = 8 - 2 \times 4\bf\implies \:y = 0}$

Hᴇɴᴄᴇ,

➢ Pair of values of the given relation are shown in the below table.

$\red{\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 4 \\ \\ \sf 1 & \sf 6 \\ \\ \sf 2 & \sf 4\\ \\ \sf 3 & \sf 2\\ \\ \sf 4 & \sf 0 \end{array}} \\ \end{gathered}}$

So, Relation is defined as

$\green{\rm :\longmapsto\:R = \{(0,8),(1,6),(2,4),(3,2),(4,0) \}}$

So,

$\red{\bf\implies \: \boxed{ \sf{ \:Domain \: of \: R = \{0,1,2,3,4 \}}}}$

and

$\red{\bf\implies \: \boxed{ \sf{ \:Range \: of \: R = \{0,1,2,3,4 \}}}}$