In each of the following cases, find $a$ and $b$.
(i) $(2 a+b, a-b)=(8,3)$
(ii) $\left(\frac{a}{4}, a-2 b\right)=(0,6+b)$
(i) We have, $(2 a+b, a-b)=(8,3)$
$$\Rightarrow \quad 2 a+b=8 \text { and } a-b=3$$
[since, two ordered pairs are equal, if their corresponding first and second elements are equal]
On substituting, $b=a-3$ in $2 a+b=8$, we get
$$\begin{aligned} & 2 a+a-3=8 \Rightarrow 3 a-3=8 \\ & \Rightarrow \quad 3 a=11 \Rightarrow a=\frac{11}{3} \end{aligned}$$
Again, $\quad$ substituting $a=\frac{11}{3}$ in $b=a-3$, we get
$\begin{array}{ll} & b=\frac{11}{3}-3=\frac{11-9}{3}=\frac{2}{3} \\ \therefore \quad & a=\frac{11}{3} \text { and } b=\frac{2}{3}\end{array}$
$\begin{array}{rlrl} & \text { (ii) We have, } & \left(\frac{a}{4}, a-2 b\right) =(0,6+b) \\ \Rightarrow & \frac{a}{4} =0 \Rightarrow a=0\end{array}$
$$\begin{aligned} \text{and}\quad a-2 b & =6+b \\ 0-2 b & =6+b \\ -3 b & =6 \end{aligned}$$
$$\begin{aligned} & \therefore \quad b=-2 \\ & \therefore \quad a=0, b=-2 \end{aligned}$$
$A=\{1,2,3,4,5\}, S=\{(x, y): x \in A, y \in A\}$, then find the ordered which satisfy the conditions given below.
(i) $x+y=5$
(ii) $x+y<5$
(iii) $x+y>8$
We have, $A=\{1,2,3,4,5\}$ and $S=\{(x, y): x \in A, y \in A\}$
(i) The set of ordered pairs satisfying $x+y=5$ is,
$$\{(1,4),(2,3),(3,2),(4,1)\}$$
(ii) The set of ordered pairs satisfying $x+y<5$ is $\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)$.
(iii) The set of ordered pairs satisfying $x+y>8$ is $\{(4,5),(5,4),(5,5)\}$.
If $R=\left\{(x, y): x, y \in W, x^2+y^2=25\right\}$, then find the domain and range of $R$.
We have,
$\begin{aligned} R & =\left\{(x, y): x, y \in W, x^2+y^2=25\right\} \\ & =\{(0,5),(3,4),(4,3),(5,0)\}\end{aligned}$
$\begin{aligned} \text{Domain of }\quad R & =\text { Set of first element of ordered pairs in } R \\ & =\{0,3,4,5\}\end{aligned}$
$\begin{aligned} \text{Range of }\quad R & =\text { Set of second element of ordered pairs in } R \\ & =\{5,4,3,0,\}\end{aligned}$
If $R_1=\{(x, y) \mid y=2 x+7$, where $x \in R$ and $-5 \leq x \leq 5\}$ is a relation. Then, find the domain and range of $R_1$.
We have, $\quad R_1=\{x, y) \mid y=2 x+7$, where $x \in R$ and $\left.-5 \leq x \leq 5\right\}$
Domain of $R_1=\{-5 \leq x \leq 5, x \in R\}$
$=[-5,5]$
$\begin{array}{ll}\because & y=2 x+7 \\ \text { When } x=-5 \text {, then } & y=2(-5)+7=-3 \\ \text { When } x=5 \text {, then } & y=2(5)+7=17\end{array}$
$$\begin{aligned} & \therefore \quad \text { Range of } R_1=\{-3 \leq y \leq 17, y \in R\} \\ & =[-3,17] \end{aligned}$$
If $R_2=\{x, y) \mid x$ and $y$ are integers and $\left.x^2+y^2=64\right\}$ is a relation, then find the value of $R_2$.
We have, $\mathrm{R}_2=\{(x, y)\} x$ and $y$ are integers and $\left.x^2+y^2=64\right\}$
Since, 64 is the sum of squares of 0 and $\pm 8$.
When $x=0$, then $y^2=64 \Rightarrow y= \pm 8$
$$\begin{aligned} x & =8 \text {, then } y^2=64-8^2 \Rightarrow 64-64=0 \\ x & =-8, \text { then } y^2=64-(-8)^2=64-64=0 \\ \therefore \quad R_2 & =\{(0,8),(0,-8),(8,0),(-8,0)\} \end{aligned}$$