Find the values of a and b, when:
(i) $(a+3, b-2)=(5,1)$
(ii) $(a+b, 2 b-3)=(4,-5)$
(iii) $\left(\frac{\mathrm{a}}{3}+1, \mathrm{~b}-\frac{1}{3}\right)=\left(\frac{5}{3}, \frac{2}{3}\right)$
(iv) $(a-2,2 b+1=(b-1, a+2)$
Since, the ordered pairs are equal, the corresponding elements are equal.
$\therefore, \mathrm{a}+3=5 \ldots$ (i) and $\mathrm{b}-2=1 \ldots$ (ii)
Solving eq. (i), we get
$a+3=5$
$\Rightarrow a=5-3$
$\Rightarrow a=2$
Solving eq. (ii), we get
$b-2=1$
$\Rightarrow b=1+2$
$\Rightarrow b=3$
Hence, the value of a = 2 and b = 3.
(ii) Since, the ordered pairs are equal, the corresponding elements are equal.
$\therefore, \mathrm{a}+\mathrm{b}=4 \ldots$ (i) and $2 \mathrm{~b}-3=-5 \ldots$ (ii)
Solving eq. (ii), we get
$2 b-3=-5$
$\Rightarrow 2 b=-5+3$
$\Rightarrow 2 b=-2$
$\Rightarrow b=-1$
Putting the value of $b=-1$ in eq. (i), we get
$a+(-1)=4$
$\Rightarrow a-1=4$
$\Rightarrow a=4+1$
$\Rightarrow a=5$
Hence, the value of a = 5 and b = -1.
(iii) Since the ordered pairs are equal, the corresponding elements are equal.
$\therefore \frac{a}{3}+1=\frac{5}{3} \ldots(\mathrm{i})$
$\& b-\frac{1}{3}=\frac{2}{3}$ .......(ii)
Solving Eq. (i), we get
$\frac{a}{3}+1=\frac{5}{3}$
$\Rightarrow \frac{a}{3}=\frac{5}{3}-1$
$\Rightarrow a=3\left(\frac{5}{3}-1\right)$
$\Rightarrow a=5-3$
$\Rightarrow a=2$
Solving eq. (ii), we get
$b-\frac{1}{3}=\frac{2}{3}$
$\Rightarrow b=\frac{2}{3}+\frac{1}{3}$
$\Rightarrow b=\frac{3}{3}$
$\Rightarrow b=1$
Hence, the value of a = 2 and b = 1.
(iv) Since, the ordered pairs are equal, the corresponding elements are equal.
$\therefore, a-2=b-1 \ldots$ (i)
$\& 2 b+1=a+2 \ldots$ (ii)
Solving eq. (i), we get
$a-2=b-1$
$\Rightarrow a-b=-1+2$
$\Rightarrow a-b=1 \ldots$ (iii)
Solving eq. (ii), we get
$2 b+1=a+2$
$\Rightarrow 2 b-a=2-1$
$\Rightarrow-a+2 b=1 \ldots$ (iv)
Adding eq. (iii) and (iv), we get
$a-b+(-a)+2 b=1+1$
$\Rightarrow a-b-a+2 b=2$
$\Rightarrow b=2$
Putting the value of b = 2 in eq. (iii), we get
$a-2=1$
$\Rightarrow a=1+2$
$\Rightarrow a=3$
Hence, the value of a = 3 and b = 2