Answer :
(1) [tex]\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}[/tex]
(2) [tex]\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{15xy}[/tex]
(3) [tex]\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{14xy}[/tex]
(4) [tex]\frac{2}{5x}+\frac{3}{7y}=\frac{6y+15x}{21xy}[/tex]
(5) [tex]\frac{7}{11x}-\frac{1}{33y}=\frac{231y-11x}{363xy}[/tex]
(6) [tex]\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{7y+42y-2x}{14xy}[/tex]
Step-by-step explanation :
(1) The given expression is: [tex]\frac{1}{x}+\frac{1}{y}[/tex]
[tex]\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}[/tex]
(2) The given expression is: [tex]\frac{1}{5x}+\frac{1}{3y}[/tex]
[tex]\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{(5x)\times (3y)}=\frac{3y+5x}{15xy}[/tex]
(3) The given expression is: [tex]\frac{1}{7x}-\frac{1}{2y}[/tex]
[tex]\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{(7x)\times (2y)}=\frac{2y-7x}{14xy}[/tex]
(4) The given expression is: [tex]\frac{2}{5x}+\frac{3}{7y}[/tex]
[tex]\frac{2}{5x}+\frac{3}{7y}=\frac{(2\times 3y)+(3\times 5x)}{(5x)\times (7y)}=\frac{6y+15x}{21xy}[/tex]
(5) The given expression is: [tex]\frac{7}{11x}-\frac{1}{33y}[/tex]
[tex]\frac{7}{11x}-\frac{1}{33y}=\frac{(7\times 33y)-(1\times 11x)}{(11x)\times (33y)}=\frac{231y-11x}{363xy}[/tex]
(6) The given expression is: [tex]\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}[/tex]
[tex]\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{(x\times 7y)+(3\times 2x\times 7y)-(2x\times x)}{(2x)\times (x)\times (7y)}=\frac{7xy+42xy-2x^2}{14x^2y}=\frac{7y+42y-2x}{14xy}[/tex]
