Simplify the following:$\frac{x^{-1}+y^{-1}}{x^{-1}}+\frac{x^{-1}-y^{-1}}{x^{-1}}$
Given :
The given expression is $\frac{x^{-1}+y^{-1}}{x^{-1}}+\frac{x^{-1}-y^{-1}}{x^{-1}}$
To do :
We have to find the value of $\frac{x^{-1}+y^{-1}}{x^{-1}}+\frac{x^{-1}-y^{-1}}{x^{-1}}$
Solution :
$\frac{x^{-1}+y^{-1}}{x^{-1}}+\frac{x^{-1}-y^{-1}}{x^{-1}} = \frac{(\frac{1}{x} + \frac{1}{y})}{\frac{1}{x}} + \frac{(\frac{1}{x} - \frac{1}{y})}{\frac{1}{x}} $
$ = \frac{\frac{x+y}{xy}}{\frac{1}{x}} + \frac{\frac{y-x}{xy}}{\frac{1}{x}}$
$= \frac{x+y}{y} + \frac{y-x}{y}$
$= \frac{x+y+y-x}{y}$
$= \frac{2y}{2}$
$= 2$
Therefore, the value of $\frac{x^{-1}+y^{-1}}{x^{-1}}+\frac{x^{-1}-y^{-1}}{x^{-1}}$ is 2.
Related Articles
- Simplify: \( \frac{11}{2} x^{2} y-\frac{9}{4} x y^{2}+\frac{1}{4} x y-\frac{1}{14} y^{2} x+\frac{1}{15} y x^{2}+\frac{1}{2} x y \).
- Solve the following pairs of equations:\( \frac{1}{2 x}-\frac{1}{y}=-1 \)\( \frac{1}{x}+\frac{1}{2 y}=8, x, y ≠ 0 \)
- Simplify: $\frac{x^{-3}-y^{-3}}{x^{-3} y^{-1}+(x y)^{-2}+y^{-1} x^{-3}}$.
- If $\frac{x+1}{y} = \frac{1}{2}, \frac{x}{y-2} = \frac{1}{2}$, find x and y.
- Solve the following system of equations:$\frac{5}{x+1} -\frac{2}{y-1}=\frac{1}{2}$$\frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2}$, where $x≠-1$ and $y≠1$
- Solve the following quadratic equation by factorization: $\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}, x ≠1,-1$
- Solve the following pairs of equations by reducing them to a pair of linear equations:(i) \( \frac{1}{2 x}+\frac{1}{3 y}=2 \)\( \frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6} \)(ii) \( \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \)\( \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 \)(iii) \( \frac{4}{x}+3 y=14 \)\( \frac{3}{x}-4 y=23 \)(iv) \( \frac{5}{x-1}+\frac{1}{y-2}=2 \)\( \frac{6}{x-1}-\frac{3}{y-2}=1 \)(v) \( \frac{7 x-2 y}{x y}=5 \)\( \frac{8 x+7 y}{x y}=15 \),b>(vi) \( 6 x+3 y=6 x y \)\( 2 x+4 y=5 x y \)4(vii) \( \frac{10}{x+y}+\frac{2}{x-y}=4 \)\( \frac{15}{x+y}-\frac{5}{x-y}=-2 \)(viii) \( \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \)\( \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8} \).
- 1. Factorize the expression \( 3 x y - 2 + 3 y - 2 x \)A) \( (x+1),(3 y-2) \)B) \( (x+1),(3 y+2) \)C) \( (x-1),(3 y-2) \)D) \( (x-1),(3 y+2) \)2. Factorize the expression \( \mathrm{xy}-\mathrm{x}-\mathrm{y}+1 \)A) \( (x-1),(y+1) \)B) \( (x+1),(y-1) \)C) \( (x-1),(y-1) \)D) \( (x+1),(y+1) \)
- Solve the following system of equations:$\frac{6}{x+y} =\frac{7}{x-y}+3$$\frac{1}{2(x+y)}=\frac{1}{3(x-y)}$
- Solve the following system of equations:$\frac{5}{x-1} +\frac{1}{y-2}=2$$\frac{6}{x-1}-\frac{3}{y-2}=1$
- If \( 2^{x}=3^{y}=6^{-z} \), show that \( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \).
- Solve the following quadratic equation by factorization: $\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$
- $\frac{x-1}{2}+\frac{2 x-1}{4}=\frac{x-1}{3}-\frac{2 x-1}{6}$.
- For $x=\frac{3}{4} $and $y=\frac{-9}{8}$, insert a rational number between:$(x+y)^{-1} and x^{-1}+y^{-1} $
- Verify: $x\times(y\times z)=(x\times y)\times z$, where $x=\frac{1}{2},\ y=\frac{1}{3}$ and $z=\frac{1}{4}$.
Kickstart Your Career
Get certified by completing the course
Get Started