Solve ( overline a ^ 1 + overline b ^ 1 )^ -1 = underline ab a | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$( a ^ { - 1 } + b ^ { - 1 } ) ^ { - 1 } = \frac { a b } { a + b }$$

Solve for a

$a\in \mathrm{R}\setminus -b,0$
$b\neq 0$

Solution Steps

Subtract $\frac{ab}{a+b}$ from both sides.

$$\left(a^{-1}+b^{-1}\right)^{-1}-\frac{ab}{a+b}=0$$

Reorder the terms.

$$-\frac{ab}{a+b}+\frac{1}{\frac{1}{a}+\frac{1}{b}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $a$ and $b$ is $ab$. Multiply $\frac{1}{a}$ times $\frac{b}{b}$. Multiply $\frac{1}{b}$ times $\frac{a}{a}$.

$$-\frac{ab}{a+b}+\frac{1}{\frac{b}{ab}+\frac{a}{ab}}=0$$

Since $\frac{b}{ab}$ and $\frac{a}{ab}$ have the same denominator, add them by adding their numerators.

$$-\frac{ab}{a+b}+\frac{1}{\frac{b+a}{ab}}=0$$

Variable $a$ cannot be equal to $0$ since division by zero is not defined. Divide $1$ by $\frac{b+a}{ab}$ by multiplying $1$ by the reciprocal of $\frac{b+a}{ab}$.

$$-\frac{ab}{a+b}+\frac{ab}{b+a}=0$$

Combine $-\frac{ab}{a+b}$ and $\frac{ab}{b+a}$ to get $0$.

$$0=0$$

Compare $0$ and $0$.

$$\text{true}$$

This is true for any $a$.

$$a\in \mathrm{R}$$

Variable $a$ cannot be equal to $0$.

$$a\in \mathrm{R}\setminus 0$$

Solve for b

$b\in \mathrm{R}\setminus -a,0$
$a\neq 0$

Solution Steps

Subtract $\frac{ab}{a+b}$ from both sides.

$$\left(a^{-1}+b^{-1}\right)^{-1}-\frac{ab}{a+b}=0$$

Reorder the terms.

$$-\frac{ab}{a+b}+\frac{1}{\frac{1}{a}+\frac{1}{b}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $a$ and $b$ is $ab$. Multiply $\frac{1}{a}$ times $\frac{b}{b}$. Multiply $\frac{1}{b}$ times $\frac{a}{a}$.

$$-\frac{ab}{a+b}+\frac{1}{\frac{b}{ab}+\frac{a}{ab}}=0$$

Since $\frac{b}{ab}$ and $\frac{a}{ab}$ have the same denominator, add them by adding their numerators.

$$-\frac{ab}{a+b}+\frac{1}{\frac{b+a}{ab}}=0$$

Variable $b$ cannot be equal to $0$ since division by zero is not defined. Divide $1$ by $\frac{b+a}{ab}$ by multiplying $1$ by the reciprocal of $\frac{b+a}{ab}$.

$$-\frac{ab}{a+b}+\frac{ab}{b+a}=0$$

Combine $-\frac{ab}{a+b}$ and $\frac{ab}{b+a}$ to get $0$.

$$0=0$$

Compare $0$ and $0$.

$$\text{true}$$

This is true for any $b$.

$$b\in \mathrm{R}$$

Variable $b$ cannot be equal to $0$.

$$b\in \mathrm{R}\setminus 0$$