What is (as+bs)(a+b)^{-1} ?

The solution to (as+bs)(a+b)^{-1} is s

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simplify (as+bs)(a+b)^{-1}

simplify

(a s + b s) (a + b)^{- 1}

s

Solution steps

(a s + b s) (a + b)^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

= (a s + b s) \frac{1}{a + b}

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( a s + b s ) \cdot 1}{a + b}

Apply rule:

a \cdot 1 = a

(a s + b s) \cdot 1 = (a s + b s)

= \frac{a s + b s}{a + b}

Factor out common term

s

= \frac{s ( a + b )}{a + b}

Cancel the common factor:

a + b

= s

e^{- 10000000000}

5.401 \cdot 1 0^{- 1}

9.5 \cdot 1 0^{- 4}

1.7 x 1 0^{- 6}