What is the answer to solvefor b,1=ae^0-be^{-0} ?

The answer to solvefor b,1=ae^0-be^{-0} is b=-1+a

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solvefor b,1=ae^0-be^{-0}

solve for

b, 1 = a e^{0} - b e^{- 0}

b = - 1 + a

Solution steps

1 = a e^{0} - b e^{- 0}

Switch sides

a e^{0} - b e^{- 0} = 1

Apply rule

a^{0} = 1, a \neq = 0

e^{0} = 1, e^{- 0} = 1

1 \cdot a - 1 \cdot b = 1

Multiply:

a \cdot 1 = a

a - 1 \cdot b = 1

Multiply:

b \cdot 1 = b

a - b = 1

Move

a

to the right side

- b = 1 - a

Divide both sides by

- 1

b = - 1 + a

3 x + 4 (x - 10) = x + 20

- 3 x - [- 6 x - (3 - x)] = 9 + (x - 1)

\frac{3 x - 2}{4} = \frac{2 x + 1}{5}

24 x - 4 = 0

72 f - 12 g = 96

What is the answer to solvefor b,1=ae^0-be^{-0} ?
The answer to solvefor b,1=ae^0-be^{-0} is b=-1+a