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

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

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solvefor a,0=ae^0+be^{-0}

solve for

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

a = - b

Solution steps

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

Switch sides

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

Apply rule

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

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

1 \cdot a + 1 \cdot b = 0

Multiply:

a \cdot 1 = a

a + 1 \cdot b = 0

Multiply:

b \cdot 1 = b

a + b = 0

Move

b

to the right side

a = - b

5 x + [- 2 x + (- x + 6)] = 18 - [- (7 x + 6) - (3 x - 24)]

- 2 - 2 x = 0

x + 5 = 37

- 5 + x = - 4

x, 3 x + y = 30

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