What is the answer to -2x-(4^0-x)/3+(x+3^0)/5 =3 ?

The answer to -2x-(4^0-x)/3+(x+3^0)/5 =3 is x=-47/22

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-2x-(4^0-x)/3+(x+3^0)/5 =3

- 2 x - \frac{4 ^{0} - x}{3} + \frac{x + 3 ^{0}}{5} = 3

x = - \frac{47}{22}

Decimal Notation

x = - 2.13636 \dots

Solution steps

- 2 x - \frac{4 ^{0} - x}{3} + \frac{x + 3 ^{0}}{5} = 3

Apply rule

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

4^{0} = 1, 3^{0} = 1

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

Multiply by LCM

- 30 x - 5 (- x + 1) + 3 (x + 1) = 45

Expand

- 30 x - 5 (- x + 1) + 3 (x + 1) : - 22 x - 2

- 22 x - 2 = 45

Move

2

to the right side

- 22 x = 47

Divide both sides by

- 22

x = - \frac{47}{22}

- 9 (- 5 g + 1) = 4 (9 g - 9) - 9

7 k + 8 = 2 k - 37

x, - x + 2 y = 7

x, 2 x + 4 y = 1

w, 2 l + 2 w = p

What is the answer to -2x-(4^0-x)/3+(x+3^0)/5 =3 ?
The answer to -2x-(4^0-x)/3+(x+3^0)/5 =3 is x=-47/22
Solve for x: -2x-(4^0-x)/3+(x+3^0)/5 =3
The solution is x=-47/22