What is the answer to solvefor y,[10y+1-1]=[x+3z^5-(-1)] ?

The answer to solvefor y,[10y+1-1]=[x+3z^5-(-1)] is y=(x+3z^5+1)/(10)

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solvefor y,[10y+1-1]=[x+3z^5-(-1)]

solve for

y, [10 y + 1 - 1] = [x + 3 z^{5} - (- 1)]

y = \frac{x + 3 z ^{5} + 1}{10}

Solution steps

[10 y + 1 - 1] = [x + 3 z^{5} - (- 1)]

Apply rule

- (- a) = a

[10 y + 1 - 1] = x + 3 z^{5} + 1

Divide both sides by

10

y = \frac{x + 3 z ^{5} + 1}{10}

4 x + 2 a - 5 x + 6 = 40

x, 6 x - 4 y = 3

\frac{( 3 )}{3} = \frac{x}{300}

y, x^{3} + 3 y - x y = 0

2 (x^{2} - x) - 2 x (x + 3) = 5 x - 1

What is the answer to solvefor y,[10y+1-1]=[x+3z^5-(-1)] ?
The answer to solvefor y,[10y+1-1]=[x+3z^5-(-1)] is y=(x+3z^5+1)/(10)