What is the answer to solvefor y,x=\sqrt[3]{2y-5}+4 ?

The answer to solvefor y,x=\sqrt[3]{2y-5}+4 is y=(x^3-12x^2+48x-59)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor y,x=\sqrt[3]{2y-5}+4

solve for

y = \frac{x ^{3} - 12 x ^{2} + 48 x - 59}{2}

Solution steps

Switch sides

Move

4

to the right side

Take both sides of the equation to the power of

3 : 2 y - 5 = x^{3} - 12 x^{2} + 48 x - 64

2 y - 5 = x^{3} - 12 x^{2} + 48 x - 64

Solve

2 y - 5 = x^{3} - 12 x^{2} + 48 x - 64 : y = \frac{x ^{3} - 12 x ^{2} + 48 x - 59}{2}

y = \frac{x ^{3} - 12 x ^{2} + 48 x - 59}{2}

Verify Solutions:

y = \frac{x ^{3} - 12 x ^{2} + 48 x - 59}{2}

True

The solution is

y = \frac{x ^{3} - 12 x ^{2} + 48 x - 59}{2}

2 = x + 1

x + x + 6 = 0

2 t + 7 = - 3

x, 6 + 5 x = x

- x^{2} - 2 x + 3 = x + 3

What is the answer to solvefor y,x=\sqrt[3]{2y-5}+4 ?
The answer to solvefor y,x=\sqrt[3]{2y-5}+4 is y=(x^3-12x^2+48x-59)/2