Solve the equation -3y-3y+y^2x*-3+2x=33x+4y

2x-3y^2\cdot x-3y-3y=33x+4y

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Answer

$-3xy^2+2x+6y-4y-33x=0$

Step by step solution

Problem

$2x-3y^2\cdot x-3y-3y=33x+4y$
1

Adding $-3-3y$ and $-3y$

$-3xy^2+2x-2\left(-3\right)y=4y+33x$
2

Multiply $-3$ times $-2$

$-3xy^2+2x+6y=4y+33x$
3

Grouping terms

$-\left(4y+33x\right)-3xy^2+2x+6y=0$
4

Multiply $\left(33x+4y\right)$ by $-1$

$-3xy^2+2x+6y-4y-33x=0$

Answer

$-3xy^2+2x+6y-4y-33x=0$

Problem Analysis

Main topic:

Polynomials

Time to solve it:

0.62 seconds

Views:

120