Solve the equation (x+3/6)/(1-1(3x)/6)=(2-1x)/(1+2x)

\frac{x+\frac{3}{6}}{1-\frac{3x}{6}}=\frac{2-x}{1+2x}

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Answer

$x+2x^2+x^{3}+4x-4x^2=0$

Step by step solution

Problem

$\frac{x+\frac{3}{6}}{1-\frac{3x}{6}}=\frac{2-x}{1+2x}$
1

Apply fraction cross-multiplication

$\left(2x+1\right)\left(\frac{1}{2}+x\right)=\left(2-x\right)\left(1-\frac{3x}{6}\right)$
2

Multiplying polynomials $2x$ and $x+\frac{1}{2}$

$\frac{1}{2}+x+x+2x^2=x\frac{3x}{6}-x-2\frac{3x}{6}+2$
3

Simplify the fraction

$\frac{1}{2}+x+x+2x^2=x\frac{3x}{6}-x-x+2$
4

Rewrite the equation

$-\left(x\frac{3x}{6}-x-x+2\right)+x+2x^2=0$
5

Adding $-x$ and $-x$

$-\left(x\frac{3x}{6}+2-2x\right)+x+2x^2=0$
6

Multiplying polynomials $2x$ and $-2x+2$

$x+2x^2+2x\cdot x\frac{3x}{6}+4x-4x^2=0$
7

When multiplying exponents with same base you can add the exponents

$x+2x^2+2\frac{3x}{6}x^2+4x-4x^2=0$
8

Simplify the fraction

$x+2x^2+1xx^2+4x-4x^2=0$
9

Any expression multiplied by $1$ is equal to itself

$x+2x^2+xx^2+4x-4x^2=0$
10

When multiplying exponents with same base you can add the exponents

$x+2x^2+x^{3}+4x-4x^2=0$

Answer

$x+2x^2+x^{3}+4x-4x^2=0$

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Problem Analysis

Main topic:

Polynomials

Time to solve it:

0.48 seconds

Views:

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