1. ## Please explainhow did he simplified this given?

Hi, Can you please show me the step-by-step process of this example. Especially the second and third line?

Here is the given:
$\displaystyle (xy-x)dx=(x^2y^2-x^2)+(y^2-1)dy$
$\displaystyle x(y-1)dx=(x^2(y^2-1)+(y^2-1))dy$
$\displaystyle =(x^2+1)(y^2-1)dy$
$\displaystyle xdx=(x^2+1)(y+1)dy$

Any help will be appreciated, thanks

2. Originally Posted by davewhite123
Hi, Can you please show me the step-by-step process of this example. Especially the second and third line?

Here is the given:
$\displaystyle (xy-x)dx=(x^2y^2-x^2)+(y^2-1)dy$
$\displaystyle x(y-1)dx=(x^2(y^2-1)+(y^2-1))dy$
$\displaystyle =(x^2+1)(y^2-1)dy$
$\displaystyle xdx=(x^2+1)(y+1)dy$

Any help will be appreciated, thanks
$\displaystyle (xy-x)dx=(x^2y^2-x^2)+(y^2-1)dy$

x is a common factor of xy and x, so $\displaystyle xy-x=x(y-1)$

$\displaystyle x^2$ is a common factor of $\displaystyle x^2y^2$ and $\displaystyle x^2$, so $\displaystyle x^2y^2-x^2=x^2(y^2-1)$

Now $\displaystyle y^2-1$ is a common factor, so $\displaystyle x^2(y^2-1)+(y^2-1)=(y^2-1)(x^2+1)$

Finally, $\displaystyle y^2-1=(y+1)(y-1)$

hence y-1 is common to both sides and can be eliminated to obtain the last line.