# Please explainhow did he simplified this given?

• Jan 31st 2010, 02:14 AM
davewhite123
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
• Jan 31st 2010, 04:11 AM
Quote:

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.