Two quick questions regarding simplification

[PCSL]

Could someone please explain how

\frac{x^2}{x^2+2}=1-\frac{2}{x^2+2}

and how

xlnx-x=xln(x-1)

Also, is there a list of types of simplifications like these two that I could study? I'm in Calc II and my own trouble comes from simple things like this that I should already know.

[Mark44]

Could someone please explain how

\frac{x^2}{x^2+2}=1-\frac{2}{x^2+2}
Start on the right side, and get a common denominator.


and how

xlnx-x=xln(x-1)
Are you sure you copied this correctly, because as you have it, it's not an identity.

xlnx - x = x(lnx - 1), by factoring x from both terms on the left side of the equation.

I'm not sure that there's a list of types of simplifications. It's probably better to go back and review the techniques of algebra, in your textbook, if you still have it, or online at a site such as Khan Academy.


Also, is there a list of types of simplifications like these two that I could study? I'm in Calc II and my own trouble comes from simple things like this that I should already know.

[symbolipoint]

Could someone please explain how

\frac{x^2}{x^2+2}=1-\frac{2}{x^2+2}

....

You can multiply the left side and the right side by x2+2.

[AJKing]

You can multiply the left side and the right side by x2+2.

Let's not make things unnecessarily complicated. He only needs to deal with one side of this equation.

[Mentallic]

Could someone please explain how

\frac{x^2}{x^2+2}=1-\frac{2}{x^2+2}

It basically takes advantage of this rule for fractions

\frac{a}{a+b}=\frac{a+b-b}{a+b}=\frac{a+b}{a+b}+\frac{-b}{a+b}=1-\frac{b}{a+b}

[symbolipoint]

Let's not make things unnecessarily complicated. He only needs to deal with one side of this equation.

Yes, I see what you mean:

From equation shown, \frac{x^2}{x^2+2}=1-\frac{2}{x^2+2}

then,
\[
\begin{array}{l}
\left( {\frac{{x^2 }}{{x^2 + 2}}} \right)\left( {x^2 + 2} \right) = \left( {1 - \frac{2}{{x^2 + 2}}} \right)\left( {x^2 + 2} \right) \\
x^2 = (x^2 + 2) - 2 \\
x^2 = x^2 \\
\end{array}
\]

[Mark44]

Yes, I see what you mean:

From equation shown, \frac{x^2}{x^2+2}=1-\frac{2}{x^2+2}

then,
\[
\begin{array}{l}
\left( {\frac{{x^2 }}{{x^2 + 2}}} \right)\left( {x^2 + 2} \right) = \left( {1 - \frac{2}{{x^2 + 2}}} \right)\left( {x^2 + 2} \right) \\
x^2 = (x^2 + 2) - 2 \\
x^2 = x^2 \\
\end{array}
\]


The trouble with this approach is that you are tacitly assuming that the two quantities of the first equation are equal when you multiply both sides by x2 + 2. The bottom line here shows that x2 = x2, which is true, but not useful. A complete proof would go on to show that since each operation applied is reversible (one-to-one, then each equation above is equivalent.

An easier and more straightforward way to do this problem is to show that the right side of the original equation can be manipulated to produce the left side.

[AJKing]

Yes, I see what you mean: [. . .]

To simplify your solution:

\frac{x^2}{x^2+2}=1-\frac{2}{x^2+2}

\frac{x^2}{x^2+2}=(1*\frac{x^2+2}{x^2+2})-\frac{2}{x^2+2}

Multiplying the numerator and denominator by the same values does not change the quotient of a fraction.

Now we simplify and the conclusion is obvious:

\frac{x^2}{x^2+2}=\frac{x^2}{x^2+2}

[symbolipoint]

Now it's clearer. The poster wanted to know HOW. I missed seeing that and attempted to solve an equation, while what was really needed was "verify the identity".