**Algebraic manipulation** involves rearranging variables to make an algebraic expression better suit your needs. During this rearrangement, the value of the expression does not change.

Algebraic expressions aren't always given in their most convenient forms. This is where algebraic manipulation comes in.

For example:

## What value of \(x\) satisfies \(5x+8 = -2x +43\)

We can rearrange this equation for \(x\) by putting the terms with \(x\) on one side and the constant terms on the other. \[\begin{align}

5x+8 &= -2x +43 \\

5x -(-2x) &= 43 -8 \\

7x &= 35 \\

x &= \frac{35}{7} \\

x &= 5 \quad_\square \end{align}\]

Algebraic manipulation is also used to simplify complicated-looking expressions by factoring and using identities. Let's walk through an example:

## If \(xy=6\) and \(x-y=-2\), find the value of

## \[ \frac{x^3+y^3}{x^2-y^2} - \frac{x^2+y^2}{x-y}.\]

It's possible to solve for \(x\) and \(y\) and plug those values into this expression, but the algebra would be very messy. Instead, we can rearrange the problem by using the factoring formula identities for \(x^3+y^3\) and \(x^2-y^2\) and then simplifying. \[\begin{align}

\frac{x^3+y^3}{x^2-y^2} - \frac{x^2+y^2}{x-y} &= \frac{(x+y)(x^2-xy+y^2)}{(x-y)(x+y)} - \frac{x^2+y^2}{x-y} \\ &= \frac{x^2-xy+y^2 -(x^2+y^2)}{x-y} \\

&= \frac{-xy}{x-y}

\end{align}\] Plugging in the values for \(xy\) and \(x-y\) gives us the answer of \(3\).\(_\square\)

## If \(x+\frac{1}{x}=8\), what is the value of \(x^3+\frac{1}{x^3}\)?

The key to solving this problem (without explicitly solving for \(x\)) is to recognize that \[\left(x+\frac{1}{x}\right)^3 = x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)\] which gives us \[\begin{align}

x^3+\frac{1}{x^3} &= \left(x+\frac{1}{x}\right)^3 - 3\left(x+\frac{1}{x}\right) \\

&= (8)^3 -3(8) \\

&= 488 \quad _\square \end{align}\]

## If \(\sqrt{2x+1}-\sqrt{x-3}=2\), what is the value of

## \[\frac{2x+8}{\sqrt{2x+1}+\sqrt{x-3}}?\]

This problem is easy once you realize that \[\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)=x+4.\] The solution is therefore \[\begin{align}

\frac{2x+8}{\sqrt{2x+1}+\sqrt{x-3}} &= \frac{2(x+4)}{\sqrt{2x+1}+\sqrt{x-3}} \\

&=\frac{2\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)}{\sqrt{2x+1}+\sqrt{x-3}} \\

&=2\left(\sqrt{2x+1}-\sqrt{x-3}\right) \\

&=2(2) \\

&=4 \quad _\square \end{align}\]

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## Comments

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TopNewestYou have posted very good solutions these kind of questions.

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The first one of applications and extensions has a mistake

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Thanks, it's fixed now.

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