# Algebraic Manipulation

## Definition

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.

## Technique

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:

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

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 $$3$$.$$_\square$$

## Application and Extensions

### 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}

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

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}

Note by Arron Kau
4 years ago

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You have posted very good solutions these kind of questions.

- 3 years, 9 months ago

The first one of applications and extensions has a mistake

- 3 years, 9 months ago

Thanks, it's fixed now.

Staff - 3 years, 9 months ago

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