Waste less time on Facebook — follow Brilliant.
×

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:

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 \(3\).\(_\square\)

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 \[\begin{align}
x^3+\frac{1}{x^3} &= \left(x+\frac{1}{x}\right)^3 - 3\left(x+\frac{1}{x}\right) \\
&= (8)^3 -3(8) \\
&= 488 \quad _\square \end{align}\]

 

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 \[\begin{align}
\frac{2x+8}{\sqrt{2x+1}+\sqrt{x-3}} &= \frac{2(x+4)}{\sqrt{2x+1}+\sqrt{x-3}} \\
&=\frac{2\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)}{\sqrt{2x+1}+\sqrt{x-3}} \\
&=2\left(\sqrt{2x+1}-\sqrt{x-3}\right) \\
&=2(2) \\
&=4 \quad _\square \end{align}\]

Note by Arron Kau
3 years, 10 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

You have posted very good solutions these kind of questions.

Anuj Shikarkhane - 3 years, 6 months ago

Log in to reply

The first one of applications and extensions has a mistake

Daniel Lim - 3 years, 6 months ago

Log in to reply

Thanks, it's fixed now.

Arron Kau Staff - 3 years, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...