Algebraic Manipulation


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 xx satisfies 5x+8=2x+435x+8 = -2x +43

We can rearrange this equation for xx by putting the terms with xx on one side and the constant terms on the other. 5x+8=2x+435x(2x)=4387x=35x=357x=5\begin{aligned} 5x+8 &= -2x +43 \\ 5x -(-2x) &= 43 -8 \\ 7x &= 35 \\ x &= \frac{35}{7} \\ x &= 5 \quad_\square \end{aligned}

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

If xy=6xy=6 and xy=2x-y=-2, find the value of

x3+y3x2y2x2+y2xy. \frac{x^3+y^3}{x^2-y^2} - \frac{x^2+y^2}{x-y}.

It's possible to solve for xx and yy 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 x3+y3x^3+y^3 and x2y2x^2-y^2 and then simplifying. x3+y3x2y2x2+y2xy=(x+y)(x2xy+y2)(xy)(x+y)x2+y2xy=x2xy+y2(x2+y2)xy=xyxy\begin{aligned} \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{aligned} Plugging in the values for xyxy and xyx-y gives us the answer of 33._\square

Application and Extensions

If x+1x=8x+\frac{1}{x}=8, what is the value of x3+1x3x^3+\frac{1}{x^3}?

The key to solving this problem (without explicitly solving for xx) is to recognize that (x+1x)3=x3+1x3+3(x+1x)\left(x+\frac{1}{x}\right)^3 = x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right) which gives us x3+1x3=(x+1x)33(x+1x)=(8)33(8)=488\begin{aligned} 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{aligned}


If 2x+1x3=2\sqrt{2x+1}-\sqrt{x-3}=2, what is the value of


This problem is easy once you realize that (2x+1+x3)(2x+1x3)=x+4.\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)=x+4. The solution is therefore 2x+82x+1+x3=2(x+4)2x+1+x3=2(2x+1+x3)(2x+1x3)2x+1+x3=2(2x+1x3)=2(2)=4\begin{aligned} \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{aligned}

Note by Arron Kau
5 years, 4 months ago

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You have posted very good solutions these kind of questions.

Anuj Shikarkhane - 5 years, 1 month ago

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The first one of applications and extensions has a mistake

Daniel Lim - 5 years ago

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Thanks, it's fixed now.

Arron Kau Staff - 5 years ago

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