Algebraic Manipulation

Algebraic manipulation refers to the manipulation of algebraic expressions, often into a simpler form or a form which is more easily handled and dealt with. It is one of the most basic, necessary and important skills in a problem solver's repertoire, as without it a problem solver would hopelessly be stuck on innumerable problems.

The skill of algebraic manipulation is acquired through practice and solving problems. This wiki serves to explain the different uses, benefits of such manipulation and the different ways to manipulate complicated algebraic expressions to an easier and more beneficial form.

It is useful to memorize the following algebraic expansions of various formulae, as you will encounter them almost everywhere as you continue on your journey of problem solving:

• \((a\pm b)^2= a^2\pm 2ab+b^2\)
• \((a+b)(a-b) = a^2 - ab + ab - b^2 = a^2 - b^2\)
• \((a+ b)^3=a^3+ b^3+ 3ab(a+ b)=a^3+ 3a^2b+ 3ab^2 + b^3\)
• \((a- b)^3=a^3- b^3- 3ab(a- b)=a^3- 3a^2b+ 3ab^2 - b^3\)
• \((a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=a^2+b^2+c^2+2ab+2ac+2bc.\)

Algebraic manipulation involves rearranging and substituting for variables to obtain an algebraic expression in a desired form. During this rearrangement, the value of the expression does not change. Algebraic expressions are not always given in their most convenient forms, and algebraic manipulation may be necessary to find the desired solution to a problem.

If \(y=8\), what is the value of \(x\) that satisfies \(5x+y = -2x +43?\)

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We first substitute \(y=8\) into the equation to obtain the equation \(5x+8 = -2x +43.\) We can rearrange this equation for \(x\) by putting the terms with \(x\) on one side and the constant terms on the other to obtain

\[\begin{align} 5x+8 &= -2x +43 \\ 5x -(-2x) &= 43 -8 \\ 7x &= 35 \\ x &= \frac{35}{7} \\ x &= 5. \ _\square \end{align}\]

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

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

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We have

\[\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)=x+4.\]

The solution is therefore

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

Algebraic manipulation is also used to simplify complicated-looking expressions by factoring and using identities.

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

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

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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 as follows:

\[\begin{align} \frac{x^3+y^3}{x^2-y^2} - \frac{x^2+y^2}{x-y} &= \frac{(x+y)(x^2-xy+y^2)}{(x-y)(x+y)} - \frac{x^2+y^2}{x-y} \\ &= \frac{x^2-xy+y^2 -(x^2+y^2)}{x-y} \\ &= \frac{-xy}{x-y}. \end{align}\]

Plugging in the values for \(xy\) and \(x-y\) gives us the answer, which is \(3\). \(_\square\)