# How to go about solving this equation?

How would I go about solving this equation?

$$\frac{2x + 9}{x} = \frac{x + 8}{4}$$

I was trying to solve it by first (note: Contains a mistake, x on left side should be 2, leaving it in, see accepted answer for details) $$x + \frac{9}{x} = \frac{x}{4} + 2$$

Then I get stuck. A friend of mine in class told me I could do this instead $$(2x + 9) * 4 = (x + 8) * x$$ $$8x + 36 = x^2 + 8x$$ $$36 = x^2$$ $$x = 6$$

So I would like to know:

1. What is the name of this rule my friend told me about
2. Is that the only/preferred way to solve this type of equations?

Thank you.

PS: This is not homework per se, it is part of the excersices in my math book. But I guess you can call that homework so I'll tag it as such.

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There are three main procedures you can use to solve an equation

$\ \ \ \$1) Simplify one or both sides of the equation.

$\ \ \ \$2) Multiply or divide both sides of the equation by hte same non-zero number.

$\ \ \ \$3) Add or subtract the same number to both sides of the equation.

Given an equation, performing one of the above operations to the equation will produce an equation that has exactly the same solutions as the original; that is, the resulting equation is equivalent to the original.

Your friend used rule 2) twice without showing the steps:

\eqalign{ &\color{maroon}{{2x+9\over x}= {x+8\over 4}}\cr \iff& 4\cdot{2x+9\over x}= {x+8\over 4}\cdot4\cr \iff& {4(2x+9)\over x}= {x+8 } \cr \Longrightarrow& x\cdot {4(2x+9)\over x}= {(x+8) } \cdot x\cr \Longrightarrow& \color{darkgreen}{ {4(2x+9) }= {x(x+8) }} \cr }

Note that the forth step can only be done if $x\ne 0$. The original equation does not have $x=0$ as a solution (the original equation isn't even well-defined for $x=0$). So, if perchance the equation at the end had $0$ as a solution, this would not be a solution of the original equation. Any non-zero solution of the last equation, though, would be a solution of the original.

The shorcut to the above could be called "cross cancellation"; you go immediately from the $\color{maroon}{\text{first equation above}}$ to the $\color{darkgreen}{\text{last equation above}}$ by recognizing that if you initially multiplied both sides of the first equation above by $4x$ (the product of the denominators) the cancellations produced will give the last equation.

An even shorter way to get to the last equation is to move denominators from one side to numerators of the other: $${2x+9\over \color{maroon}{x}}= {x+8\over \color{darkgreen}{4}} \Rightarrow \color{darkgreen}{4}{(2x+9) }= {(x+8 )} \color{maroon}{x}$$ I advise, though, to not use these sort of "tricks". Instead, when solving equations, think about how you get from one stage to another and what rule allows you to do it.

As for your second question, there are other ways to solve this equation (slightly different; but eventually you'll have to deal with clearing the $x$ in the denominator). We could take your approach, starting with $$2+{9\over x}={x\over 4}+2.$$ (note you made a mistake here). Then subtract 2 from both sides $${9\over x}={x\over 4}.$$ Note $x=0$ is not a solution. Now multiply both sides by $4x$ to produce $$36={x^2 }.$$ Solving for $x$ gives $x=6$ or $x=-6$.

You could also bring everything over to one side, write it as one fraction, and set the numerator equal to zero. The solutions to the original equation will be the values of $x$ where both the numerator of this fraction is 0 and the denominator is non-zero. Note this is more or less the first method. \eqalign{ &{2x+9\over x}= {x+8\over 4}\cr & {2x+9\over x}- {x+8\over 4}=0\cr & {4\over 4}\cdot {2x+9\over x}-{x\over x}\cdot {x+8\over 4}=0 \cr & {4(2x+9) - x(x+8)\over 4x }=0 \cr & {4(2x+9) }- {x(x+8) =0 } \cr }

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Thank you for a great answer! :) – Storm M Feb 19 '12 at 11:20

its basic equation rules, what you did was finding a common denominator. Also, the square roots of $x^2 = 36$ are $6$ and $-6$. because if you multiply $-6\cdot-6$ it equals to $36$ also

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Thank you. I think I need to go back and study some of these basic rules. – Storm M Feb 18 '12 at 12:03
Quadratic equations will not always fall out quite so neatly - you might also want to investigate the technique of 'completing the square', which gets more complicated equations into a form you can solve. – Mark Bennet Feb 18 '12 at 13:25