# Evaluate and simplify f(b/3)

#### mbrez2

assume that f(x)= x+7/x^2+4 for every real number x. evaluate and simplify f(b/3).

#### Plato

MHF Helper
assume that f(x)= x+7/x^2+4 for every real number x. evaluate and simplify f(b/3).
This is basic simple algebra: $\displaystyle{\frac{{\frac{b}{3} + 7}}{{{{\left( {\frac{b}{3}} \right)}^2} + 4}}}$

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#### mbrez2

Thanks for the response. That is what I got originally. Unfortunately, the online platform I'm entering the answer into is marking it wrong. I guess I'll have to play around with it for a little bit. Thanks again.

#### Plato

MHF Helper
Thanks for the response. That is what I got originally. Unfortunately, the online platform I'm entering the answer into is marking it wrong. I guess I'll have to play around with it for a little bit. Thanks again.
The OP does contain some ambiguity. Review the question to be sure if you need to add grouping symbols.

#### JeffM

If the question asked you to find and simplify, then Plato was leaving you to do the simplification. The computer may not accept anything but a fully simplified answer, which is NOT

$\dfrac{\dfrac{b}{3} + 7}{\left ( \dfrac{b}{3} \right )^2 + 4}.$

#### mbrez2

so is the simplified version of Plato's answer 7/(b/3)+4?

#### JeffM

so is the simplified version of Plato's answer 7/(b/3)+4?
No, not at all, and please use grouping symbols properly. You mean 7/{(b/3) + 4}. You need to review PEMDAS.

$\dfrac{\dfrac{b}{3} + 7}{\left ( \dfrac{b}{3} \right )^2 + 4}$ $\ne$ $\dfrac{\cancel {\dfrac{b}{3}} + 7}{\left ( \cancel {\dfrac{b}{3}} * \dfrac{b}{3} \right ) + 4} = \dfrac{7}{\dfrac{b}{3} + 4}.$ THIS IS WRONG.

For example

$\dfrac{3 + 12}{3 + 2} = \dfrac{15}{5} = 3\ NOT\ \dfrac{\cancel 3 + 12}{\cancel 3 + 2} = \dfrac{12}{2} = 6.$

You must turn both numerator and denominator into complete fractions, and then invert the denominator to multiply by the numerator.

$\dfrac{\dfrac{b}{3} + 7}{\left ( \dfrac{b}{3} \right )^2 + 4} = \dfrac{\dfrac{b}{3} + \dfrac{21}{3}}{\dfrac{b^2}{9} + 4} =$

$\dfrac{\dfrac{b + 21}{3}}{\dfrac{b^2}{9} + \dfrac{36}{9}} = \dfrac{\dfrac{b + 21}{3}}{\dfrac{b^2 + 36}{3 * 3}} =\dfrac{b + 21}{\cancel 3} * \dfrac{\cancel 3 * 3}{b^2 + 36} = \dfrac{3(b + 21)}{b^2 + 36}.$

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