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a and b are real numbers and satisfy a*b^2 = 27/5 and a^2*b = 1080. Compute a + 5b.

 Jun 24, 2021
 #1
avatar+32401 
+3

As follows:

 

 Jun 24, 2021
 #2
avatar+139 
+2

solving it this way is easier, i reckon.

 

$ \begin{cases} ab^2 = \frac{27}{5} \\  a^2b = 1080  \end{cases} $

 

lets work this -- $ ab^2=\frac{27}{5} $ ; take the $b^2$ to the other side and you get

 

$ a=\frac{27}{5b^2}   $

 

plugging that in the other equation you get \( (\frac{27}{5b^2})^2\times b=1080  \)

that is equal to $   \frac{729}{25b^3} \times b=1080 $

$ 729=27000b^3  $

$  b^3=\frac{27}{1000}$

$ b=\sqrt[3]{\frac{27}{1000}}   $

$  b=\frac{3}{10} $

 

back to our a equaltion:  $  ab^2 = \frac{27}{5} \ \ \ \Rightarrow \ \ \ \   a (\frac{3}{10})^2 = \frac{27}{5}  $ 

$a\frac{9}{100}=\frac{27}{5}$

$  \cancel{100}a\frac{9}{\cancel{100}}=\frac{27\times100}{5}  $

$  9a=540 $

$  a=60 $

 

now to our last condition $a+5b   \ \ \ \Rightarrow \ \ \  60+ 5(\frac{3}{10}) $

$  60+  \frac{5\times 3}{10} $

$  60+ \frac{3}{2}  $

$ \frac{60\times 2+3}{2} $

$  \frac{123}{2} $

 

:D

 Jun 24, 2021

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