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# system

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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
+32401
+3

As follows:

Jun 24, 2021
#2
+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