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# How can i do this? i don't know how to answer this cause fractions..

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168
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+42

x= -8/3 , 2(3x^2+24x+48)+1(x-6)=2(1)

TO EACH OTHER?

Virax1o1  Jun 1, 2017
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#1
+5931
+2

If    $$x=\,-\,\frac83$$   , does    $$2(3x^2+24x+48)+1(x-6)=2(1)$$   ?

Replace every   " x "   with   $$-\,\frac83$$   in the second equation.

$$2(3(-\frac83)^2+24(-\frac83)+48)+1((-\frac83)-6)\stackrel{?}{=}2(1) \\~\\ 2(3(\frac{64}{9})-64+48)+(-\frac{26}3)\stackrel{?}{=}2 \\~\\ 2(\frac{64}{3}-16)-\frac{26}3\stackrel{?}{=}2 \\~\\ 2(\frac{16}{3})-\frac{26}3\stackrel{?}{=}2 \\~\\ \frac{32}{3}-\frac{26}3\stackrel{?}{=}2 \\~\\ \frac63\stackrel{?}{=}2 \\~\\ 2=2\quad\text{ true!}$$

hectictar  Jun 1, 2017
#2
+42
+1

Thank you very much!!

Virax1o1  Jun 1, 2017

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