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# Help

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What is this 8 < 4x + 4 < 2(x + 10)

Dec 3, 2018

#1
+1035
-1

Split this:

$$8<4x+4, 4x+4<2x+20$$

$$4x+4<2x+20, 2x<16, x<8$$

Combining $$1 and \(x<8$$ we get

\(1

You are very welcome!

:P

Dec 3, 2018
edited by CoolStuffYT  Dec 3, 2018
#2
+107554
+1

Edit:

The LaTex is not displaying properly - I do not know whay, maybe it will fix itself

\( 8 < 4x + 4 < 2(x + 10)\\ 8 < 4x + 4 \quad and \quad 4x + 4 < 2(x + 10)\\ 4 < 4x \qquad and \qquad 4x + 4 < 2x + 20\\ 1 < x \qquad \;\;and \qquad \;\;\quad2x < 16\\ 1 < x \qquad \;\;and \qquad \;\;\quad x < 8\\~\\ 1

Dec 3, 2018
edited by Melody  Dec 3, 2018
edited by Melody  Dec 3, 2018
#4
+1035
0

Yes, that is pretty odd, my answer looked weird.

CoolStuffYT  Dec 3, 2018
#3
+107656
+1

8 < 4x + 4 < 2(x + 10)

We have two inequalities to consider

8 < 4x+ 4                                                   4x +4  < 2(x + 10)

Subtract 4 from both sides                        divide through by 2

4 < 4x                                                         2x + 2 < x + 10

Divide both sides by 4                               Subtract 2, x from both sides

1 < x    (1)                                                           x  < 8   (2)

Combining (1) and (2) we have

1 < x < 8

Dec 3, 2018