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# Check

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72
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I've been stuck on these for a while so I just want to make sure I understand them. Please check. Sorry.

Jan 15, 2019

#1
+4096
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12. Taking the x's and the constants to the other side, we get $$y=\frac{-6-x}{3x-1}.$$

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Jan 15, 2019
#2
+99585
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12

3xy  + x  = y - 6      rearrange as

y - 3xy  = x + 6       factor out y

y ( 1 - 3x)  = x + 6   divide both sides by 1 - 3x

y =   [ x + 6 ] / [ 1 - 3x ]

13

y^2 + 6y + 9  = x + 8      factor the left side

(y + 3)^2  =  x + 8          take both roots

y + 3 =  ±√ [ x + 8 ]        subtract 3 from both sides

y = - 3 ±√ [ x + 8 ]

18  looks good

19

-(3x - 2)^2  + 3(3x - 2) + 10

- [ 9x^2 - 12x + 4]  + 9x - 6 + 10

-9x^2 + 12x - 4  + 9x - 6 + 10

-9x^2 + 21x - 10 + 10

-9x^2 + 21x

Good job    !!!!!

Jan 15, 2019
edited by CPhill  Jan 15, 2019
#3
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Thank you for helping!! Sorry for the late reply. Computer shut down.

Guest Jan 15, 2019
#4
+99585
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OK......

CPhill  Jan 15, 2019