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Calculus help? Anonymous 17/09/28(Thu)18:49 No. 16553 ID: ebea66

File 150661734419.jpg - (3.64MB , 4608x3456 , 20170928_124631.jpg )

Hey guys, quick noob math question if I may.. in the picture i posted you see the equation y-1=(-1/2)(x-(-4))

Its then simplified to 2y -2 = -x-4
And then again to 2y+x+2=0

..my question is could someone breakdown the simplifying into steps? Im not sure the process they used to arrive at their final answer

Anonymous 17/09/28(Thu)19:08 No. 16554 ID: ebea66

Oh god the math

Anonymous 17/09/30(Sat)03:24 No. 16555 ID: 012486

y - 1 = (-1/2)(x - (-4))
(y - 1)2 = 2(-1/2)(x - (-4))
2y - 2 = (-2/2)(x - (-4))
2y - 2 = (-1)(x - (-4))
2y - 2 = -x + (-4)
2y - 2 = -x - 4
2y - 2 - (-x - 4) = -x - 4 - (-x - 4)
2y - 2 + x + 4 = 0
2y + x + 2 = 0

Anonymous 17/11/25(Sat)02:28 No. 16583 ID: 4b785f

This ain't no gawt daing calcuebutts!

Anonymous 18/01/21(Sun)17:25 No. 16616 ID: 10df8c

This isn't even calculus. This is very basic week 1 algebra, and I recommend revisiting it before you attempt "calculus".
The first step was that they distributed the -(1/2) term across the right hand side of the equation and then multiplied the entire equation by 2. Then they simply brought all terms to one side to set the equation equal to zero.

Anonymous 18/04/10(Tue)15:03 No. 16645 ID: f47f1a

That picture is upside-down BTW.

re Anonymous 18/06/13(Wed)12:52 No. 16660 ID: 1bb067

Equation of a line if slope (m) and a point (x1,y1) along which line passes is given
Equation is given by -

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