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i have been trying to find the slope for years: -4x+3y=-41

 Apr 19, 2016
 #1
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\(-4x+3y=-41\)

 

\(3y=-41+4x\)

 

\(y=\frac{-41+4x}{3}\)

 

\(y=\frac{-41}{3}+\frac{4}{3}x\)

 

\(y=-\frac{41}{3}+\frac{4}{3}x\)

 

\(y=\frac{4}{3}x-\frac{41}{3}\)

 

\(slope=\frac{4}{3};\)  up \(4,\) right \(3\) or down \(4,\) left \(3\)

 Apr 19, 2016
edited by gibsonj338  Apr 19, 2016
 #2
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Find the derivative of the following via implicit differentiation:
d/dx(-4 x+3 y) = d/dx(-41)
Differentiate the sum term by term and factor out constants:
3 d/dx(y)-4 d/dx(x) = d/dx(-41)
The derivative of x is 1:
3 (d/dx(y))-1 4 = d/dx(-41)
The derivative of y is y'(x):
-4+3 y'(x) = d/dx(-41)
The derivative of -41 is zero:
-4+3 y'(x) = 0
Add 4 to both sides:
3 y'(x) = 4
Divide both sides by 3:
Answer: |  y'(x) = 4/3

 Apr 19, 2016

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