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Linear equations - standard form, slope intercept form, point-slope form











































(a) Rewrite the following linear equation in the standard form, slope intercept form, and point-slope form. (b) Identify the slope and the y intercept.


2y = 4x + 16




Comments for Linear equations - standard form, slope intercept form, point-slope form

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Jun 27, 2012
Linear equations
by: Staff


The answer:


Summary for formats:


Standard Form: Ax + By = C
A, B, and C are constants


Slope Intercept Form: y = mx + b
m = slope
b = y intercept


Point-Slope Form: (y - y₁) = m (x - x₁)
this format uses a single known point (x₁,y₁) and the slope m (which is also a known value)



2y = 4x + 16



Standard Form: Ax + By = C

2y = 4x + 16

2y - 4x = 4x - 4x + 16

2y - 4x = 0 + 16

>>> 2y - 4x = 16, Standard Form


Slope Intercept Form: y = mx + b

2y = 4x + 16

2y / 2 = (4x + 16) / 2

y * (2/ 2) = (4x + 16) / 2

y * (1) = (4x + 16) / 2

y = (4x + 16) / 2

y = 4x / 2 + 16 / 2

y = (4/2)x + 16 / 2

>>> y = 2x + 8, Slope Intercept Form

>>> m (the slope) = 2

>>> b (the y intercept) = 8



Point-Slope Form: (y - y₁) = m (x - x₁)

2y = 4x + 16

Compute two points (x₁,y₁) and (x₂,y₂). Both points are needed to compute the slope “m”.

If x₁ = 0

2y₁ = 4x₁ + 16

2y₁ = 4*0 + 16

2y₁ = 0 + 16

2y₁ = 16

2y₁ / 2 = 16 / 2

y₁*(2 / 2) = 16 / 2

y₁*(1) = 16 / 2

y₁ = 16 / 2

y₁ = 8

(x₁,y₁) = (0, 8)



If x₂ = 1

2y₂ = 4x₂ + 16

2y₂ = 4*1 + 16

2y₂ = 4 + 16

2y₂ = 20

2y₂ / 2 = 20 / 2

y₂*(2 / 2) = 20 / 2

y₂*(1) = 20 / 2

y₂ = 20 / 2

y₂ = 10

(x₂,y₂) = (1, 10)


Slope “m”

m = (y₂ - y₁) / (x₂ - x₁)

m = (10 - 8) / (1 - 0)

m = 2 / 1
m = 2

(y - y₁) = m (x - x₁)

>>> (y - 8) = 2 (x - 0), Point-Slope Form





Thanks for writing.

Staff
www.solving-math-problems.com


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