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Hi! I need help finding the slope of this line and the x and y intercepts.

0 votes
(1/6)x + 4/3 = - (1/2)y.
asked Mar 13, 2014 in PRE-ALGEBRA by dkinz Apprentice

1 Answer

+1 vote

The equation is (1/6)x + (4/3) = - (1/2)y.

 (- 1/2)y = (1/6)x + (4/3)

Multiply -2 to each side.

(- 2) (- 1/2)y = (- 2)(1/6)x + (4/3)(- 2)

Cancel common terms.

y = (- 1/3)x - (8/3).

Compare the above equation with slope - intercept - form y =mx + b.

where m is slope and b is y - intercept.

Slope (m) = - 1/3.

Therefore,slope of this line is - 1/3.

Line equation with intercets :  x /a + y /b = 1.

where a is x - intercept and  b is y - intercept.

(1/6)x + (4/3) = - (1/2)y

Subtract 4/3 from each side.

(1/6)x + (4/3) - (4/3) = - (1/2)y - (4/3)

(1/6)x = - (1/2)y - (4/3)

Add (1/2)y to each side.

(1/6)x + (1/2)y = - (1/2)y - (4/3) + (1/2)y

(1/6)x + (1/2)y = - 4/3

Divide each side by negative 4/3.

[(1/6)/(- 4/3)]x + [(1/2)/(- 4/3)]y = [(- 4/3)/(- 4/3)]

Cancel common terms.

(- 1/8)x + (- 3/8)y = 1

x /(- 8) + y /(- 8/3) = 1.

Compare the above equation with x /a + y /b = 1.

The x and y intercepts are - 8 and - 8/3.

answered Mar 18, 2014 by dozey Mentor
edited Mar 18, 2014 by dozey

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