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Math: Solve systems of equations?

0 votes
1) -x + 2y = 2
x - 2y = 2

2) -x - 4y = 8
x - 4y = 8
asked Jun 27, 2013 in ALGEBRA 2 by angel12 Scholar

1 Answer

0 votes

1)Given first line is -x + 2y -2 = 0

The slope of the first line is m = (-xcofficient)/(ycofficient) = -(-1)/2 = 1/2

Given second line is x - 2y - 2 = 0

The slope of the second line is m = (-xcoff)/(ycoff) = (-1)/(-2) = 1/2

Therefore the slopes of the two lines are equal

Therefore the lines have no solution

2)Given first line is  -x - 4y - 8 = 0

The slope of the first line is  (-xcoff)/(ycoff) =(-(-1)/(-4) = (-1/4)

Given second line is x - 4y - 8 = 0

The slope of the second line is (-xcoff)/(ycoff) = (-1)/(-4) = 1/4

Therefore the slopes of the two lines are opposite hence the lines are perpendicular

The solution of the equation is

-x - 4y = 8 ------->1

x - 4y = 8 -------->2

Add equation 1 and 2

The obtained line is -8y = 16 ====> y = (-16)/8 ===> y = -2

Substitute y = -2 in equation 1

Therefore -x -4(-2) = 8 ===> -x - 8 = 8 ===> -x = 16 ===> x = -16

Therefore the solution is x = -16 , y = -2

answered Jun 27, 2013 by jouis Apprentice

(2).

Substitute y = -2 in equation 1

Therefore -x -4(-2) = 8 ===> -x + 8 = 8 ===> x = 0

Therefore the solution is x = 0, and y = -2

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