Thursday

July 31, 2014

July 31, 2014

Posted by **Annika Rae** on Sunday, January 5, 2014 at 12:18am.

2) Given the systems x+y+z=-14 and 2x+y+6z=16, how many number of solutions to the system are there?

- Math -
**Reiny**, Sunday, January 5, 2014 at 10:31am1. strange question

Without looking at your last part of the given x and z values, just solving the system ....

from the 2nd:

2y = 6x+1

y = (6x+1)/2

from the 3rd:

z = x-3

back into the 1st:

x + 3y - z = -3

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

times 2

2x + 18x + 3 - 2x + 6 = -6

18x = -15

x = -15/18 =-5/6

then y = (6(-5/6) + 1)/2 = -2

z = -5/6 - 3 = -23/6

So there is one uniques solution of (-5/6, -2, -23/6)

and the value of y is -2

Now when you give an x value and a z value different from the actual solution, it will depend on which equation you sub into.

There would be 3 different answers for the value of y

2. Each equation represents a plane. Since the 2 planes are not parallel, they will intersect in a line.

There will be an infinite number of points on that line, thus there will be an infinite number of solutions for your system.

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