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# Helppppp Please ( solving a system)

3x+y+z=13
2x+2y-3z+3
3x+y-z=7

A) (2,-4,11)
B) (3,4,0)
C) (2,4,3)
D)(6,-3,4)

One way to solve a system of three variables is to create a system of two variables by using combinations of the three equations in such a way to eliminate the third variable. It is important that you use all three equations to do this. For example I will multiply the first equation by 3 and add it to the second equation:

3(3x + y +z = 13) = 9x + 3y + 3z = 39

9x + 3y + 3z = 39
+ 2x + 2y -3z = 3
11x + 5y + 0z = 42

To get a second equation, I will multiply the third equation by -3 and add it to the second equation:

-3(3x+y-z=7) = -9x - 3y  + 3z = -21

-9x - 3y + 3z = -21
+ 2x + 2y - 3z = 3
-7x - y + 0z = -18

Now we have a system of two equations:
11x + 5y = 42
-7x - y = -18

Use either the substitution method or elimination method to solve this system for x and y. Once you have x and y, plug those in to any of the original equations to solve for z. Plug in this (x,y,z) coordinate into all three of the equations, to make sure that it satisfies each constraint.