Question 683915
x+y=0


y=-x


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5x-3y+z=0


5x-3(-x)+z=0


5x+3x+z=0


8x+z=0


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-15x+9y-3z=0 


-15x+9(-x)-3z=0 


-15x-9x-3z=0 


-24x-3z=0 


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So we have the new system of equations


8x+z=0
-24x-3z=0



3(8x+z)=3*0
-24x-3z=0


24x+3z=0
-24x-3z=0
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0x+0y=0


0=0


Since that equation is ALWAYS true regardless of what x, y and z are, this means that there are an infinite number of solutions.


Solve the first equation for x to get


8x+z = 0


8x = -z


x = -z/8


Since y = -x, we know that y = -(-z/8) ----> y = z/8



So in the end, x = -z/8, y = z/8 and z is a free variable (ie it can be any number)


Therefore, the final answer as an ordered triple is <img src="http://latex.codecogs.com/gif.latex?\large \left(-\frac{z}{8},\frac{z}{8},z \right )" title="\large \left(-\frac{z}{8},\frac{z}{8},z \right )" />


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Note: what does <img src="http://latex.codecogs.com/gif.latex?\large \left(-\frac{z}{8},\frac{z}{8},z \right )" title="\large \left(-\frac{z}{8},\frac{z}{8},z \right )" /> mean exactly? Well...



*** If z = 8, then <img src="http://latex.codecogs.com/gif.latex?\large \left(-\frac{z}{8},\frac{z}{8},z \right )" title="\large \left(-\frac{z}{8},\frac{z}{8},z \right )" /> becomes <img src="http://latex.codecogs.com/gif.latex?\large \left(-\frac{8}{8},\frac{8}{8},8 \right )" title="\large \left(-\frac{z}{8},\frac{z}{8},z \right )" /> which turns into <img src="http://latex.codecogs.com/gif.latex?\large \left(-1,1,8 \right )" title="\large \left(-\frac{z}{8},\frac{z}{8},z \right )" /> which is one of the infinitely many solutions. If you plug x = -1, y = 1 and z = 8 into the original system, you'll find that all 3 satisfy all 3 equations.





*** If z = 16, then <img src="http://latex.codecogs.com/gif.latex?\large \left(-\frac{z}{8},\frac{z}{8},z \right )" title="\large \left(-\frac{z}{8},\frac{z}{8},z \right )" /> becomes <img src="http://latex.codecogs.com/gif.latex?\large \left(-\frac{16}{8},\frac{16}{8},16 \right )" title="\large \left(-\frac{z}{8},\frac{z}{8},z \right )" /> which turns into <img src="http://latex.codecogs.com/gif.latex?\large \left(-2,2,16 \right )" title="\large \left(-\frac{z}{8},\frac{z}{8},z \right )" /> which is another solution of the infinitely many solutions. If you plug x = -2, y = 2 and z = 16 into the original system, you'll find that all 3 satisfy all 3 equations.



You can plug in ANY real number for 'z' to get an ordered triple that is a solution to the original system of equations.