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Question 683915: Please help me with this problem.
Solve the system. If the system has infinitely many solutions,write the set with z arbitrary.
5x-3y+z=0
x+y=0
-15x+9y-3z=0
Thanks
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 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  "\large \left(-\frac{z}{8},\frac{z}{8},z \right )")
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Note: what does mean exactly? Well...
*** If z = 8, then becomes  "\large \left(-\frac{z}{8},\frac{z}{8},z \right )") which turns into  "\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 becomes  "\large \left(-\frac{z}{8},\frac{z}{8},z \right )") which turns into  "\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.
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