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Question 1179614: Is it possible to get infinitely many solutions and an x, y, and z value for the following problem. Solve the system of equations using elimination.
a. -4x+5y-3z=17
b. -3x-2y-4z=-1
c. 5x+5y+4z=12
a*-2= 8x-10y+6z=-34
c*2= 10x+10y+8z=24
equation d: 18x+14z=-10
-4x+5y-3z=17
-5x-5y-4z=-12
equation e: -9x-7z=5
e*2 -18x-14z=10
d 18x+14z=-10
result 0=0 infinitely many
I solved another way and got x=1, x=3, and x=-2
Found 2 solutions by Alan3354, MathTherapy:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Is it possible to get infinitely many solutions and an x, y, and z value for the following problem. Solve the system of equations using elimination.
a. -4x+5y-3z=17
b. -3x-2y-4z=-1
c. 5x+5y+4z=12
a*-2= 8x-10y+6z=-34
c*2= 10x+10y+8z=24
equation d: 18x+14z=-10
-4x+5y-3z=17
-5x-5y-4z=-12
equation e: -9x-7z=5
e*2 -18x-14z=10
d 18x+14z=-10
result 0=0 infinitely many
I solved another way and got x=1, x=3, and x=-2
Assuming you meant x, y, and z:
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Sub your numbers and see if they work.
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a. -4x+5y-3z=17
-4*1 + 5*3 - 3*-2 = -4 + 15 + 6 = 17
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b. -3x-2y-4z=-1
-3*1 -2*3 - 4*-2 = -3 -6 +8 = -1
c. 5x+5y+4z=12
5*1 + 5*3 + 4*-2 = 5 + 15 - 8 = 12
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Your solution is correct, so there is a unique solution to the system.
It's not possible to get an infinite # of solutions.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! Is it possible to get infinitely many solutions and an x, y, and z value for the following problem. Solve the system of equations using elimination.
a. -4x+5y-3z=17
b. -3x-2y-4z=-1
c. 5x+5y+4z=12
a*-2= 8x-10y+6z=-34
c*2= 10x+10y+8z=24
equation d: 18x+14z=-10
-4x+5y-3z=17
-5x-5y-4z=-12
equation e: -9x-7z=5
e*2 -18x-14z=10
d 18x+14z=-10
result 0=0 infinitely many
I solved another way and got x=1, x=3, and x=-2
You multiplied eqs (i) and (iii) to get eq (iv). You also subtracted eq (iii) from eq (i) to get eq (v). They're BOTH the same
equation, so that's why you ended up with equations that're the same, and ultimately, infinitely many solutions; but, that's INCORRECT.
You probably multiplied eq (i) by 2, and eq (ii) by 5, or any other combination to get eqs in x and z, and from thereon to solve
for x, y, and z. That's how you got the correct answer, which by the way should be: 
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