3x-4y+5z=5
-x+2y-3z=-3
3x-2y+z=1
The answer is said to be 2p/ P+1 and P and I am wondering how they came to this solution, can someone please explain in steps how this answer was achieved?
-x+2y-3z=-3
3x-2y+z=1
The answer is said to be 2p/ P+1 and P and I am wondering how they came to this solution, can someone please explain in steps how this answer was achieved?
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By elimination or substitution, same as with two variables. But the answers, if any, are going to be numbers. What is P? That can't be the answer.
You could subtract the 3rd equation from the 1st to eliminate the 3x term. That gives you
-2y + 4z = 4
Then you could multiply the 2nd equation by 3:
-3x + 6y - 9z = -9
and add it to the first equation, again to eliminate the 3x.
2y - 4z = -4
Uh-oh. Those are not independent equations. If I multiply the 1st equation by -1, I get the second equation. That means there are either 0 solutions or infinitely many. If I added these two equations to try to eliminate another variable, I'd get 0 = 0. Since that's a true statement, then there are infinitely many solutions. (If you end up with a contradiction like 0 = 4, that means there are no solutions.)
OK, that means I can pick any value I want for any of the variables, then solve for the other two in terms of it. So obviously the answer manual picked z = P. P is any real number at all. There will be a solution for z = 0.0001 or z = -2345 pi, or z = sqrt(99). Anything.
Going back to one of the above equations, 2y - 4z = -4
2y - 4P = -4
2y = -4 + 4P
y = -2 + 2P = 2(P - 1). Not what you said.
Plugging these into any of the original equations
-x + 2y - 3z = -3
-x = -3 - 2y + 3z
x = 3 + 2y - 3z = 3 + 2(-2 + 2P) - 3P
= 3 + -4 + 4P - 3P
= -1 + P. Again, not what you wrote. So either P is not the chosen value for z, or I made an algebra mistake (always possible). But that's the general idea.
Edit: Ah, I see it. x = P - 1, y = 2(P - 1). Let's use Q for the value of X. So Q = P - 1 and y = 2Q. That makes z = Q + 1. I was confused by the order you gave the solutions. It matters which of those expressions is x, which is y, and which is z. You gave them in the order y, z, x.
You could subtract the 3rd equation from the 1st to eliminate the 3x term. That gives you
-2y + 4z = 4
Then you could multiply the 2nd equation by 3:
-3x + 6y - 9z = -9
and add it to the first equation, again to eliminate the 3x.
2y - 4z = -4
Uh-oh. Those are not independent equations. If I multiply the 1st equation by -1, I get the second equation. That means there are either 0 solutions or infinitely many. If I added these two equations to try to eliminate another variable, I'd get 0 = 0. Since that's a true statement, then there are infinitely many solutions. (If you end up with a contradiction like 0 = 4, that means there are no solutions.)
OK, that means I can pick any value I want for any of the variables, then solve for the other two in terms of it. So obviously the answer manual picked z = P. P is any real number at all. There will be a solution for z = 0.0001 or z = -2345 pi, or z = sqrt(99). Anything.
Going back to one of the above equations, 2y - 4z = -4
2y - 4P = -4
2y = -4 + 4P
y = -2 + 2P = 2(P - 1). Not what you said.
Plugging these into any of the original equations
-x + 2y - 3z = -3
-x = -3 - 2y + 3z
x = 3 + 2y - 3z = 3 + 2(-2 + 2P) - 3P
= 3 + -4 + 4P - 3P
= -1 + P. Again, not what you wrote. So either P is not the chosen value for z, or I made an algebra mistake (always possible). But that's the general idea.
Edit: Ah, I see it. x = P - 1, y = 2(P - 1). Let's use Q for the value of X. So Q = P - 1 and y = 2Q. That makes z = Q + 1. I was confused by the order you gave the solutions. It matters which of those expressions is x, which is y, and which is z. You gave them in the order y, z, x.
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y = 2 x and z = 1 + x