So x = 9, which then gives us that y = 12 once we have substituted our x value into equation (5) or (6).
We can then substitute these values into any of the original equations to find that z = 3.
So, x = 9, y = 12 and z = 3.We can verify that this is the unique solution to the four given equations by direct substitution.
In your (1)+(2) line, how did you get 8x?
By the way it's worth noting that you might not get just one solution (x,y,z) as these are equations of planes, and it's possible to get lines of intersection. To see what I mean I'll post some configurations of three planes.
So x = 9, which then gives us that y = 12 once we have substituted our x value into equation (5) or (6).
We can then substitute these values into any of the original equations to find that z = 3.
So, x = 9, y = 12 and z = 3.We can verify that this is the unique solution to the four given equations by direct substitution.
Yeh swear (1)+(2) is different to what you have but your answers work, wtf. Yeh also in the question it asks it as if there will be different solutions, but it asks you to find them all.
Apologies for my initial, incorrect attempt at solving this problem. I would advise looking up Gaussian elimination (if you have not yet come across this method for solving simultaneous equations). It quickly becomes apparent that there is no unique solution to the problem (as the equations are dependent, producing a row of zeroes after reducing the augmented matrix to row-echelon form). You then solve the equations in parametric form to find a set of solutions to the problem. I hope this helps!
Apologies for my initial, incorrect attempt at solving this problem. I would advise looking up Gaussian elimination (if you have not yet come across this method for solving simultaneous equations). It quickly becomes apparent that there is no unique solution to the problem (as the equations are dependent, producing a row of zeroes after reducing the augmented matrix to row-echelon form). You then solve the equations in parametric form to find a set of solutions to the problem. I hope this helps!
aha no problem. I've not heard of that stuff ill have to google it, cheers
I see so they wanted a more general solution, do you have pointers to finding general solutions like this?
See the diagram in I hate maths' post. It's possible that instead of a single solution you have a whole line or even a whole plane of solutions. If this is the case you will not be able to solve for all three variables - you will end up expressing one variable in terms of another (or possibly two) once you've eliminated. Usually when you express your solution set you put the three variables in terms of a parameter k if it's a line, or two parameters if it's a plane.
Apologies for my initial, incorrect attempt at solving this problem. I would advise looking up Gaussian elimination (if you have not yet come across this method for solving simultaneous equations). It quickly becomes apparent that there is no unique solution to the problem (as the equations are dependent, producing a row of zeroes after reducing the augmented matrix to row-echelon form). You then solve the equations in parametric form to find a set of solutions to the problem. I hope this helps!
Since this question is supposedly from an old MAT paper, I am almost certain that this is super overkill.
See the diagram in I hate maths' post. It's possible that instead of a single solution you have a whole line or even a whole plane of solutions. If this is the case you will not be able to solve for all three variables - you will end up expressing one variable in terms of another (or possibly two) once you've eliminated. Usually when you express your solution set you put the three variables in terms of a parameter k if it's a line, or two parameters if it's a plane.