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# Simultaneous equation - 3-unknowns watch

1. How do you solve problems like this?

1.
x(y+z)=35
y(z+x)=32
z(x+y)=27

2.
x+y+xy=10
y+z+yz=11
z+x+zx=14

3.
xy/(x+y)=1-z <-EDIT
yz/(y+z)=2-x
zx/(z+x)=2-y

Whole solutions would be appreciated so I can learn from the method. Thanks.
2. This is an honest answer, not a joke, but I'd put it in the calculator. I wouldn't bother trying to do it myself.

Perhaps it's frowned upon in "real" maths, but I can't be bothered.

Just an idea!
3. x(y+z) = 35 => x = 35/(y+z)
You can then put this into the other two equations to eliminate x and then you'll have two simulataneous equations in y and z which should be easy enough to solve (GCSE level).
4. (Original post by OllyThePhilosopher)
How do you solve problems like this?

1.
x(y+z)=35 (1)
y(z+x)=32 (2)
z(x+y)=27 (3)
Subtract equation (2) from equation (1) to get z(x-y) = 3. Add to and subtract from equation 3 to get xz and yz, and use another equation to get xy. You'll get something like (but I've made up these numbers) xy = 4, xz = 3, yz = 9, which you should be able to solve pretty easily (e.g. from the first and second equations, y/z = 4/3, which you can then multiply by the final equation).
5. (Original post by calcium878)
x(y+z) = 35 => x = 35/(y+z)
You can then put this into the other two equations to eliminate x and then you'll have two simulataneous equations in y and z which should be easy enough to solve (GCSE level).
This'll be fine too. Any approach is fair game as long as you make sure you're not dividing by anything weird. We know that x, y, z, x+y, x+z and y+z are non-zero, but we don't know that, say, x+y+z is non-zero. So make sure that you don't divide by it at any point and you'll be fine.
6. For Q1, I would treat as simultaneous equations in xy, yz and zx. Solve for xy, yz, zx and then find x, y and z.

For Q2, x+y+xy = 10 => (x+1)(y+1) = 11. You get two other similar equations, solve for x+1, y+1 and z+1.

Q3 looks like it might have a typo; are you sure it's right?
7. or you could put in into a augmented matrix and do a series of row reductions
8. (Original post by EierVonSatan)
or you could put in into a augmented matrix and do a series of row reductions
I feel like I've missed out, having never been taught the augmented matrix method
9. (Original post by calcium878)
I feel like I've missed out, having never been taught the augmented matrix method
lol you really haven't
10. (Original post by calcium878)
I feel like I've missed out, having never been taught the augmented matrix method
You haven't. You've missed out on the same old boring way of solving simultaneous equations but making them not look like simultaneous equations.
11. (Original post by OllyThePhilosopher)
How do you solve problems like this?

1.
x(y+z)=35
y(z+x)=32
z(x+y)=27
1) what two numbers multiply to give 35? It cant be 2, and most likely is 5x7

2) what two numbers multiply to give 32? 2x16, 4x8

3) what two numbers multiply to give 27? 9x3 mostly

Assuming x =/= y =/= z

3 = z(y-x), then either z =1, y-x = 3 or the other way round, but eqn 3 if you let z=1, then x+y has to equal 27, so its more likely z=3, y-x=1

I arrive at z=3,y=4,x=5 from using eqn 1

My reaosing is a bit flawed here as I've assumed all variables to be +ve and not =0, so dont take my advise =p
12. Thanks everyone
13. Following your edit, Q3 is solvable using the same method I suggested for Q1 (after multiplying out denominators).

(Except I think you'll find the equations are inconsistent so there are no solutions!).

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