# Thread: Simultaneous equations - 3 variables

1. ## Simultaneous equations - 3 variables

Hi!

I've got myself into this rather unhelpful question. I have tried many different ways to solve it, but all of them have turned out to fail miserably .The question is:

Solve,
$\displaystyle xy+5(x+y)=47$
$\displaystyle yz+5(y+z)=65$
$\displaystyle zx+5(z+x)=55$

I thought it would be reasonably simple to solve, however, (for me at least!) it is harder than it looks!

Could anybody help me a little here?
Thanks very much!

2. ## Re: Simultaneous equations - 3 variables

2 integer solutions:
a = 3, b = 4, c = 5
a = -13, b = -14, c = -15

3. ## Re: Simultaneous equations - 3 variables

$\displaystyle xy+5(x+y)=47$ (1)
$\displaystyle yz+5(y+z)=65$ (2)
$\displaystyle zx+5(z+x)=55$ (3)

Solving (1) for $\displaystyle x$ yields

$\displaystyle x=\frac{47-5y}{y+5}$ (4)

And solving (2) for $\displaystyle z$,

$\displaystyle z=\frac{65-5y}{y+5}$ (5)

Then, substituting (4) and (5) into (3):

$\displaystyle \left(\frac{65-5y}{y+5}\right)\left(\frac{47-5y}{y+5}\right)+5\left(\frac{65-5y}{y+5}+\frac{47-5y}{y+5}\right)=55$

$\displaystyle \Rightarrow\frac{5\left(5y^2-112y+611\right)}{(y+5)^2}+\frac{10(56-5y)}{y+5}=55$

$\displaystyle \Rightarrow5\left(5y^2-112y+611\right)+10(56-5y)(y+5)=55(y+5)^2$

$\displaystyle \Rightarrow80y^2+800y-4480=0$

$\displaystyle \Rightarrow80(y+14)(y-4)=0$

$\displaystyle \Rightarrow y=-14\mathrm{\quad or\quad}y=4$

And substituting both of these into (4) and (5) gives us $\displaystyle x$ and $\displaystyle z$.

4. ## Re: Simultaneous equations - 3 variables

Oh right thanks! I must have made a mistake when I did a similar thing.

Thanks very much!