# math

Solve the simultaneous equations
(x +3)(y+ 5)=24
(y +5)(z-7)=48
(z+ 7)(x+ 3)=32

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1. The set of given equations is not easy to solve, but if there was a typo, and the equations were:
(x +3)(y+ 5)=24
(y +5)(z+7)=48
(z+ 7)(x+ 3)=32
Then the equation can be solved readily because of symmetry, namely
(x+3)=4, (y+5)=6, (z+7)=8

Can you confirm if there was a typo?

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2. No typo

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3. In a similar way, we can substitute
X=x+3
Y=y+5
Z=z+7, and
Z-14=z-7
The equations become:
XY=24 ...(1a)
Y(Z-14)=48 ...(2a)
ZX=32 ...(3a)
Divide (3a) by (1a) to get
Z/Y=32/24
Z=4Y/3 ...(4)
Substitute (4) in (2a) to get:
Y(4Y/3-14)=48
Solve the quadratic equation to get
Y=(±3sqrt(113)+21)/4
Substitute Y into 1a and 3a to get the remaining solutions of X, Y and Z.

Solve for x,y,z from X,Y,Z from the initial substitutions
X=x+3
Y=y+5
Z=z+7

I get
x=(11-3*sqrt(113)-11)/(sqrt(113)+7) or
x=-(3*sqrt(113)+11)/(sqrt(113)-7)
...

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