A) 3
B) 2
C) 1
D) 0
Correct Answer: B
Solution :
Consider the second equation |
\[xy+yz=y(x+z)=31=1\times 31\] |
It leads to\[y=1,x+z=31.\]. By substituting them into the first equation, it follows that |
\[x\left( 1+31-x \right)=255,\] |
\[{{x}^{2}}-32x+255=0,\] |
\[\left( x-15 \right)\left( x-17 \right)=0,\] |
\[\therefore {{x}_{1}}=15,{{x}_{2}}=17.\] |
Thus, the solutions are |
\[x=15,y=1,z=16\] |
and \[x=17,y=1,\,z=\text{ }14.\] |
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