Let $t$ be TNYWR.

There is exactly one real number $x$ with the property that both $x^{2}-t x+36=0$ and $x^{2}-8 x+t=0$.

What is the value of $x$ ?

## The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca  2022 Canadian Team Mathematics Contest 
Suppose $x$ is the real number such that $x^{2}-t x+36=0$ and $x^{2}-8 x+t=0$.

Therefore, we get that $x^{2}-t x+36=x^{2}-8 x+t$ which can be rearranged to get $36-t=$ $t x-8 x=(t-8) x$.

Dividing by $t-8$ gives $x=\frac{36-t}{t-8}$.

Substituting $t=15$ gives $x=\frac{36-15}{15-8}=\frac{21}{7}=3$.

ANSWER: $(4,15,3)$