Let $a,b,c$ be real numbers satisfying:

Given that the product $abc$ can take two values $\frac{r}{s}$ and $\frac{t}{u}$ in lowest terms, find $r+s+t+u$.

Step 1: Multiply each equation by the remaining variable to introduce the term $3abc$:

Step 2: Isolate the quadratic terms from the original equations and substitute them:

• From original (2): $3bc=5c-2⟹6bc=10c-4$. Substituting into (1):

• From original (3): $3ca=4a-2⟹5ac=\frac{20a-10}{3}$. Substituting into (2):

• From original (1): $3ab=6b-2⟹4ab=\frac{24b-8}{3}$. Substituting into (3):

Step 3: Equate the three expressions for $9abc$:

This allows us to express $a,b,c$ in terms of $\lambda$:

Step 4: Substitute $a$ and $b$ back into the first original equation $3ab+2=6b$:

Step 5: We find $abc$ for both roots:

• If $\lambda =6$: $a=\frac{8}{7},b=\frac{7}{9},c=\frac{3}{4}⟹abc=\frac{2}{3}$.

• If $\lambda =4$: $a=1,b=\frac{2}{3},c=\frac{2}{3}⟹abc=\frac{4}{9}$.

Step 6: Comparing to $\frac{r}{s}$ and $\frac{t}{u}$, we have $r=2,s=3,t=4,u=9$. These are in lowest terms. Find the sum: