Three times t1, t2 and t3 (in seconds) satisfy t2 + t3 = 6.72, t1 + t3 = 9.34 and t1 + t2 = 8.56. Find each time.

Step-by-step solution

1. 1

   Exploit the symmetry: add all three equations. Each unknown appears in exactly two of the three equations, so adding them gives

   $2\left(t_1+t_2+t_3\right)=6.72+9.34+8.56=24.62$

   This one step replaces the usual elimination work.

2. 2

   Halve to get the grand total.

   $t_1+t_2+t_3=\frac{24.62}{2}=12.31\ \text{s}$

3. 3

   Recover each unknown by subtracting the equation it is missing from. $t_1$ is absent from the first equation, so

   $t_1=12.31-6.72=5.59\ \text{s}$

4. 4

   Do the same for the other two.

   $t_2=12.31-9.34=2.97\ \text{s},\qquad t_3=12.31-8.56=3.75\ \text{s}$

   $\boxed{t_1=5.59\ \text{s},\quad t_2=2.97\ \text{s},\quad t_3=3.75\ \text{s}}$

5. 5

   Check all three original equations. $t_2+t_3=2.97+3.75=6.72$ ✓; $t_1+t_3=5.59+3.75=9.34$ ✓; $t_1+t_2=5.59+2.97=8.56$ ✓. All three hold exactly, with no rounding needed — a sign that the data was constructed consistently.

6. 6

   Note the general rule. For any system of pairwise sums $s_1=t_2+t_3$, $s_2=t_1+t_3$, $s_3=t_1+t_2$, the solution is always $t_i=\tfrac{s_1+s_2+s_3}{2}-s_i$. The method fails only if the three sums cannot come from real numbers — which cannot happen here, since any three reals produce three pairwise sums.