The ages of two people add up to 46 years. Four years ago, the first person was twice as old as the second. Find their present ages.

Step-by-step solution

1. 1

   Name the unknowns as present ages. Let $S$ and $D$ be the present ages of the first and second person. Choosing present ages (rather than past ages) as the variables is what keeps the second condition simple to write.

2. 2

   Turn the first sentence into an equation.

   $S+D=46$

3. 3

   Turn the second sentence into an equation. Four years ago the ages were $S-4$ and $D-4$, and the first was twice the second:

   $S-4=2(D-4)$

   The bracket matters: subtracting $4$ from both ages before doubling is the step most often mishandled.

4. 4

   Simplify and substitute. Expanding, $S-4=2D-8$, so $S=2D-4$. Putting this into $S+D=46$:

   $(2D-4)+D=46\ \Longrightarrow\ 3D=50\ \Longrightarrow\ D=\frac{50}{3}=16\tfrac23$

5. 5

   Back-substitute for the other age.

   $S=46-\frac{50}{3}=\frac{138-50}{3}=\frac{88}{3}=29\tfrac13$

6. 6

   Check both original conditions. Sum: $\tfrac{88}{3}+\tfrac{50}{3}=\tfrac{138}{3}=46$ ✓. Four years ago: $\tfrac{88}{3}-4=\tfrac{76}{3}$ and $2\left(\tfrac{50}{3}-4\right)=2\cdot\tfrac{38}{3}=\tfrac{76}{3}$ ✓. Both hold exactly, so the fractional ages - unusual but perfectly consistent - are the genuine solution.