Some equations have a group in parentheses, like $3(x + 4) = 27$. You can solve them with two steps, or distribute first. Both roads lead to the same answer.
Key idea
For $p(x + q) = r$: either divide both sides by $p$ first and then undo the $q$, or distribute first ($px + pq = r$) and solve the two-step equation.
Why it works
In $3(x + 4)$, the order of operations says: add 4 to $x$, then multiply the whole group by 3. Undo in reverse: divide by 3, then subtract 4. Distributing first is also fine, because $3(x + 4)$ and $3x + 12$ are equal for every $x$.
Worked examples
Worked example: Divide first
Solve $3(x + 4) = 27$.
- Divide both sides by 3: $x + 4 = 9$.
- Subtract 4: $x = 5$.
- Check: $3(5 + 4) = 3 \times 9 = 27$ ✓.
Answer: $x = 5$
Worked example: Two ways with a negative
Solve $-2(x - 5) = 18$.
- Divide first: $x - 5 = 18 \div (-2) = -9$, so $x = -4$.
- Or distribute first: $-2x + 10 = 18$, so $-2x = 8$ and $x = -4$. Same answer.
- Check: $-2(-4 - 5) = -2 \times (-9) = 18$ ✓.
Answer: $x = -4$
Worked example: A fraction outside
Solve $\frac{3}{4}(x + 8) = 9$.
- Multiply both sides by the reciprocal $\frac{4}{3}$: $x + 8 = 9 \times \frac{4}{3} = 12$.
- Subtract 8: $x = 4$.
- Check: $\frac{3}{4}(12) = 9$ ✓.
Answer: $x = 4$
Common mistake
Distributing to the first term only ($3(x + 4) = 27$ becoming $3x + 4 = 27$), or subtracting 4 before dividing ($3(x + 4) = 27$ becoming $3x = 23$). The 4 is inside the group, so it is multiplied by 3 too. Check your answer in the original equation.