Equations with brackets

Two cinema deals, two phone plans, two supermarket offers: as soon as you want to know when two things cost the same, you end up with an equation that has x on both sides and often brackets too. With the right order of steps you'll solve these just as confidently as the simple ones.

Tidy up first, then solve

With an equation like 2 · (x + 3) = 10, you tidy up first: remove the brackets and combine like parts. Once the bracket is gone, the equation reads 2x + 6 = 10. If another part sits next to the bracket, as in 4x + 2 · (x + 1) = 20, you remove the bracket first (4x + 2x + 2 = 20) and then combine the x parts (6x + 2 = 20). Only when both sides are as simple as possible do you start solving. If you rearrange in the middle of the mess, it's easy to lose a sign or a whole part.

The balance and x on both sides

After tidying up, the balance takes over: whatever you do on one side, you must do on the other, or it tips. For 2x + 6 = 10 you subtract 6 on both sides (2x = 4) and divide by 2 (x = 2). If x appears on both sides, as in 5x + 4 = 2x + 19, first bring the x terms to one side: subtract 2x on both sides, leaving 3x + 4 = 19. From there everything runs as usual.

Don't forget the check

At the end, substitute your solution into the original equation, not into an intermediate line. For x = 2 in 2 · (x + 3) = 10 you calculate 2 · (2 + 3) = 2 · 5 = 10. Both sides agree, so the solution is correct. The check takes ten seconds and catches almost every slip.