Variables on both sides, step by step

5x + 3 = 2x + 18 is the first equation that needs a decision before it needs arithmetic.

The question

10x + 13 = 12x + 3

10x + 13=12x + 3
10x + 13 = 12x + 3
13 = 2x + 3
2x + 3 = 13
2x + 3=13
2x + 3=13
2x=10
2x=10
x=5
10x + 13 = 12x + 3
63 = 63
  1. This one has x on BOTH sides, so nothing can be undone yet — subtracting 13 would still leave an x on the right. Collect the x terms first.
  2. Take 10x off BOTH sides. That clears x from the left, leaving 13 = 2x + 3. An equals sign does not care which side is which, so write it as 2x + 3 = 13. That is an equation you have already met.
  3. Two things were done to x: multiplied by 2, then 3 added. Undo them in the REVERSE order — whatever happened last comes off first.
  4. The 3 went on last, so it comes off first — and off BOTH sides, or the two halves stop being equal. 13 − 3 = 10.
  5. That leaves 2x, which means 2 lots of x. Divide both sides by 2: 10 ÷ 2 = 5.
  6. Check by putting it back into the ORIGINAL, both sides: 10(5) + 13 = 63, and 12(5) + 3 = 63 — both 63, so x = 5.

How it works.

  1. 01

    This one has x on BOTH sides, so nothing can be undone yet — subtracting 13 would still leave an x on the right. Collect the x terms first.

  2. 02

    Take 10x off BOTH sides. That clears x from the left, leaving 13 = 2x + 3. An equals sign does not care which side is which, so write it as 2x + 3 = 13. That is an equation you have already met.

  3. 03

    Two things were done to x: multiplied by 2, then 3 added. Undo them in the REVERSE order — whatever happened last comes off first.

  4. 04

    The 3 went on last, so it comes off first — and off BOTH sides, or the two halves stop being equal. 13 − 3 = 10.

  5. 05

    That leaves 2x, which means 2 lots of x. Divide both sides by 2: 10 ÷ 2 = 5.

  6. 06

    Check by putting it back into the ORIGINAL, both sides: 10(5) + 13 = 63, and 12(5) + 3 = 63 — both 63, so x = 5.