When nothing cancels
Take 3x + 2y = 12 and x + y = 5. Elimination needs a letter with the same coefficient, the number in front of it, in both equations. Here x has 3 and 1, and y has 2 and 1, so no letter matches.
Adding the equations gives 4x + 3y = 17, and subtracting them gives 2x + y = 7. Both still have x and y in them, so neither removes a letter.
Multiply a whole equation
Multiply every term of the second equation by 2, on both sides: 2 × x + 2 × y = 2 × 5, which is 2x + 2y = 10. Now the y terms match the 2y in the first equation.
This is allowed because doubling both sides of a true equation keeps it true. Every pair of values that makes x + y = 5 true also makes 2x + 2y = 10 true, and no other pair does. The new equation has exactly the same solutions as the old one.
2x + 2y = 10 has the same solutions as x + y = 5, so it draws this same gold line, and the crossing stays where x = 2 and y = 3.
Subtract, then substitute
Both equations now have +2y, so subtract: (3x + 2y) − (2x + 2y) = 12 − 10. The 2y terms cancel, and 3x − 2x = x, so x = 2.
Put x = 2 into the easier equation, x + y = 5: 2 + y = 5, so y = 3. Check in the other equation: 3 × 2 + 2 × 3 = 6 + 6 = 12. The solution is x = 2 and y = 3.
Scaling both equations
Sometimes neither equation can be scaled to match the other. In 2x + 3y = 13 and 3x + 2y = 12, no whole number turns 2x into 3x. Scale both instead, to a common multiple: multiply the first equation by 3 and the second by 2, so that both have 6x. That gives 6x + 9y = 39 and 6x + 4y = 24.
Subtract: 9y − 4y = 5y and 39 − 24 = 15, so 5y = 15 and y = 3. Then 2x + 3 × 3 = 13 gives 2x = 4, and x = 2. Check: 3 × 2 + 2 × 3 = 12.
The usual mistakes
Multiplying only the left side: doubling x + y = 5 and writing 2x + 2y = 5. Every term on both sides is multiplied, so the right side becomes 10.
Subtracting before anything matches. (3x + 2y) − (x + y) = 2x + y still has both letters, so no letter has been eliminated. Scale first, then subtract.