The expression 2x − 3 has the same value as y wherever the first equation holds. It therefore replaces one whole occurrence of y. In 3y, the three multiplies that complete value, so substitution produces 3(2x − 3), not just three times the first term.

Think of the brackets as preserving the boundary of the replaced quantity. They show which operations already applied to y and must now apply to its replacement. Write the unchanged surrounding expression before simplifying. This makes the first possible error easy to locate.

Start with y = 2x − 3 and 3y + x = 19. Substitute into the second equation to get 3(2x − 3) + x = 19. Distribute three across both terms: 6x − 9 + x = 19. Combine like terms to get 7x − 9 = 19, then 7x = 28 and x = 4.

Recover y from the first original equation: y = 2 × 4 − 3 = 5. The solution is the pair (x, y) = (4, 5), not only x = 4. Retaining the two-variable answer matters because the task asks for values satisfying both original relationships.

The incorrect expression 6x − 3 differs from the correct 6x − 9 by six. It treats only 2x as multiplied by three and leaves the subtraction untouched. The later algebra may be accurately performed on this wrong equation, but it no longer solves the original problem.

A useful check is to try a simple x-value in the two expressions. At x = 4, 3(2x − 3) equals fifteen, while 6x − 3 equals twenty-one. Because these are different values, removing the brackets that way was not an equivalent simplification.

Suppose y = 5 − x and x − 2y = −4. Substitution gives x − 2(5 − x) = −4. Expanding the product gives x − 10 + 2x = −4, so 3x = 6 and x = 2. Then y = 3.

The minus sign belongs to the multiplier −2, which changes both terms in the bracket. Writing x − 10 − 2x would distribute the sign incorrectly. Keep the substitution line separate, then check the expansion term by term. Finally, 2 − 2 × 3 = −4 confirms the second condition.

If y = (x + 1)/2 and 3y − x = 5, substitute the full fraction: 3((x + 1)/2) − x = 5. Multiplying the whole equation by two gives 3(x + 1) − 2x = 10. Expanding leads to x + 3 = 10, hence x = 7 and y = 4.

Do not lose the denominator’s scope or multiply only part of an equation when clearing it. The numerator x + 1 stays together. Once again, both original equations check: 4 = (7 + 1)/2 and 3 × 4 − 7 = 5. A correct fractional replacement follows the same whole-expression principle.

Use the original equations, not only the transformed equation where the mistake may have occurred. Substitute both proposed values into each original condition independently. A pair fitting one equation alone is not yet a solution of the simultaneous system.

If a check fails, inspect the first substitution line before restarting every calculation. The error may be the replacement boundary, the multiplier, an expansion sign or arithmetic later on. Keeping those stages visible allows a focused correction instead of an unexplained second attempt.

Question 1: solve y = x + 2 and 2y + x = 16. Question 2: solve x = 3y − 1 and 2x + y = 19. Question 3: in the main example, which expression replaces 3y? Question 4: does x = 2, y = 3 satisfy y = 5 − x and x − 2y = −4?

Answers: 1) x = 4, y = 6. 2) y = 3, x = 8. 3) 3(2x − 3), before expansion. 4) Yes: 3 = 5 − 2 and 2 − 6 = −4. Show the first replacement line and check both equations; do not rely on an answer pair alone.

Bring the original equations and the line where y disappeared. A precise note such as ‘multiplies only the first term of the replacement’ identifies the reasoning step. In a small-group 3-pax lesson, students can compare correct and incorrect replacements before completing whole systems.

For parents considering Secondary 2 Mathematics tuition in Punggol, ask ‘What complete expression is replacing this letter?’ Once that is clear, leave the student to expand, solve and verify. The goal is to preserve the expression’s meaning through substitution, not to make every calculation depend on an adult’s reminder.