Simplifying and solving are different jobs
Simplifying rewrites an expression without changing its value. For example, 3x + 2x − 4 becomes 5x − 4 because 3x and 2x are like terms. It does not produce a value of x. Solving an equation finds the values that make two expressions equal: 5x − 4 = 11 gives x = 3. Check the command word before choosing what your final line should contain.
An identity holds for every allowed value of its variable. The statement 2(x + 3) = 2x + 6 is an identity over the real numbers. A conditional equation such as 2x + 6 = 14 is true only at its solution. This distinction matters when the variable disappears during your working: that disappearance does not automatically mean x = 0.
Solve equations with a variable on both sides
Consider 5x − 4 = 2x + 11. Subtract 2x from both sides to collect the variable terms: 3x − 4 = 11. Add 4 to both sides: 3x = 15. Divide by 3: x = 5. The check uses the original equation: the left side is 25 − 4 = 21, and the right side is 10 + 11 = 21.
Two other outcomes are possible. In 4x + 2 = 4x + 9, subtracting 4x leaves the false statement 2 = 9, so there is no solution. In 4x + 2 = 2(2x + 1), expanding and simplifying leaves 2 = 2, so every real x works. Read the remaining statement before reporting a solution.
Choose when to expand brackets
In 3(x − 2) = 15, dividing by 3 first leaves x − 2 = 5, so x = 7. Expanding first gives 3x − 6 = 15, then 3x = 21 and x = 7. Both routes preserve equality. Dividing first is shorter here because the entire bracket is multiplied by the same nonzero number.
Distribution must include every term and its sign: −2(x − 3) = −2x + 6. A substitution can expose an incorrect expansion. At x = 4, the original expression gives −2, and −2x + 6 also gives −2. The incorrect expression −2x − 6 gives −14. One matching test value can catch a mistake, but it does not by itself prove an identity for every x.
Clear fractions without losing a term
For x/3 + 2 = 5, multiply both sides by 3. The full left side is multiplied, giving x + 6 = 15, so x = 9. Multiplying only x/3 would create x + 2 = 15, which is a different equation. Check by substitution: 9/3 + 2 = 5.
When a variable occurs in a denominator, record values that make that denominator zero before changing the equation. In 1/x = 2, x cannot be 0; the solution is x = 1/2. More complex rational equations may produce candidates that violate an original restriction, so keep the starting equation for the final check.
Connect a straight-line graph to its equation
In y = 2x + 1, the y-intercept is 1: setting x = 0 gives the point (0, 1). The slope is 2, meaning y increases by 2 when x increases by 1. Points (1, 3) and (2, 5) satisfy the same relationship. Read the axis scales before judging the line by its visual steepness.
For a simple system, y = x + 1 and y = 2x − 2 intersect where x + 1 = 2x − 2. This gives x = 3 and y = 4. Check (3, 4) in both original equations. A point that works in only one equation does not solve the system.
Ask about the first transition you cannot justify
Compare the lines 5x − 4 = 2x + 11 and 3x = 7. Collecting the x terms was a sensible choice, but the constants were combined incorrectly. Subtracting 2x gives 3x − 4 = 11; adding 4 then gives 3x = 15. Repair that transition before checking the remaining arithmetic. The corrected solution is x = 5.
A different hazard appears when dividing by an expression containing the unknown. For x(x − 3) = 0, dividing by x silently excludes x = 0. Use the zero-product property instead: x = 0 or x − 3 = 0, giving x = 0 or x = 3. Both values satisfy the original equation. Before dividing, decide whether the divisor could be zero.
When you ask Lirno to inspect your working, include the original equation and the two lines around the suspected error. Ask which operation changes the solution set. That question distinguishes an arithmetic slip from a transformation that loses a valid solution.
