Quick questions on Solving linear equations for HSC Maths Standard 2

An equation is balanced like a set of scales: the left side weighs exactly the same as the right. The single rule that keeps a solution valid is

When the variable is inside a bracket, like $5(p + 3) = 65$, you have two equally valid first moves. You can expand the bracket, multiplying the outside number through every term inside ($5p + 15 = 65$), and then solve the two-step equation. Or, when the number on the other side divides exactly by the multiplier, you can divide both sides by that multiplier first ($p + 3 = 13$) to peel the bracket off in one move. The trap is multiplying only the first term inside the bracket: the multiplier hits every term, so $5(p + 3)$ is $5p + 15$, not $5p + 3$.

Fractions are easiest to remove entirely before you solve. First find the lowest common denominator (LCD), the smallest number that every denominator divides into. Then multiply every term on both sides by the LCD. Each denominator cancels and you are left with a fraction-free equation.

When the variable appears on both sides, like $5x + 8 = 2x + 23$, first gather all the variable terms on one side and all the numbers on the other. Subtract the smaller variable term from both sides so the variable stays positive (here subtract $2x$, leaving $3x + 8 = 23$), then finish as a two-step equation. Keeping the variable term positive avoids a sign slip later; if you do end with a negative coefficient, dividing by that negative is still fine, just watch the sign.

The "do the same to both sides" rule is exactly how a balance scale behaves, and watching it makes the method obvious. The four panels below solve $3x + 5 = 20$ on a balance, where each $x$ tile is one unknown weight and each numbered tile is that many units. The scale starts level because the two sides are equal, and every move keeps it level.

A linear equation has exactly one solution, and you can always verify it. Put your answer back into the original equation (not a tidied-up line, in case the tidying went wrong) and evaluate each side separately; if the left side equals the right side, the solution is correct. This costs a few seconds and catches almost every arithmetic slip, which is why it is worth doing in the exam. For an equation with the variable on both sides, check both sides come out to the same number, as in $5x + 8 = 2x + 23$ where $x = 5$ gives $33$ on each side.