Solve linear equations using inverse operations, including one-step, two-step and multi-step equations, equations with brackets and fractions, and equations with the variable on both sides

What this dot point is asking

NESA wants you to find the unknown value in a linear equation. A linear equation is one where the variable (the letter you are solving for) appears only to the power of $1$: no $x^2$, no $\sqrt{x}$, and no $x$ sitting alone in a denominator. You solve it by inverse operations, which means doing the opposite of whatever was done to the variable. You peel away one operation at a time and do the same thing to both sides, so the equation stays balanced, until the variable sits alone. The arithmetic is short, but marks are lost in the same few places. The usual slips are forgetting to apply an operation to every term, mishandling a bracket or a fraction, dropping a negative sign, or not gathering the variable correctly when it appears on both sides. The deeper idea is that an equation says two things are equal, like a level scale, and the only moves allowed are ones that keep it level.

The answer

Inverse operations: undo in reverse order

Solving a linear equation is unwrapping a parcel. Look at $3x + 5 = 20$. To build the left side from $x$, you would first multiply by $3$, then add $5$. To get back to $x$ you reverse that, undoing the last operation first: subtract $5$, then divide by $3$. Each operation has an opposite:

• addition is undone by subtraction, and subtraction by addition;
• multiplication is undone by division, and division by multiplication.

The diagram above shows both directions. The top row is how the expression was built up from $x$, and the bottom row is how you solve, peeling the operations off in reverse. Working the order backwards is what makes a two-step equation routine. Deal with the $+5$ or $-5$ before the $\times 3$, because addition and subtraction sit on the outside of the parcel.

Keep the equation balanced

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

whatever you do to one side, do to the other side as well.

If you subtract $5$ from the left you must subtract $5$ from the right. If you divide the left by $3$ you must divide the right by $3$. Do the same thing to both sides and the scale stays level, so the two sides stay equal and the value of the variable does not change. The balance-scale sequence further down shows this happening for real. Remove the same weight from both pans and the scale is still level; share both pans into the same number of equal groups and it is still level.

One-step, two-step and multi-step

The number of operations wrapped around the variable tells you how many inverse steps you need:

• One-step: the variable has a single operation on it, like $x + 7 = 12$ (undo with $-7$) or $4x = 20$ (undo with $\div 4$).
• Two-step: two operations, like $3x + 5 = 20$. Undo the addition or subtraction first, then the multiplication or division.
• Multi-step: more layers, or brackets, or fractions, or the variable on both sides. You first tidy the equation (expand brackets, clear fractions, gather the variable) and then it collapses into a one- or two-step equation you already know how to finish.

The plan never changes: tidy, then peel the operations off the variable in reverse order, doing the same to both sides each time.

Brackets: expand or divide

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: clear them with the lowest common denominator

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. For $\dfrac{x}{3} + \dfrac{x}{4} = 14$ the LCD of $3$ and $4$ is $12$; multiplying all three terms by $12$ turns it into $4x + 3x = 168$, an easy two-step. There are two traps to avoid. The first is multiplying only some of the terms by the LCD, when in fact every term, including any whole number, must be multiplied. The second is picking a common denominator that is not the lowest; this still works but leaves bigger numbers to simplify.

The variable on both sides

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.

How exam questions ask about linear equations

The wording tells you which kind of equation you are facing and how much tidying it needs:

• "Solve the equation..." is the direct instruction. Identify whether it is one-step, two-step or multi-step, tidy if needed, then peel the operations off in reverse order.
• "Solve for $x$" or "find the value of $x$" mean the same thing: get the variable by itself.
• An equation with brackets ("solve $4(m + 3) = 20$") signals expand-or-divide first.
• An equation with fractions ("solve $\dfrac{x}{2} + \dfrac{x}{6} = 1$") signals clear the fractions with the LCD first.
• The variable on both sides ("solve $7p + 2 = 6p - 6$") signals gather the variable on one side first.
• "Express your answer as a fraction / mixed number" is a presentation instruction: do not round, leave the exact fraction.
• A worded or applied problem ("after how many weeks is the cost the same", "find the width of the rectangle") asks you to form the equation from the words first, then solve it. The marks split between setting up the correct equation and solving it, so always write the equation line.
• "Check your solution" or "show that $x = \ldots$ satisfies the equation" asks you to substitute the value back in and show both sides come out equal.

Solving by balancing: a stage-by-stage scale

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.

Stage 1, set up the scale. The left pan holds three $x$ tiles and a $5$ tile (that is $3x + 5$), and the right pan holds a $20$ tile. The scale balances because $3x + 5 = 20$.

Stage 2, take $5$ from both sides. Remove the $5$ tile from the left pan and remove $5$ from the right pan ($20$ becomes $15$). The same weight has gone from each side, so the scale is still level: $3x = 15$.

Stage 3, divide both sides by $3$. Share each pan into three equal groups. One group on the left is a single $x$ tile, and the matching third of the right pan is worth $5$. Doing the same division to both sides keeps the balance: $x = 5$.

Stage 4, read and check the solution. One $x$ tile balances a $5$ tile, so $x = 5$. Substitute back to be sure: $3 \times 5 + 5 = 15 + 5 = 20$, which equals the right pan, confirming the solution.

Always check by substituting back

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.