EQUATIONS

Add 8 to both sides. This is needed because the equation must stay 'balanced' — whatever is done to one side must be done to the other, to keep both sides equal.

$$x-8+8=15+8 \Rightarrow x=23$$

Subtract 3 from both sides: $$\frac{x}{5}=6$$

Multiply both sides by 5: $$x=30$$

$10+5=10 \Rightarrow 15=10$ — this is FALSE. Adding to only one side breaks the equality.

$10+5=10+5 \Rightarrow 15=15$ — TRUE, the balance is maintained.

Since an equation represents a true balance between two equal quantities, any operation applied to only one side will generally destroy that balance (as shown), while applying the SAME operation to both sides preserves it — this is why balance must always be maintained on both sides throughout solving.

Subtracting 6 from BOTH sides gives $x+6-6=14-6$, which simplifies to $x=8$. The '6' effectively disappears from the left side and reappears as '$-6$' on the right — this is exactly what 'moving it and changing the sign' describes, just using balance-operation language instead of a shortcut phrase.

The shortcut can become confusing or lead to errors in more complex equations (e.g. those with fractions, brackets, or the variable on both sides), where it's not always obvious what should 'move'. Understanding the underlying balance principle ensures the method can be applied correctly and confidently to ANY equation, not just simple ones.

Removing 15 kg from both sides keeps the see-saw balanced: $x=40-15=25$ kg.

Just as removing the same amount of weight from both sides of a physically balanced see-saw keeps it balanced, performing the same mathematical operation on both sides of an equation keeps the equality true — the see-saw is a physical illustration of exactly the same balance principle used in algebra.

Add 3 to both sides: $$\frac{x}{4}=9$$

Multiply both sides by 4: $$x=36$$

$2x=14 \Rightarrow x=7$.

$x+3=10 \Rightarrow x=7$ — matches. Both orderings work because each step, regardless of order, is a VALID balanced operation (applied equally to both sides) — as long as every step maintains the equality, the final isolated value of $x$ must be the same, since balanced operations never change WHICH value of $x$ makes the equation true, only how the equation is temporarily expressed along the way.

While it IS valid to perform multiple balanced operations at once (as long as they're applied identically to both sides), doing so increases the risk of ARITHMETIC ERRORS — combining several sign changes and term movements in one step makes it harder to spot mistakes, and harder for a reader (or the student themselves, checking their own work) to verify each individual step was done correctly. Breaking the process into smaller, clearly labelled steps reduces error risk, especially while still building confidence with the method.

Step 1 (subtract $3x$ from both sides): $2x-2=8$. Step 2 (add 2 to both sides): $2x=10$. Step 3 (divide by 2): $x=5$. Each step is a single, clearly checkable balanced operation.

$2x+10=3x-3$. Subtract $2x$ from both sides: $10=x-3$. Add 3 to both sides: $13=x$, i.e. $x=13$.

$2(13+5)=2(18)=36$. $3(13-1)=3(12)=36$ ? — both sides equal 36.

Removing 15g from both sides: $x=47-15=32$g.

Contents mass $=x-\text{empty container mass}=32-9=23$g. Since $23\text{g}\ne25\text{g}$ (and is actually LESS), the contents do NOT meet the required 25g — the pharmacist would need to add approximately 2 more grams of contents.