Exercises — Addition, subtraction, multiplication, division of fractions

Same denominator already — the pieces are the same size (sevenths), so just count them. $\frac{2}{7}+\frac{3}{7}=\frac{2+3}{7}=\frac{5}{7}$ What / Why: we added numerators only. The bottom stays $7$ because "5 sevenths" is still measured in sevenths — the piece size never changes when you add same-size pieces. See the number line: two jumps of $\frac{1}{7}$ then three more land on $\frac{5}{7}$.

Multiplication → march straight across (see Fractions - meaning and equivalent fractions for the "fraction-of-a-fraction" picture). $\frac{2}{5}\times \frac{3}{4}=\frac{2\times 3}{5\times 4}=\frac{6}{20}=\frac{3}{10}$ Why the last step: $\frac{6}{20}=\frac{3\cdot 2}{10\cdot 2}=\frac{3}{10}$ by the golden rule (divide top and bottom by the common factor $2$, which is nonzero).

Division → Keep · Change · Flip. Keep $\frac{5}{6}$, change $÷$ to $\times$, flip the divisor $\frac{5}{2}\to \frac{2}{5}$. The flip is legal because the divisor's numerator $5\ne 0$. The flipped fraction is the reciprocal — see Reciprocals and multiplicative inverse. $\frac{5}{6}÷\frac{5}{2}=\frac{5}{6}\times \frac{2}{5}=\frac{5\times 2}{6\times 5}=\frac{10}{30}=\frac{1}{3}$

Same denominator (fifths), so subtract the tops directly: $\frac{2}{5}-\frac{3}{5}=\frac{2-3}{5}=\frac{-1}{5}=-\frac{1}{5}$ What / Why: taking away more pieces than you have pushes you past $0$ into the negatives — a perfectly valid rational number. On the number line this lands one fifth to the left of $0$. The minus sign belongs to the whole fraction: $\frac{-1}{5}=\frac{1}{-5}=-\frac{1}{5}$ are all the same point.

Bottoms differ, so the pieces are not the same size — we must force a common bottom. Since $6$ is a multiple of $3$, use $6$ (this is $\text{LCM}(3,6)=6$; see LCM and HCF). $\frac{2}{3}=\frac{2\cdot 2}{3\cdot 2}=\frac{4}{6},\frac{1}{6}=\frac{1}{6}$ Now both are sixths — same size, so count: $\frac{4}{6}+\frac{1}{6}=\frac{5}{6}$

$\text{LCM}(10,15)=30$ (smallest number both divide). Why LCM not $10\times 15=150$? Same value, smaller numbers to carry. $\frac{7}{10}=\frac{7\cdot 3}{30}=\frac{21}{30},\frac{2}{15}=\frac{2\cdot 2}{30}=\frac{4}{30}$ Same-size pieces ⇒ subtract tops: $\frac{21-4}{30}=\frac{17}{30}$ Cannot simplify: $17$ is prime and doesn't divide $30$.

Flip the divisor $\frac{9}{8}\to \frac{8}{9}$ (legal: its numerator $9\ne 0$), then multiply. Cancel early to stay small: $\frac{3}{4}\times \frac{8}{9}=\frac{3\times 8}{4\times 9}$ Cancel $8$ with $4$ ($=2$) and $3$ with $9$ ($=3$) — this is the golden rule dividing top and bottom by common factors: $=\frac{1\times 2}{1\times 3}=\frac{2}{3}$ Sanity check: we divided by $\frac{9}{8}$, which is bigger than $1$, so the answer should shrink below $\frac{3}{4}$ — and $\frac{2}{3}<\frac{3}{4}$. ✓

First turn the mixed number into a single fraction. $2\frac{1}{3}$ means $2+\frac{1}{3}$; two wholes are $\frac{6}{3}$, so: $2\frac{1}{3}=\frac{6}{3}+\frac{1}{3}=\frac{7}{3}$ Common bottom of $3$ and $6$ is $6$: $\frac{7}{3}=\frac{14}{6},\frac{14}{6}-\frac{5}{6}=\frac{9}{6}=\frac{3}{2}=1\frac{1}{2}$

Order matters: multiplication happens before addition. Do the product first. $\frac{2}{3}\times \frac{3}{4}=\frac{2\times 3}{3\times 4}=\frac{6}{12}=\frac{1}{2}$ Now add with common bottom $2$: $\frac{1}{2}+\frac{1}{2}=1$ What if you had added first (wrong)? $\frac{1}{2}+\frac{2}{3}=\frac{7}{6}$, then $\times \frac{3}{4}=\frac{7}{8}\ne 1$. Different answer — so order is not optional.

We are subtracting a bigger fraction from a smaller one, so expect a negative answer. Common bottom of $4$ and $6$ is $\text{LCM}(4,6)=12$: $\frac{1}{4}=\frac{3}{12},\frac{5}{6}=\frac{10}{12}$ Subtract tops (keep the sign of the difference): $\frac{3-10}{12}=\frac{-7}{12}=-\frac{7}{12}$ On the number line this sits between $0$ and $-1$, to the left of $0$ — a valid rational number.

Brackets first. Common bottom of $4,2$ is $4$: $\frac{3}{4}-\frac{2}{4}=\frac{1}{4}$ Now divide → flip the divisor $\frac{1}{8}\to \frac{8}{1}$ (its numerator $1\ne 0$): $\frac{1}{4}\times \frac{8}{1}=\frac{8}{4}=2$ Meaning check: "how many $\frac{1}{8}$-chunks fit into $\frac{1}{4}$?" Two eighths make a quarter, so 2 — matches. ✓

Multiplication before addition. Cancel while multiplying (golden rule): $\frac{5}{6}\times \frac{3}{10}=\frac{5\times 3}{6\times 10}=\frac{15}{60}=\frac{1}{4}$ Then add (common bottom $4$, already matching): $\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$

Division before subtraction. Flip $\frac{4}{9}\to \frac{9}{4}$ (numerator $4\ne 0$), cancel: $\frac{2}{3}\times \frac{9}{4}=\frac{2\times 9}{3\times 4}=\frac{18}{12}=\frac{3}{2}$ Now subtract with common bottom of $2$ and $6$, which is $6$: $\frac{3}{2}=\frac{9}{6},\frac{9}{6}-\frac{1}{6}=\frac{8}{6}=\frac{4}{3}$

A stacked fraction is a division: numerator $÷$ denominator. Simplify each part to one fraction first (and note the bottom must not simplify to $0$ — here it won't). Top: $\frac{2}{3}+\frac{1}{6}=\frac{4}{6}+\frac{1}{6}=\frac{5}{6}$. Bottom: $\frac{3}{4}-\frac{1}{2}=\frac{3}{4}-\frac{2}{4}=\frac{1}{4}$. So the bar means: $\frac{5}{6}÷\frac{1}{4}=\frac{5}{6}\times \frac{4}{1}=\frac{20}{6}=\frac{10}{3}$

Division binds tighter than $+$ and $-$, so evaluate $\frac{1}{2}÷\frac{3}{4}$ first (divisor numerator $3\ne 0$): $\frac{1}{2}÷\frac{3}{4}=\frac{1}{2}\times \frac{4}{3}=\frac{4}{6}=\frac{2}{3}$ Now left to right: $1-\frac{2}{3}+\frac{1}{3}$. Common bottom $3$ (write $1=\frac{3}{3}$): $\frac{3}{3}-\frac{2}{3}+\frac{1}{3}=\frac{3-2+1}{3}=\frac{2}{3}$

"How many $\frac{3}{4}$-cup chunks fit into $\frac{9}{2}$ cups?" is a division — see Ratios and proportion for the "how many fit" viewpoint. $\frac{9}{2}÷\frac{3}{4}=\frac{9}{2}\times \frac{4}{3}=\frac{9\times 4}{2\times 3}=\frac{36}{6}=6$ Exactly $6$ — no remainder. So 6 full batches, and sugar left over $=0$ cups. Check by multiplying back: $6\times \frac{3}{4}=\frac{18}{4}=\frac{9}{2}$ ✓ — all the sugar is used.