Translate a description piece by piece, matching each phrase to an operation:

Word problems often hide an equation inside a real-world statement. Whenever a sentence says two quantities are "the same as" or "equal to" each other, that's the cue to write an equals sign and form an actual equation rather than just an expression.

Treat the equation like a balanced set of scales — whatever is done to one side must be done to the other. Expand any brackets first, then collect the unknowns on one side and the numbers on the other.

Solve $5(x-3) = 2x+9$:

1. Expand the bracket: $5x - 15 = 2x + 9$.
2. Subtract $2x$ from both sides: $3x - 15 = 9$.
3. Add $15$ to both sides: $3x = 24$.
4. Divide by $3$: $x = 8$.

Simultaneous equations have two unknowns, and exactly one pair $(x,y)$ makes both equations true together. Elimination removes one letter by combining the equations:

• Same signs on the matching term $\Rightarrow$ subtract the equations.
• Different signs on the matching term $\Rightarrow$ add the equations.

Solve $3x+2y=18$ and $5x-2y=14$:

1. The $y$-terms are $+2y$ and $-2y$ — different signs, so add the two equations to eliminate $y$: $8x = 32$.
2. Solve for $x$: $x = 4$.
3. Substitute $x=4$ into the first equation: $3(4)+2y=18 \implies y=3$.
4. Solution: $x=4,\ y=3$.

Sometimes neither letter is ready to eliminate straight away. Multiply one or both equations by a suitable number first, so that one pair of matching terms does have equal (or opposite) coefficients, then proceed exactly as before.

Solve $2x+3y=16$ and $x+2y=9$:

1. Multiply the second equation by $2$ so the $x$-terms match: $2x+4y=18$.
2. Both equations now start with $2x$ (same sign), so subtract them: $(2x+3y)-(2x+4y) = 16-18$, giving $-y=-2$, so $y=2$.
3. Substitute $y=2$ into the original second equation: $x+2(2)=9 \implies x=5$.
4. Solution: $x=5,\ y=2$.

The subject is the letter sitting alone on one side of the formula. Use inverse operations to peel away everything else until the new letter stands by itself — at Core level, the subject appears only once, with no powers or roots to undo.

Make $r$ the subject of $C=2\pi r$: since $r$ is multiplied by $2\pi$, divide both sides by $2\pi$:

The formula for the speed of an object is $v = u+at$. Make $t$ the subject.

Solution

1. Subtract $u$ from both sides to isolate the term containing $t$:

1. Divide both sides by $a$ to leave $t$ completely on its own:

The final line is the answer only once the new subject is alone on one side — stopping at $at = v-u$ has not fully answered the question.