Before Calculus: 10 Algebra Skills You Should Know

Negative numbers: a short repair before algebra

If negative signs interrupt your algebra, pause here. This optional section practises the numbers first, then returns to an expression with a letter. You can go straight to basic algebra if these steps are already comfortable.

In $-4$, the minus sign belongs to a negative number. In $5-4$, it tells you to subtract. Parentheses help separate the two uses: $5-(-4)$ means subtract the number $-4$. Zero is neither positive nor negative.

1 Add and subtract on a number line

Start at the first number. Adding a positive number moves right; adding a negative number moves left. Move by the added number’s distance from zero. To subtract a number, add its opposite instead: the opposite of $-4$ is $4$.

Worked example 1: the same start, different directions

$$-2-4=-2+(-4)=-6.$$

Subtracting $4$ is adding $-4$. Start at $-2$ and move 4 units left.

$$-2-(-4)=-2+4=2.$$

Subtracting $-4$ undoes adding $-4$, so move 4 units right instead.

Watch the operation: two negative numbers do not always give a positive answer. For addition, $-2+(-4)=-6$; do not use the multiplication rule here.

2 Multiply and divide: choose the sign separately

For two nonzero numbers, first multiply or divide their distances from zero. Then choose the sign: matching signs give a positive result; different signs give a negative result.

Why can two negative factors give a positive product? Multiplying by $-1$ takes the opposite: the opposite of $-4$ is $4$. For example, $(-3)(-4)$ is the opposite of $3(-4)=-12$, so it equals $12$.

Worked example 2: use multiplication to check division

$$(-3)(-4)=12,\qquad 12\div(-3)=-4.$$

The product is positive because the factors have matching signs. The quotient is negative because its two numbers have different signs. Check division by multiplying: $(-3)(-4)=12$.

Zero times any number is zero. Zero divided by a nonzero number is zero. Division by zero is undefined.

3 Keep the substituted number inside parentheses

Calculate inside parentheses first, then powers, then multiplication and division from left to right, then addition and subtraction from left to right.

Worked example 3: substitute a negative value

Evaluate $2x^2-3$ when $x=-2$:

$$2(-2)^2-3=2(4)-3=8-3=5.$$

The whole input $-2$ is squared: $(-2)^2=(-2)(-2)=4$.

The base matters: $-2^2=-(2^2)=-4$, but $(-2)^2=4$. Without parentheses, the square applies before the leading minus. Similarly, $-a^2$ means $-(a^2)$; $(-a)^2$ squares the entire base $-a$.

Try four short questions

Choose a question, try it on paper, then open the hint or solution.

1. Calculate $-5+8$.

   Show a hint

   Start at $-5$. Adding $8$ moves 8 units right.

   Show the solution

   $-5+8=3$. Five units take you to zero; three more take you to $3$.

2. Calculate $4-(-3)$.

   Show a hint

   Replace subtracting $-3$ with adding its opposite.

   Show the solution

   $4-(-3)=4+3=7$. Start at $4$ and move 3 units right.

3. Calculate $(-18)\div3\times(-2)$.

   Show a hint

   Multiplication and division have equal priority. Work from left to right.

   Show the solution

   $(-18)\div3=-6$, then $(-6)(-2)=12$. The answer is $12$. Do not multiply $3$ by $-2$ first: the expression has no parentheses grouping those factors.

4. When $a=-3$, evaluate $-a^2$ and $(-a)^2$.

   Show a hint

   Rewrite the first as $-(a^2)$. In the second, the entire expression $-a$ is the base.

   Show the solution

   $-a^2=-((-3)^2)=-9$. For the second, $-a=-(-3)=3$, so $(-a)^2=3^2=9$. The parentheses change what is squared.

Return to basic algebra and try a negative input again. If fractions are your next obstacle, use fractions, one step at a time.

Starting with basic algebra

If letters in an equation still feel unfamiliar, start here. You will read an expression, combine matching terms and solve an equation one step at a time. You need addition, subtraction, multiplication and division. A variable is a letter that represents a number. In these examples, the variables represent real numbers.

Use a pencil and try each question before opening its answer. Checking an answer is part of the solution.

Stuck on a negative sign? Review operations with negative numbers, then return to this step.

1 Read the expression before calculating

The expression $3x+2$ means “multiply $x$ by 3, then add 2.” The writing $3x$ means $3\times x$. An expression can be evaluated once a value of $x$ is given. An equation, such as $3x+2=14$, states that two expressions have the same value.

Worked example: evaluate $3x+2$ when $x=4$.

$$3(4)+2=12+2=14.$$

The parentheses show where the value of $x$ went. For a negative input, keep them: when $x=-2$, $3(-2)+2=-6+2=-4$.

Try it: evaluate $5a-3$ when $a=2$.

Retry: evaluate $2b+5$ when $b=-3$.

2 Combine terms that match

Like terms have the same variable part. Four lots of $x$ plus two lots of $x$ make six lots of $x$. A constant is a number without a variable.

Worked example:

$$\begin{aligned} 4x+3+2x-5&=(4x+2x)+(3-5)\\ &=6x-2. \end{aligned}$$

You can combine $4x$ and $2x$, and you can combine 3 and $-5$. You cannot combine $6x-2$ into $4x$: the two terms have different variable parts.

Try it: simplify $7y-2+3y+6$.

Retry: simplify $8p-3p+1-4$.

3 Keep the two sides equal

An equation is like a balance. Adding the same number to both sides, subtracting the same number from both sides, or multiplying or dividing both sides by the same nonzero number keeps that balance.

Start with $x+5=12$. Subtract 5 from both sides to get $x=7$. Check: $7+5=12$.

Worked example: solve $3x-4=11$.

$$\begin{aligned} 3x-4+4&=11+4\\ 3x&=15\\ x&=5. \end{aligned}$$

We added 4 to undo subtracting 4, then divided both sides by 3 to undo multiplication by 3. Check in the original equation: $3(5)-4=11$.

Try it: solve $2x+7=19$ and check your answer.

Retry: solve $-2x=10$.

4 Undo parentheses and division carefully

In $2(x+3)=14$, the whole expression $x+3$ is multiplied by 2. Dividing both sides by 2 first gives $x+3=7$, so $x=4$. Check: $2(4+3)=14$.

Worked example: solve $\frac{x-1}{3}=4$.

$$\begin{aligned} x-1&=12\\ x&=13. \end{aligned}$$

Multiplying both sides by 3 undoes division by 3. Adding 1 then gives 13. Check: $(13-1)/3=12/3=4$. The fraction bar divides the whole numerator $x-1$.

Try it: solve $\frac{x+2}{4}=3$.

Retry: solve $3x+5=2x+12$.

One short application

A rectangle is 7 cm long and has perimeter 22 cm. Its width is $w$ cm. The perimeter is the total distance around the rectangle, so

$$2(7)+2w=22.$$

Subtract 14 to get $2w=8$, then divide by 2: $w=4$. The width is 4 cm. Check by adding all four sides: $7+4+7+4=22$ cm.

Build an equation from a word problem

Sometimes the hard part comes before solving: the equation is not given. Start by naming one unknown quantity, express the other quantities using it, then use a stated total or equality to connect them. The equation is a model of the situation. Solving it finds a candidate answer; checking the original story decides whether that answer fits.

1. Name the unknown and its unit. State what the letter measures or counts. Decide whether it can be any positive real number, or must be a whole-number count.
2. Translate the relationship. “4 cm longer than the width” means add 4 to the width measured in centimetres. “Twice the width, plus 1 cm” means multiply the width by 2, then add 1.
3. Write what is equal. For a perimeter, add all four sides. For a total price, add the price of each group. Both sides must describe the same kind of quantity in the same units.
4. Solve and check the story. Use balanced operations, recover every requested quantity, and check both the relationship and the total.

Worked example: neither side length is given

A rectangle has perimeter 32 cm. Its length is 4 cm greater than its width. Find both dimensions.

Let $w$ be the numerical value of the width in centimetres. Then the length, in centimetres, is $w+4$. The allowed values have $w>0$; lengths need not be whole numbers. The number 4 represents a difference of 4 cm, not four copies of the width.

There are two lengths and two widths. Their numerical values in centimetres add to 32:

$$2w+2(w+4)=32.$$

This is why the parentheses matter: each of the two lengths includes the extra 4 cm. Expand and solve:

$$2w+2w+8=32$$

$$4w+8=32$$

$$4w=24$$

$$w=6.$$

The width is 6 cm and the length is $6+4=10$ cm. Both are positive. The difference is $10-6=4$ cm, and the perimeter is $2(6)+2(10)=32$ cm. Checking only the perimeter would not check the “4 cm greater” condition.

Common setup error: $2w+2w+4=32$ counts the extra 4 cm only once. Read the two lengths separately as $(w+4)+(w+4)$ before combining terms. The expression $w(w+4)$ instead describes area in square centimetres; it cannot equal a perimeter in centimetres.

Guided retry: translate the new relationship

A second rectangle has perimeter 44 cm. Its length is 1 cm greater than twice its width. Let $w$ again be the numerical value of its width in centimetres, with $w>0$. Write the length in terms of $w$, build the perimeter equation, and find both dimensions.

Show a setup hint

“Twice the width” gives $2w$; adding 1 cm gives $2w+1$. It does not give $2(w+1)$, which adds 1 before doubling. Now use two widths and two lengths.

Show the solution and checks

$$2w+2(2w+1)=44$$

$$6w+2=44$$

$$6w=42$$

$$w=7.$$

The width is 7 cm and the length is $2(7)+1=15$ cm. The relationship holds: $15=2(7)+1$. The perimeter is $2(7)+2(15)=44$ cm, and both dimensions are positive.

Independent problem: a total in a different setting

A school event sells exactly 22 tickets. Each regular ticket costs $8 CAD and each reduced-price ticket costs $5 CAD. The ticket revenue is exactly $140 CAD. There are no other ticket categories, fees or discounts. How many of each ticket were sold? Define one unknown, express the other count using it, write an equation, and check both totals.

Show the solution and checks

Let $r$ be the number of regular tickets. The reduced-price count is $22-r$: the two counts must add to 22. The domain is the integers from 0 to 22 inclusive, because tickets are counted individually and neither count can be negative.

Price per ticket multiplied by the number of tickets gives the revenue from that group. Expressing the revenue in CAD gives

$$8r+5(22-r)=140$$

$$8r+110-5r=140$$

$$3r+110=140$$

$$3r=30$$

$$r=10.$$

There are 10 regular tickets and $22-10=12$ reduced-price tickets. Check the count: $10+12=22$. Check the revenue: $8(10)+5(12)=80+60=140$ CAD. Both counts are nonnegative integers. Here $5r$ would price another group of $r$ tickets, rather than the remaining $22-r$ tickets.

A non-integer count would not be fixed by rounding: it would mean that the exact ticket totals and prices do not fit the model. Recheck the setup and the stated data.

Before moving on: explain what each term counts or measures. A correct calculation cannot repair an equation that describes the wrong situation.

For a word problem you’re stuck on, algebra and precalculus tutoring can begin with choosing the unknown and building the equation together from an example like these. We can start with the basics and build your confidence step by step. Share your level and topic so we can confirm support and availability before booking. Sessions are currently in English.

Choose what to practise next

• If fractions interrupt your work, continue to Work confidently with fractions in this same guide.
• If signs change when you open parentheses, continue to Expand without losing signs.
• If equations are still uncertain, continue to Solve linear equations with balanced operations.
• If you can explain each step above, try the Points, slope and straight-line graphs lesson.
• For lengths and angles, continue to Right triangles: lengths and angles.

The later skills in this guide build towards precalculus. Work through them when they appear in your current course; you do not need logarithms or composition to practise the first equations above.

Algebra and precalculus tutoring

Linear inequalities: describe all the allowed values

An equation asks where two expressions are equal. An inequality compares them. Here $x$ is a real number: $x<3$ means less than 3, while $x\leq3$ means less than or equal to 3. Similarly, $x>3$ means greater than 3 and $x\geq3$ includes 3. A solution set contains every value that makes the original inequality true.

Use the equation-balancing skills above. Adding or subtracting the same number on both sides keeps the inequality direction. Multiplying or dividing both sides by a positive number also keeps it. Multiplying or dividing both sides by a negative number reverses it.

Worked example 1: an endpoint that is excluded

Solve $x+2<5$. Subtract 2 from both sides:

$$x<3.$$

The answer is all real numbers less than 3, not just one number. An open circle at 3 excludes the endpoint; the ray points left toward smaller values. Check an inside value: $x=2$ gives $4<5$, which is true. An outside value, $x=4$, gives $6<5$, which is false. At the endpoint, $x=3$ gives $5<5$, also false. These checks illustrate the set; the reversible algebra establishes the full solution.

Why a negative multiplier reverses the comparison

Start with $2<5$. Multiplying both numbers by $-1$ gives $-2>-5$: reflection across zero reverses their left-to-right order on the number line. More generally, if $a<b$, then $b-a>0$. For a negative number $c$, the difference $cb-ca=c(b-a)$ is negative, so $ca>cb$. Here $a,b,c$ are real numbers and $c<0$. Division by a negative number is multiplication by its reciprocal, which is negative, so the same reversal applies. Equality remains included when the symbol is $\leq$ or $\geq$.

Worked example 2: reverse the sign and include the endpoint

Solve $-2x+1\leq5$. Subtract 1, then divide both sides by $-2$:

$$\begin{aligned}-2x&\leq4\\x&\geq-2.\end{aligned}$$

The answer is all real numbers greater than or equal to $-2$. A filled circle includes $-2$, and the ray points right. Check $x=0$ inside the set: $1\leq5$ is true. Check $x=-3$ outside it: $7\leq5$ is false. At $x=-2$, $5\leq5$ is true, so the endpoint belongs.

Practise one decision at a time

For questions 1–3, $x$ is real. Give the solution set, say whether the endpoint is included, and check one value inside and one outside. In question 4, use the stated counting domain.

1. Solve $x+4\leq7$.

2. Solve $x/3>2$.

3. Solve $5-2x>11$.

4. A workshop charges a fixed fee of 5 CAD plus 3 CAD per kit. Your total budget is 20 CAD. Let $n$ be the number of kits, a nonnegative integer. Write the budget inequality and find every allowed value of $n$.

Before moving on: explain why the comparison changes when dividing by a negative number, then connect the symbol to the endpoint and direction on the number line.

Optional extension: absolute value is a distance

Use this extension after the linear-inequality examples. Allow about 15 extra minutes. All variables below are real numbers unless a physical meaning adds a restriction.

The absolute value |u| is the distance from u to zero on a number line. A distance is never negative: |−3|=3, |3|=3 and |0|=0. The bars do not mean “remove every minus sign”: first evaluate the whole expression inside them. For example, at x=1, $|x-4|=|1-4|=|-3|=3$.

More generally, |x−a| is the distance between x and a. If that distance is at most a positive number r, then x must lie inside the interval from a−r to a+r, including both ends:

$$|x-a|\leq r\quad\Longleftrightarrow\quad a-r\leq x\leq a+r,\qquad r>0.$$

The two inequalities mean and: both must hold. By contrast, |x−a|>r means the distance is greater than r, so x<a−r or x>a+r. These are two outside regions, not the interval between them. Replace “at most” by “less than” to exclude inside endpoints; replace “greater than” by “at least” to include outside endpoints.

Worked example: the expression is twice the distance

Solve $|2x-6|\leq4$. Since $2x-6=2(x-3)$ and $|2(x-3)|=2|x-3|$, the distance from x to 3 can be at most 2. We can also derive the endpoints without guessing:

$$-4\leq2x-6\leq4.$$

Add 6 to all three parts, then divide every part by positive 2:

$$2\leq2x\leq10\qquad\Longrightarrow\qquad1\leq x\leq5.$$

The solution interval is $[1,5]$. Square brackets include the endpoints. Check them in the original statement: $|2(1)-6|=4$ and $|2(5)-6|=4$. The inside value x=3 gives 0; the outside value x=6 gives 6 and fails. These checks support the algebra but do not replace the reasoning for every value.

Check the right-hand side before splitting into cases

A negative number cannot be a distance. Therefore |u|=−2 and |u| ≤ −2 have no solutions, while |u| ≥ −2 is true for every value for which u is defined. At zero, |u|=0 or |u| ≤ 0 requires u=0; |u|<0 is impossible; |u|>0 requires u ≠ 0; and |u| ≥ 0 is always true on the original domain. Keep any restrictions from the expression inside the bars.

Try the four questions · Return to linear inequalities

Four steps of distance-and-inequality practice

1. Find the centre, then decide about endpoints

Solve $|x+2|<3$. State the centre, both bounds and whether the endpoints belong.

2. Keep the two outside branches separate

Solve $|3-2x|\geq5$. Explain why “and” would give the wrong solution set.

3. Diagnose an impossible distance

Without splitting into two algebraic branches, solve both $|x-4|=-2$ and $|x-4|\geq-2$. Then solve $|x-4|=0$.

4. Translate a tolerance into an interval

In this original model, a strip should be 18 cm long. It is accepted when its length differs from 18 cm by at most 0.4 cm, including the limit. Let L be its numerical length in centimetres, with L>0. Write an absolute-value condition and the accepted interval. Decide whether lengths 17.58 cm and 18.35 cm are accepted. Treat these stated lengths as exact for this exercise.

Before moving on: say whether the allowed values are inside, outside, a single point, all of the original domain, or none. Then check the endpoints in the original condition.