Algebra

Level · Middle & high school · exam foundations Chapters · 11 Explorers · 5 interactive Feeds into · SAT · ACT · IGCSE · A-Level

Linear Equations & Lines

A linear relationship changes at a constant rate — its graph is a straight line. Mastering lines is the foundation for everything from SAT word problems to the slope fields of calculus.

● Definition 1.1 — Linear equation

A linear equation in one variable has the form $ax+b=0$ with $a\neq 0$ ; its single solution is $x=-\dfrac{b}{a}$ . In two variables, $y=mx+b$ describes a line of slope $m$ and $y$ -intercept $b$ .

The slope measures steepness as rise over run, $m=\dfrac{\Delta y}{\Delta x}=\dfrac{y_2-y_1}{x_2-x_1}$ . Drag the sliders to see how $m$ tilts the line and $b$ slides it up or down — the amber dots mark where it crosses the axes.

⊙ Explorer · Slope & interceptinteractive

Slope m = 1

Intercept b = 1

y = x + 1

slope m—

y-intercept—

x-intercept—

Parallel lines

L_1

and

L_2

share identical slope

m = 0.5

. Perpendicular line

L_3

has the negative reciprocal slope

m = -2

, intersecting

L_1

at right angles at

(0,1)

Three forms of a line

Slope–intercept

y=mx+b

Point–slope

y-y_1=m\,(x-x_1)

Standard

Ax+By=C

Worked example 1.A — Solve

Solve $3x+5=20$ .

Subtract 5, then divide by 3:

3x=15 \quad\Longrightarrow\quad x=5.

Worked example 1.B — Equation of a line

Find the line through $(1,2)$ and $(5,10)$ .

Slope first, then point–slope with $(1,2)$ :

m=\frac{10-2}{5-1}=2, \qquad y-2=2(x-1)\;\Longrightarrow\; y=2x.

⚠ Common traps — Undefined slopes & reversing inequality signs

Vertical vs horizontal lines: A horizontal line $y = b$ has zero slope ( $m = 0$ ). A vertical line $x = a$ has an undefined slope (division by zero: $\Delta x = 0$ ) and cannot be written in slope-intercept form $y = mx + b$ .

Perpendicular slopes: Perpendicular lines have negative reciprocal slopes: $m_2 = -\dfrac{1}{m_1}$ (meaning $m_1 \cdot m_2 = -1$ ). Simply negating without inverting is a frequent mistake.

Inequality sign reversal: Whenever you multiply or divide both sides of an inequality by a negative number, always reverse the inequality symbol (e.g., $-2x < 8 \implies x > -4$ ).

Find the equation of the line that passes through

(4, -1)

and is perpendicular to the line

2x - 3y = 9

. Express your answer in slope-intercept form.

1. Find the slope of the given line by converting to slope-intercept form:

-3y = -2x + 9 \implies y = \frac{2}{3}x - 3.

The slope of the given line is $m_1 = \dfrac{2}{3}$ .

2. The perpendicular slope is the negative reciprocal:

m_\perp = -\frac{3}{2}.

3. Use point-slope form with $(x_1, y_1) = (4, -1)$ :

y - (-1) = -\frac{3}{2}(x - 4) \implies y + 1 = -\frac{3}{2}x + 6 \implies y = -\frac{3}{2}x + 5.

Solve the linear inequality

-3(2x - 4) \le 4x - 8

and express the solution in interval notation.

Expand the left side: $-6x + 12 \le 4x - 8.$ Collect all $x$ terms on one side: $-10x \le -20.$ Divide both sides by $-10$ and reverse the inequality symbol: $x \ge 2.$ In interval notation: $[2, \infty)$ .

📝 Chapter Quiz

Questions Level

Quadratic Functions

Quadratics introduce curvature. Their U-shaped graphs — parabolas — appear in projectile motion, optimization, and a large share of every standardized test.

● Definition 2.1 — Quadratic function

A quadratic function has the form $f(x)=ax^2+bx+c$ with $a\neq 0$ . Its graph is a parabola opening upward when $a>0$ and downward when $a<0$ , with axis of symmetry $x=-\dfrac{b}{2a}$ .

▲ Theorem 2.2 — The quadratic formula

The solutions of $ax^2+bx+c=0$ are

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

The discriminant $\Delta=b^2-4ac$ reveals the roots: $\Delta>0$ gives two real roots, $\Delta=0$ one repeated root, and $\Delta<0$ no real roots.

In the explorer, $a$ , $b$ , and $c$ reshape the parabola live. The amber vertex sits on the dashed axis of symmetry, and when the discriminant is non-negative the amber roots appear exactly where the curve meets the $x$ -axis.

⊙ Explorer · The parabola

ax^2+bx+c

interactive

a

b

-1

c

-3

y = x² − x − 3

discriminant Δ—

vertex—

roots—

The discriminant

\Delta = b^2 - 4ac

governs real roots:

\Delta > 0

intersects the

x

-axis twice,

\Delta = 0

touches the axis tangentially at one root, and

\Delta < 0

never crosses.

Three forms of a quadratic

Standard

f(x)=ax^2+bx+c

Vertex

(h,k)

f(x)=a(x-h)^2+k

Factored

f(x)=a(x-r_1)(x-r_2)

Worked example 2.A — Factoring

Solve $x^2-5x+6=0$ .

Find two numbers multiplying to $6$ and adding to $-5$ : namely $-2$ and $-3$ .

(x-2)(x-3)=0 \quad\Longrightarrow\quad x=2 \ \text{or}\ x=3.

Worked example 2.B — Quadratic formula

Solve $2x^2+3x-2=0$ .

Here $a=2,\ b=3,\ c=-2$ , so $\Delta=9+16=25$ :

x=\frac{-3\pm\sqrt{25}}{4}=\frac{-3\pm 5}{4}=\tfrac12 \ \text{or}\ -2.

⚠ Common traps — Vertex sign errors & the square root property

Vertex form sign: In $f(x) = a(x - h)^2 + k$ , the vertex is at $(h, k)$ , not $(-h, k)$ . For example, $f(x) = (x - 3)^2 + 5$ has its vertex at $(3, 5)$ , whereas $f(x) = (x + 3)^2 + 5$ has its vertex at $(-3, 5)$ .

Plus-or-minus roots: When solving $(x - h)^2 = d$ , always include both roots: $x - h = \pm\sqrt{d}$ . Omitting the negative root loses half of the solution set.

Squaring negative coefficients in $b^2 - 4ac$ : When $b$ is negative, $b^2$ is always positive: $(-5)^2 = 25$ , not $-25$ . A sign error here completely alters the discriminant.

A projectile is launched from ground level with height modelled by

h(t) = -16t^2 + 64t + 80

feet after

t

seconds. Find (a) the maximum height reached, and (b) the time when it hits the ground.

1. The maximum occurs at the vertex time $t = -\dfrac{b}{2a}$ :

t = -\frac{64}{2(-16)} = \frac{64}{32} = 2\text{ seconds}.

Substitute $t = 2$ to get maximum height:

h(2) = -16(2^2) + 64(2) + 80 = -64 + 128 + 80 = 144\text{ feet}.

2. Ground impact occurs when $h(t) = 0$ :

-16t^2 + 64t + 80 = 0 \implies t^2 - 4t - 5 = 0 \implies (t - 5)(t + 1) = 0.

Since time must be non-negative, $t = 5$ seconds.

Find all solutions to

2x^2 - 6x + 5 = 0

using the quadratic formula, and determine whether the roots are real or complex.

Identify coefficients: $a = 2$ , $b = -6$ , $c = 5$ .

Evaluate the discriminant: $\Delta = b^2 - 4ac = (-6)^2 - 4(2)(5) = 36 - 40 = -4.$ Because $\Delta < 0$ , the quadratic has two complex conjugate roots.

Apply the quadratic formula: $x = \frac{-(-6) \pm \sqrt{-4}}{2(2)} = \frac{6 \pm 2i}{4} = \frac{3 \pm i}{2} = \frac{3}{2} \pm \frac{1}{2}i.$

📝 Chapter Quiz

Questions Level

Exponents & Polynomials

Exponent rules and polynomial structure are the algebraic “grammar” that keeps longer manipulations clean and error-free.

▲ Reference — Laws of exponents

$a^m a^n=a^{m+n}$ , $\quad \dfrac{a^m}{a^n}=a^{m-n}$ , $\quad (a^m)^n=a^{mn}$ ,

$a^0=1\ (a\neq 0)$ , $\quad a^{-n}=\dfrac{1}{a^n}$ , $\quad a^{1/n}=\sqrt[n]{a}.$

Graphs of even power

y = x^2

(symmetric about the

y

-axis) and odd power

y = x^3

(origin symmetry). Both pass through the common points

(0,0)

and

(1,1)

Special products & factoring

Square of a sum

(a+b)^2=a^2+2ab+b^2

Square of a difference

(a-b)^2=a^2-2ab+b^2

Difference of squares

a^2-b^2=(a-b)(a+b)

Worked example 3.A — Simplify

Simplify $\dfrac{x^5 \cdot x^2}{x^3}$ .

Add exponents on top, then subtract:

\frac{x^{5+2}}{x^3}=x^{7-3}=x^4.

Worked example 3.B — Factor

Factor $x^2-9$ .

A difference of squares with $a=x,\ b=3$ :

x^2-9=(x-3)(x+3).

Worked example 3.C — Negative & fractional exponents

Simplify the expression completely with positive exponents: $\dfrac{(8x^6y^{-3})^{2/3}}{2x^{-1}y^4}$ .

Step 1: Apply the power of a product rule to the numerator:

(8x^6y^{-3})^{2/3} = 8^{2/3} \cdot x^{6 \cdot 2/3} \cdot y^{-3 \cdot 2/3} = (\sqrt[3]{8})^2 \cdot x^4 \cdot y^{-2} = 4x^4y^{-2}.

Step 2: Divide by the denominator $2x^{-1}y^4$ :

\frac{4x^4y^{-2}}{2x^{-1}y^4} = \frac{4}{2} \cdot x^{4 - (-1)} \cdot y^{-2 - 4} = 2x^5y^{-6} = \frac{2x^5}{y^6}.

⚠ Common traps — Exponents over addition & negative signs

Exponents do NOT distribute across sums: $(a+b)^n \neq a^n+b^n$ . For example, $(x+3)^2 = x^2+6x+9 \neq x^2+9$ . The middle cross-term $2ab$ must never be omitted.

Parentheses with negative bases: $-3^2 = -(3^2) = -9$ , whereas $(-3)^2 = (-3)(-3) = +9$ . Without grouping parentheses, exponentiation takes precedence over the negative sign.

Negative exponents mean reciprocals: $a^{-n} = \dfrac{1}{a^n}$ is always a positive number when $a > 0$ . For example, $2^{-3} = \dfrac{1}{8} > 0$ , not $-8$ or $-6$ .

Evaluate the numerical value of

16^{-3/4} + \left(\dfrac{27}{8}\right)^{-2/3}

1. Evaluate the first term: $16^{-3/4} = \dfrac{1}{16^{3/4}} = \dfrac{1}{(\sqrt[4]{16})^3} = \dfrac{1}{2^3} = \dfrac{1}{8}$ .

2. Evaluate the second term: inverting the fraction removes the negative exponent:

\left(\frac{27}{8}\right)^{-2/3} = \left(\frac{8}{27}\right)^{2/3} = \frac{(\sqrt[3]{8})^2}{(\sqrt[3]{27})^2} = \frac{2^2}{3^2} = \frac{4}{9}.

3. Combine with a common denominator of $72$ :

\frac{1}{8} + \frac{4}{9} = \frac{9}{72} + \frac{32}{72} = \frac{41}{72}.

Simplify the algebraic expression

\dfrac{(2x^3y^{-2})^3}{(4x^{-1}y^4)^2}

leaving only positive exponents.

Apply power of a power to numerator and denominator separately:

\text{Numerator} = 2^3 \cdot (x^3)^3 \cdot (y^{-2})^3 = 8x^9y^{-6}.

\text{Denominator} = 4^2 \cdot (x^{-1})^2 \cdot (y^4)^2 = 16x^{-2}y^8.

Divide coefficients and subtract exponents:

\frac{8}{16} \cdot x^{9 - (-2)} \cdot y^{-6 - 8} = \frac{1}{2}x^{11}y^{-14} = \frac{x^{11}}{2y^{14}}.

📝 Chapter Quiz

Questions Level

Functions & Transformations

A function is a rule that assigns exactly one output to each input. Once you know a parent graph, simple changes to its formula shift, stretch, and flip it — no re-plotting required.

● Definition 4.1 — Function, domain, range

A function $f$ pairs each input $x$ with one output $f(x)$ . The domain is the set of allowed inputs; the range is the set of resulting outputs.

▲ Theorem 4.2 — The transformation template

Starting from a parent function $f$ , the graph of

g(x)=a\,f(x-h)+k

is stretched vertically by $|a|$ (and reflected if $a<0$ ), shifted right by $h$ , and shifted up by $k$ .

Pick a parent function and move the sliders — the faint grey curve is the original, the indigo curve is the transformed graph, and the amber dot tracks its key point at $(h,k)$ .

⊙ Explorer · Shifting & stretchinginteractive

Parent function

f(x)

Stretch / reflect a = 1

Horizontal shift h = 0

Vertical shift k = 0

g(x) = (x)²

key point (h, k)—

Worked example 4.A — Describe the transformation

How is $y=(x-3)^2+2$ related to $y=x^2$ ?

Matching $a=1,\ h=3,\ k=2$ : the parabola shifts right $3$ and up $2$ , placing the vertex at $(3,2)$ .

Worked example 4.B — Domain

Find the domain of $f(x)=\sqrt{x-1}$ .

The radicand must be non-negative: $x-1\geq 0$ , so the domain is $[1,\infty)$ .

▲ Functions and their transformations

A function assigns exactly one output to each input. Transformations shift, reflect, stretch, or compress the graph of a parent function.

Transformation from parent

f(x) = x^2

g(x) = -(x-3)^2 + 4

: reflected across the

x

-axis, shifted right

3

units, and shifted up

4

units.

Vertical shift

y=f(x)+k

: up

k

units

Horizontal shift

y=f(x-h)

: right

h

units

Reflection

y=-f(x)

: over

x

-axis;

y=f(-x)

: over

y

-axis

Stretch/compress

y=af(x)

: vertical;

y=f(bx)

: horizontal

Worked example 4.A — Describing transformations

Describe the transformation from $y=x^2$ to $y=-(x-3)^2+4$ .

Right 3, reflect over $x$ -axis, up 4. Vertex at $(3,4)$ , opens downward.

⚠ Common traps — Inside vs outside transformations & domain limits

Counter-intuitive horizontal shifts: Inputs inside the parentheses operate in reverse: $f(x - c)$ shifts RIGHT by $c$ , and $f(x + c)$ shifts LEFT by $c$ . Outside shifts $f(x) + c$ move UP as expected.

Domain exclusions: Two universal algebraic rules dictate real domain boundaries: (1) denominators cannot equal zero ( $\frac{1}{g(x)} \implies g(x) \neq 0$ ), and (2) expressions under even radicals must be non-negative ( $\sqrt{g(x)} \implies g(x) \ge 0$ ).

Composition is not commutative: $(f \circ g)(x) = f(g(x))$ means evaluate $g$ first and plug into $f$ . In general, $f(g(x)) \neq g(f(x))$ .

Given

f(x) = \dfrac{2}{x - 1}

and

g(x) = \sqrt{x + 5}

, find (a) the value of

(f \circ g)(4)

, and (b) the domain of

g(x)

1. Evaluate the inner function first: $g(4) = \sqrt{4 + 5} = \sqrt{9} = 3$ .

2. Substitute into $f$ :

(f \circ g)(4) = f(g(4)) = f(3) = \frac{2}{3 - 1} = \frac{2}{2} = 1.

3. Domain of $g(x) = \sqrt{x + 5}$ : The radicand must be non-negative:

x + 5 \ge 0 \implies x \ge -5 \quad\text{or in interval notation } [-5, \infty).

Find the inverse function

f^{-1}(x)

for

f(x) = \dfrac{3x + 4}{2x - 5}

, and state the domain of

f^{-1}

Let $y = \dfrac{3x + 4}{2x - 5}$ . Swap variables: $x = \frac{3y + 4}{2y - 5}.$ Multiply both sides by $2y - 5$ : $x(2y - 5) = 3y + 4 \implies 2xy - 5x = 3y + 4.$ Group all $y$ -terms on one side: $2xy - 3y = 5x + 4 \implies y(2x - 3) = 5x + 4.$ Divide to isolate $y$ : $f^{-1}(x) = \frac{5x + 4}{2x - 3}.$ Domain of $f^{-1}$ : $x \neq \frac{3}{2}$ (all real numbers except $x = 1.5$ ).

📝 Chapter Quiz

Questions Level

Exponential & Logarithmic Functions

Exponential functions model growth and decay — populations, interest, radioactivity. Logarithms are their inverses, and the tool that solves for an unknown exponent.

● Definition 5.1 — Exponential function

An exponential function has the form $f(x)=a\,b^{x}$ with $b>0,\ b\neq 1$ . It models growth when $b>1$ and decay when $0<b<1$ , with horizontal asymptote $y=0$ and $y$ -intercept $a$ .

▲ Theorem 5.2 — Logarithms invert exponentials

$y=\log_b x \iff b^{y}=x$ . The laws follow directly:

\log_b(MN)=\log_b M+\log_b N,\qquad \log_b\!\big(M^{p}\big)=p\,\log_b M.

Move the sliders to see growth flip to decay as the base crosses $1$ . Tick the box to mirror the curve over $y=x$ and reveal its inverse logarithm.

⊙ Explorer · Growth, decay & inverseinteractive

Coefficient a = 1

Base b = 2

Show inverse

y=\log_b x

and the line

y=x

y = 1·2ˣ

behavior—

y-intercept—

Both exponential curves share the

y

-intercept

(0,1)

and horizontal asymptote

y=0

. When base

b > 1

the curve grows exponentially; when

0 < b < 1

it decays toward zero.

Exponential form

f(x)=ab^x,\;b>0,\,b\neq1

Log definition

\log_b x=y \Leftrightarrow b^y=x

Log laws

\log(mn)=\log m+\log n;\;\log\frac{m}{n}=\log m-\log n

Change of base

\log_b a=\dfrac{\ln a}{\ln b}

Worked example 5.A — Solve an exponential equation

Solve $2^{x}=32$ .

Write $32$ as a power of $2$ : $32=2^{5}$ . Equal bases force equal exponents:

2^{x}=2^{5}\ \Longrightarrow\ x=5.

Worked example 5.B — Logarithms

Evaluate $\log_2 8$ and expand $\log\!\big(xy^2\big)$ .

Since $8=2^{3}$ , $\log_2 8=3$ . By the log laws,

\log\!\big(xy^2\big)=\log x+2\log y.

⚠ Common traps — Non-distributive logs & asymptote boundaries

Logs never distribute over sums: $\log(A + B) \neq \log A + \log B$ . The log product rule is $\log(AB) = \log A + \log B$ . Likewise, $\dfrac{\log A}{\log B} \neq \log(A - B)$ ; rather, $\dfrac{\log A}{\log B} = \log_B A$ .

Domain of logarithm is strictly positive: $\log_b(x)$ requires $x > 0$ . You cannot take the logarithm of zero or a negative number in real algebra. Always check for extraneous non-positive values when solving log equations.

Exponentials never equal zero: $b^x > 0$ for all real $x$ (when base $b > 0$ ). An equation like $4^x = 0$ or $5^x = -3$ has no real solution.

Solve

5 \cdot 2^{3x - 1} = 80

for

x

1. Isolate the exponential expression by dividing both sides by $5$ :

2^{3x - 1} = \frac{80}{5} = 16.

2. Express $16$ as a power of the base $2$ :

16 = 2^4 \implies 2^{3x - 1} = 2^4.

3. Equate exponents and solve:

3x - 1 = 4 \implies 3x = 5 \implies x = \frac{5}{3}.

A culture of bacteria doubles in population every 3 hours. If the initial count is 250 bacteria, write an exponential model

B(t)

for the population after

t

hours, and find the population after 12 hours.

Initial population $a = 250$ , doubling factor $b = 2$ , time interval = 3 hours:

B(t) = 250 \cdot 2^{t/3}.

For $t = 12$ hours, the number of doubling periods is $12/3 = 4$ :

B(12) = 250 \cdot 2^4 = 250 \cdot 16 = 4000\text{ bacteria}.

📝 Chapter Quiz

Questions Level

Systems & Inequalities

A system of equations asks where two conditions hold at once. Graphically, the solution is the point where the lines cross — and sometimes there is no such point.

● Definition 6.1 — Solution of a linear system

A solution satisfies every equation simultaneously. Two lines either cross once (one solution), run parallel (no solution), or coincide (infinitely many).

Solve by substitution (isolate one variable and plug in) or elimination (add multiples of the equations to cancel a variable). Drag the four sliders: the amber dot is the live intersection, and parallel slopes report “no solution”.

⊙ Explorer · Two lines, one solutioninteractive

Line 1 slope m₁ = 1

Line 1 intercept b₁ = -1

Line 2 slope m₂ = -1

Line 2 intercept b₂ = 2

solution—

Worked example 6.A — Elimination

Solve $\begin{cases} 2x+y=7 \\ x-y=2 \end{cases}$ .

Add the equations to cancel $y$ : $3x=9 \Rightarrow x=3$ . Back-substitute: $3-y=2 \Rightarrow y=1$ .

(x,y)=(3,1).

Worked example 6.B — No solution

Solve $\begin{cases} y=2x+1 \\ y=2x-3 \end{cases}$ .

Equal slopes but different intercepts: the lines are parallel and never meet — no solution.

▲ Systems of equations and inequalities

Solve systems by substitution, elimination, or matrices. Graph inequalities with dashed or solid boundary lines and shade the solution region.

The unique solution to the linear system

y = x + 1

and

y = -\frac{1}{2}x + 4

is their intersection point

(2,3)

, where both equations hold true simultaneously.

Substitution

Solve one equation for a variable, substitute into the other

Elimination

Add/subtract equations to cancel a variable

Cramer's rule (2×2)

x=\dfrac{D_x}{D},\; y=\dfrac{D_y}{D}

D=ad-bc

Inequality shading

Test a point; if true, shade that side

Worked example 6.A — Elimination

Solve: $x+y=5$ , $2x-y=4$ .

Add: $3x=9 \Rightarrow x=3,\;y=2$ .

⚠ Common traps — Distributing negatives in elimination & boundary lines

Distributing the negative sign: When subtracting one equation from another during elimination, remember to distribute the negative sign to every term on both sides. Subtracting the left side while adding the constant on the right side is an extremely common error.

Zero-equality outcomes: If variables cancel out entirely: (1) a true statement like $0 = 0$ means the equations represent the same line (infinitely many solutions); (2) a false statement like $0 = 7$ means the lines are parallel (no solution).

Dashed vs solid boundary lines: In linear inequalities, strict inequalities ( $<$ or $>$ ) require a dashed boundary line (points on the line are not solutions). Non-strict inequalities ( $\le$ or $\ge$ ) use a solid line.

For what value of

k

will the system have no solution?

\begin{cases} 3x - 2y = 8 \\ 6x + ky = 5 \end{cases}

Solution:

1. A linear system has no solution if the lines are parallel (identical slopes, different $y$ -intercepts).

2. Express the first line in slope-intercept form:

-2y = -3x + 8 \implies y = \frac{3}{2}x - 4 \quad\implies m_1 = \frac{3}{2}.

3. Express the second line in slope-intercept form:

ky = -6x + 5 \implies y = -\frac{6}{k}x + \frac{5}{k} \quad\implies m_2 = -\frac{6}{k}.

4. Equate slopes to make lines parallel:

\frac{3}{2} = -\frac{6}{k} \implies 3k = -12 \implies k = -4.

When $k = -4$ , the intercepts differ ( $-4 \neq -\frac{5}{4}$ ), so the lines are strictly parallel with no solution.

Solve the nonlinear system

y = x^2 - 3x + 2

and

y = 2x - 4

algebraically. State all intersection points

(x, y)

Equate the expressions for $y$ : $x^2 - 3x + 2 = 2x - 4 \implies x^2 - 5x + 6 = 0.$ Factor the quadratic: $(x - 2)(x - 3) = 0 \implies x = 2 \text{ or } x = 3.$ Find corresponding $y$ -values using $y = 2x - 4$ :

For $x = 2$ : $y = 2(2) - 4 = 0$ , yielding intersection point $(2, 0)$ .
For $x = 3$ : $y = 2(3) - 4 = 2$ , yielding intersection point $(3, 2)$ .

The system intersects at $(2, 0)$ and $(3, 2)$ .

📝 Chapter Quiz

Questions Level

Sequences & Series

A sequence is an ordered list of numbers following a rule; a series is their sum. Two patterns dominate: adding a constant (arithmetic) and multiplying by a constant (geometric).

● Definition 7.1 — Arithmetic sequence

Each term adds a common difference $d$ : $a_n=a_1+(n-1)d$ . The sum of the first $n$ terms is

S_n=\frac{n}{2}\big(a_1+a_n\big).

● Definition 7.2 — Geometric sequence

Each term multiplies by a common ratio $r$ : $a_n=a_1\,r^{\,n-1}$ . The first $n$ terms sum to $S_n=a_1\dfrac{1-r^{n}}{1-r}$ , and when $|r|<1$ the infinite sum converges:

S_\infty=\frac{a_1}{1-r}.

Comparison of arithmetic growth (

a_n = 2n

, straight dashed line with constant difference

d=2

) and geometric growth (

a_n = 2^{n-1}

, multiplying by ratio

r=2

Arithmetic — term

a_n=a_1+(n-1)d

Arithmetic — sum

S_n=\dfrac{n}{2}(a_1+a_n)

Geometric — term

a_n=a_1\cdot r^{n-1}

Geometric — sum

S_n=a_1\cdot\dfrac{1-r^n}{1-r}

Worked example 7.A — Arithmetic sum

Find the sum of the first $20$ terms of $3,7,11,\dots$

Here $a_1=3,\ d=4$ , so $a_{20}=3+19(4)=79$ . Then

S_{20}=\frac{20}{2}\big(3+79\big)=10\cdot 82=820.

Worked example 7.B — Infinite geometric sum

Evaluate $8+4+2+1+\cdots$

With $a_1=8$ and $r=\tfrac12$ (and $|r|<1$ ):

S_\infty=\frac{8}{1-\tfrac12}=16.

⚠ Common traps — The (n - 1) offset & geometric convergence

The $(n-1)$ index shift: The formula is $a_n = a_1 + (n-1)d$ , NOT $a_1 + nd$ . The first term $a_1$ has zero common differences added. For example, the 10th term adds $9d$ , not $10d$ .

Infinite sum condition $|r| < 1$ : $S_\infty = \dfrac{a_1}{1-r}$ applies ONLY to converging series where $-1 < r < 1$ . If $r = 2$ , the terms grow without bound and the series has no sum.

Sequence term $a_n$ vs series sum $S_n$ : Do not mix up the $n$ -th term $a_n$ (the value of the single item at position $n$ ) with the partial sum $S_n$ (the total sum of all terms from $1$ to $n$ ).

Find the sum of all integers between

10

and

100

(inclusive) that are divisible by

4

1. Identify the sequence: The first multiple of $4 \ge 10$ is $a_1 = 12$ . The last multiple $\le 100$ is $a_n = 100$ . The common difference is $d = 4$ .

2. Find the number of terms $n$ :

a_n = a_1 + (n-1)d \implies 100 = 12 + (n-1)4 \implies 88 = 4(n-1) \implies n - 1 = 22 \implies n = 23.

3. Calculate the sum using $S_n = \dfrac{n}{2}(a_1 + a_n)$ :

S_{23} = \frac{23}{2}(12 + 100) = \frac{23}{2}(112) = 23 \cdot 56 = 1288.

In a geometric sequence, the 3rd term is 18 and the 6th term is 486. Find the common ratio

r

, the first term

a_1

, and the sum of the first 5 terms

S_5

Use the general formula $a_n = a_1 r^{n-1}$ :

a_3 = a_1 r^2 = 18, \quad a_6 = a_1 r^5 = 486.

Divide $a_6$ by $a_3$ :

\frac{a_1 r^5}{a_1 r^2} = r^3 = \frac{486}{18} = 27 \implies r = 3.

Solve for $a_1$ :

a_1 (3^2) = 18 \implies 9a_1 = 18 \implies a_1 = 2.

Compute $S_5$ using the geometric series sum formula:

S_5 = \frac{a_1(r^5 - 1)}{r - 1} = \frac{2(3^5 - 1)}{3 - 1} = \frac{2(243 - 1)}{2} = 242.

📝 Chapter Quiz

Questions Level

Factoring Polynomials

Factoring rewrites a polynomial as a product of simpler factors — the key step in solving equations, finding roots, and simplifying rational expressions.

▲ Reference — Factoring patterns

Always take out the greatest common factor (GCF) first. Difference of squares: $a^{2}-b^{2}=(a-b)(a+b)$ . Monic trinomial: $x^{2}+(p+q)x+pq=(x+p)(x+q)$ . For $ax^2 + bx + c$ , use the $ac$ -method.

Roots & Linear Factors

Setting each linear factor to zero gives the x-intercepts of the polynomial: x - 2 = 0 → x = 2 and x - 3 = 0 → x = 3.

GCF

factor common term

Difference of squares

(a-b)(a+b)

Trinomial

(x+p)(x+q)

Perfect square

(a\pm b)^{2}

Worked example 8.A

Factor $x^{2}-5x+6$ .

(x-2)(x-3).

Worked example 8.B

Factor completely: $3x^3 - 75x$ .

First extract the greatest common factor (GCF), $3x$ :

3x(x^2 - 25).

Recognize $x^2 - 25$ as a difference of two squares $a^2 - b^2 = (a-b)(a+b)$ :

3x(x - 5)(x + 5).

Worked example 8.C

Factor by grouping: $2x^3 + 6x^2 + 5x + 15$ .

Group into pairs:

(2x^3 + 6x^2) + (5x + 15).

Factor out common terms from each group:

2x^2(x + 3) + 5(x + 3).

Factor out the common binomial factor $(x + 3)$ :

(x + 3)(2x^2 + 5).

Common Pitfall — Sum of Squares & Stopping Early

Two frequent mistakes in factoring: 1. Stopping too early: After factoring out a GCF or one factor, always inspect the remaining polynomial to see if it factors further (e.g. $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$ ). 2. Sum of squares trap: Difference of squares $a^2 - b^2$ factors into $(a - b)(a + b)$ , but a sum of squares $a^2 + b^2$ is prime over real numbers — it does NOT equal $(a + b)^2$ !

Factor the quadratic expression

6x^2 + 11x - 10

completely.

Use the $ac$ -method. We seek two integers whose product is $a \cdot c = 6(-10) = -60$ and whose sum is $b = 11$ .

The factor pair of $-60$ that sums to $11$ is $15$ and $-4$ (since $15 \times (-4) = -60$ and $15 + (-4) = 11$ ).

Rewrite the middle term and factor by grouping:

6x^2 + 15x - 4x - 10 = 3x(2x + 5) - 2(2x + 5).

Factor out the binomial $(2x + 5)$ :

(2x + 5)(3x - 2).

Factor the polynomial completely by grouping:

2x^3 - 3x^2 - 8x + 12

Group terms in pairs:

(2x^3 - 3x^2) - (8x - 12).

Factor out the greatest common factor from each pair:

x^2(2x - 3) - 4(2x - 3).

Factor out the common binomial $(2x - 3)$ :

(x^2 - 4)(2x - 3).

Recognize $(x^2 - 4)$ as a difference of squares and factor completely:

(x - 2)(x + 2)(2x - 3).

📝 Chapter Quiz

Questions Level

Radicals & Exponents

Roots can be written as radicals or as fractional exponents; both obey the laws of exponents.

▲ Reference — Radical laws

$\sqrt{ab}=\sqrt{a}\,\sqrt{b}$ , $a^{1/n}=\sqrt[n]{a}$ , and $a^{m/n}=\sqrt[n]{a^{m}}$ . Rationalise a denominator by multiplying by the conjugate.

Square Root Function Graph

The square root function y = √x is defined only for x ≥ 0 (domain) and produces only non-negative outputs y ≥ 0 (range).

Product

\sqrt{ab}=\sqrt{a}\sqrt{b}

Fractional

a^{m/n}=\sqrt[n]{a^{m}}

Simplify

\sqrt{a^{2}b}=a\sqrt{b}

Rationalise

× conjugate

Worked example 9.A

Simplify $\sqrt{72}$ .

\sqrt{36\cdot2}=6\sqrt{2}.

Worked example 9.B

Rationalize the denominator: $\dfrac{6}{3 - \sqrt{3}}$ .

Multiply numerator and denominator by the conjugate $3 + \sqrt{3}$ :

\frac{6(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})} = \frac{18 + 6\sqrt{3}}{3^2 - (\sqrt{3})^2} = \frac{18 + 6\sqrt{3}}{9 - 3} = \frac{18 + 6\sqrt{3}}{6} = 3 + \sqrt{3}.

Common Pitfall — Splitting Radicals Over Addition & Extraneous Roots

Two crucial rules for radicals: 1. Never split over addition: $\sqrt{a^2 + b^2} \neq a + b$ (for example, $\sqrt{9 + 16} = \sqrt{25} = 5$ , NOT $3 + 4 = 7$ ). You can only split radicals over multiplication and division. 2. Extraneous solutions: Squaring both sides of an equation can introduce false roots. You MUST substitute candidate solutions back into the original un-squared equation.

Solve the equation

\sqrt{2x + 7} = x + 2

for all real values of

x

Square both sides of the equation:

(\sqrt{2x + 7})^2 = (x + 2)^2 \implies 2x + 7 = x^2 + 4x + 4.

Rearrange into standard quadratic form:

x^2 + 2x - 3 = 0 \implies (x + 3)(x - 1) = 0 \implies x = 1 \quad \text{or} \quad x = -3.

Now check both candidate roots in the original equation:

For $x = 1$ : LHS = $\sqrt{2(1) + 7} = \sqrt{9} = 3$ ; RHS = $1 + 2 = 3$ . (Valid ✓)
For $x = -3$ : LHS = $\sqrt{2(-3) + 7} = \sqrt{1} = 1$ ; RHS = $-3 + 2 = -1$ . Since $1 \neq -1$ , $x = -3$ is extraneous.

The only real solution is $x = 1$ .

Simplify and rationalize the denominator:

\dfrac{\sqrt{3}}{\sqrt{5} - \sqrt{2}}

Multiply numerator and denominator by the conjugate $\sqrt{5} + \sqrt{2}$ :

\frac{\sqrt{3}(\sqrt{5} + \sqrt{2})}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})}.

Expand denominator using difference of squares $(a-b)(a+b) = a^2 - b^2$ :

(\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3.

Expand numerator and simplify:

\frac{\sqrt{15} + \sqrt{6}}{3}.

📝 Chapter Quiz

Questions Level

Rational Expressions

A rational expression is a ratio of polynomials; simplify by factoring and cancelling common factors.

▲ Reference — Working with rationals

Multiply $\tfrac{a}{b}\cdot\tfrac{c}{d}=\tfrac{ac}{bd}$ ; divide by multiplying by the reciprocal; add over a common denominator. Exclude any value making a denominator zero.

Rational Function & Asymptotes

The rational function y = 1/x has a vertical asymptote at x = 0 (where denominator is zero) and a horizontal asymptote at y = 0.

Simplify

cancel factors

Multiply

\tfrac{ac}{bd}

Divide

× reciprocal

Domain

b\neq 0

Worked example 10.A

Simplify $\dfrac{x^{2}-9}{x-3}$ .

x+3,\quad x\neq 3.

Worked example 10.B

Combine into a single rational expression: $\dfrac{3}{x - 2} + \dfrac{2}{x + 1}$ .

The common denominator is $(x - 2)(x + 1)$ :

\frac{3(x + 1) + 2(x - 2)}{(x - 2)(x + 1)} = \frac{3x + 3 + 2x - 4}{(x - 2)(x + 1)} = \frac{5x - 1}{(x - 2)(x + 1)}, \quad x \neq 2, -1.

Common Pitfall — Cancelling Terms Instead of Factors

The single most common error in algebra is the illegal cancellation trap: in $\dfrac{x + 5}{x}$ , you cannot cancel the $x$ 's to get $5$ ! You can only divide out common multiplicative factors, never additive terms. Always factor both numerator and denominator completely into products before cancelling.

Solve the equation

\dfrac{x}{x - 3} - \dfrac{2}{x + 1} = \dfrac{6}{x^2 - 2x - 3}

for

x

Factor the denominator on the RHS: $x^2 - 2x - 3 = (x - 3)(x + 1)$ . The LCD is $(x - 3)(x + 1)$ . Note the domain restriction: $x \neq 3$ and $x \neq -1$ .

Multiply every term by the LCD:

x(x + 1) - 2(x - 3) = 6 \implies x^2 + x - 2x + 6 = 6.

Simplify and solve:

x^2 - x = 0 \implies x(x - 1) = 0 \implies x = 0 \quad \text{or} \quad x = 1.

Both values are within the allowed domain (neither equals 3 or -1). The solutions are $x = 0$ and $x = 1$ .

Simplify the complex rational expression:

\dfrac{\frac{1}{x} - \frac{1}{y}}{\frac{1}{x^2} - \frac{1}{y^2}}

, where

x \neq y

and

xy \neq 0

Factor the denominator as a difference of squares:

\frac{1}{x^2} - \frac{1}{y^2} = \left(\frac{1}{x} - \frac{1}{y}\right)\left(\frac{1}{x} + \frac{1}{y}\right).

Cancel the common factor $\left(\frac{1}{x} - \frac{1}{y}\right)$ :

\frac{1}{\frac{1}{x} + \frac{1}{y}} = \frac{1}{\frac{y + x}{xy}} = \frac{xy}{x + y}.

📝 Chapter Quiz

Questions Level

Statistics & Data Analysis

Summarise data with measures of centre and spread, and read trends from tables and plots.

▲ Reference — Summary statistics

The mean is the average, the median the middle value, the mode the most frequent, and the range the spread. A line of best fit models a linear trend in a scatter plot.

Five-Number Summary & Box Plot

A box plot visually divides data into quartiles. The box spans from Q1 = 25 to Q3 = 65 (IQR = 40) with the median marked at 40.

Mean

\tfrac{\sum x}{n}

Median

middle value

Range

max − min

Trend

line of best fit

Worked example 11.A

Find the mean of $3,\,6,\,9,\,12$ .

\frac{30}{4}=7.5.

Worked example 11.B

Find the median and interquartile range (IQR) for the ordered dataset:

4,\ 7,\ 10,\ 14,\ 18,\ 22,\ 25.

The dataset has $n = 7$ values. The median is the 4th value: $\text{Median} = 14$ .

Lower half: $4, 7, 10 \implies Q_1 = 7$ .

Upper half: $18, 22, 25 \implies Q_3 = 22$ .

The interquartile range is $\text{IQR} = Q_3 - Q_1 = 22 - 7 = 15$ .

Common Pitfall — Mean vs. Median with Outliers

A high-frequency question on standardized math exams tests how adding extreme values alters summary statistics: Mean and range are sensitive to outliers and will shift substantially. Median and IQR are resistant — their values change very little (or not at all) when extreme outliers are added. In right-skewed data, $\text{mean} > \text{median}$ ; in left-skewed data, $\text{mean} < \text{median}$ .

A distribution of test scores has first quartile

Q_1 = 58

and third quartile

Q_3 = 82

. Using the standard

1.5 \times \text{IQR}

rule, what is the lowest score that would NOT be considered an outlier?

First compute the interquartile range (IQR):

\text{IQR} = Q_3 - Q_1 = 82 - 58 = 24.

Calculate the lower fence below which values are classified as low outliers:

\text{Lower Fence} = Q_1 - 1.5(\text{IQR}) = 58 - 1.5(24) = 58 - 36 = 22.

Any score at or above $22$ is not a low outlier. The lowest non-outlier score is $22$ .

A set of five exam scores has a mean of 84. Four of the scores are 78, 86, 92, and 80. Find the 5th score, and find the new mean if the lowest score is removed.

Total sum of the 5 scores = $5 \times 84 = 420$ .

Sum of the 4 known scores = $78 + 86 + 92 + 80 = 336$ .

5th score = $420 - 336 = 84$ .

Removing the lowest score (78) leaves four scores: 80, 84, 86, 92.

New sum = $420 - 78 = 342$ . New mean = $342 / 4 = 85.5$ (an increase of 1.5 points).

📝 Chapter Quiz

Questions Level

Practice set

Fifty mixed problems spanning all seven chapters, ordered roughly from linear equations to series and tagged by difficulty. Click any problem to reveal a full worked solution.

Solve $4x-7=2x+9$ .

Basic

Collect like terms: $4x-2x=9+7$ , so $2x=16$ .

x=8.

Find the slope of the line through $(1,2)$ and $(5,10)$ .

Basic

m=\frac{10-2}{5-1}=\frac{8}{4}=2.

Solve $x^2-7x+12=0$ .

Basic

Factor: two numbers with product $12$ , sum $-7$ are $-3,-4$ .

(x-3)(x-4)=0 \Rightarrow x=3 \ \text{or}\ 4.

Solve $2x^2+5x-3=0$ using the quadratic formula.

Intermediate

$a=2,\ b=5,\ c=-3$ , so $\Delta=25+24=49$ :

x=\frac{-5\pm 7}{4}=\tfrac12 \ \text{or}\ -3.

Simplify $\dfrac{(x^3y^2)(x^2y^4)}{x\,y^3}$ .

Intermediate

Add exponents in the numerator, then subtract the denominator's:

x^{3+2-1}\,y^{2+4-3}=x^4 y^3.

Factor $9x^2-25$ .

Intermediate

Difference of squares with $a=3x,\ b=5$ :

9x^2-25=(3x-5)(3x+5).

Find the vertex and axis of symmetry of $y=x^2-6x+5$ .

Advanced

Axis: $x=-\dfrac{b}{2a}=\dfrac{6}{2}=3$ . Then $y=9-18+5=-4$ .

\text{Vertex } (3,-4), \quad \text{axis } x=3.

Describe the transformation from $y=x^2$ to $y=(x-2)^2-5$ .

Basic

With $h=2,\ k=-5$ : shift right $2$ and down $5$ .

\text{Vertex moves to } (2,-5).

Solve $3^{x}=81$ .

Basic

Write $81=3^{4}$ ; equal bases give equal exponents:

3^{x}=3^{4}\Rightarrow x=4.

A10

Solve $\begin{cases} x+y=10 \\ x-y=4 \end{cases}$ .

Intermediate

Add the equations: $2x=14\Rightarrow x=7$ . Then $7+y=10\Rightarrow y=3$ .

(x,y)=(7,3).

A11

Find the 6th term and the infinite sum of $27,9,3,\dots$

Intermediate

Geometric with $a_1=27,\ r=\tfrac13$ . The 6th term is $a_6=27\big(\tfrac13\big)^{5}=\tfrac{27}{243}=\tfrac19$ . Since $|r|<1$ ,

S_\infty=\frac{27}{1-\tfrac13}=\frac{27}{2/3}=40.5.

A12

Solve $5x-3=2x+12$ .

Basic

$3x=15$ :

x=5.

A13

Solve $2(x-4)=10$ .

Basic

Divide by 2: $x-4=5$ .

x=9.

A14

Find the slope of the line through $(2,3)$ and $(6,11)$ .

Basic

m=\frac{11-3}{6-2}=\frac{8}{4}=2.

A15

Write the equation of the line through $(0,-2)$ with slope $3$ .

Intermediate

Slope–intercept with $b=-2$ :

y=3x-2.

A16

Find the $x$ -intercept of $2x-5y=10$ .

Intermediate

Set $y=0$ : $2x=10$ .

(5,0).

A17

Are $y=2x+1$ and $y=2x-4$ parallel, perpendicular, or neither?

Intermediate

Equal slopes $(2=2)$ , different intercepts:

\text{parallel}.

A18

What is the slope of any line perpendicular to $y=\tfrac{2}{3}x+1$ ?

Intermediate

Negative reciprocal of $\tfrac23$ :

-\frac{3}{2}.

A19

Solve $x^2=49$ .

Basic

x=\pm 7.

A20

Solve $x^2-9x+20=0$ .

Basic

Factor: $(x-4)(x-5)=0$ .

x=4 \ \text{or}\ 5.

A21

Solve $x^2+6x+9=0$ .

Intermediate

Perfect square $(x+3)^2=0$ :

x=-3 \ \text{(double root)}.

A22

How many real roots does $2x^2-4x+5=0$ have?

Intermediate

Discriminant $\Delta=16-40=-24<0$ :

\text{no real roots}.

A23

Solve $3x^2-12=0$ .

Basic

$x^2=4$ :

x=\pm 2.

A24

Find the vertex of $y=x^2+4x+1$ .

Intermediate

$x=-\tfrac{b}{2a}=-2$ , then $y=4-8+1=-3$ :

(-2,-3).

A25

Solve $x^2-4x-1=0$ .

Advanced

Quadratic formula, $\Delta=16+4=20$ :

x=\frac{4\pm\sqrt{20}}{2}=2\pm\sqrt5.

A26

Complete the square: $x^2+6x+5$ .

Advanced

Half of 6 is 3; add and subtract $9$ :

x^2+6x+5=(x+3)^2-4.

A27

Simplify $x^4\cdot x^3$ .

Basic

x^{4+3}=x^7.

A28

Simplify $\dfrac{x^5}{x^2}$ .

Basic

x^{5-2}=x^3.

A29

Simplify $\big(x^2\big)^3$ .

Basic

x^{2\cdot 3}=x^6.

A30

Evaluate $2^{-3}$ .

Basic

2^{-3}=\frac{1}{2^3}=\frac18.

A31

Expand $(x+4)^2$ .

Intermediate

(x+4)^2=x^2+8x+16.

A32

Factor $x^2-16$ .

Intermediate

Difference of squares:

(x-4)(x+4).

A33

Factor $x^2+7x+12$ .

Intermediate

Two numbers multiplying to 12, adding to 7 are 3 and 4:

(x+3)(x+4).

A34

If $f(x)=2x^2-1$ , find $f(3)$ .

Basic

2(9)-1=17.

A35

Find the domain of $f(x)=\sqrt{x-3}$ .

Intermediate

Need $x-3\ge 0$ :

[3,\infty).

A36

Find the domain of $f(x)=\dfrac{1}{x-2}$ .

Intermediate

The denominator cannot be zero:

x\neq 2.

A37

Describe the transformation from $y=|x|$ to $y=|x+2|-3$ .

Intermediate

$h=-2,\ k=-3$ :

\text{left 2, down 3}.

A38

Is $f(x)=x^3$ even, odd, or neither?

Intermediate

$f(-x)=-x^3=-f(x)$ :

\text{odd}.

A39

Solve $5^{x}=125$ .

Basic

$125=5^3$ :

x=3.

A40

Evaluate $\log_3 81$ .

Intermediate

$81=3^4$ :

\log_3 81=4.

A41

Evaluate $\log_2\!\big(\tfrac18\big)$ .

Intermediate

$\tfrac18=2^{-3}$ :

\log_2\!\big(\tfrac18\big)=-3.

A42

Expand $\log\!\big(x^3 y\big)$ .

Intermediate

\log\!\big(x^3 y\big)=3\log x+\log y.

A43

Solve $2^{\,x+1}=16$ .

Advanced

$16=2^4$ , so $x+1=4$ :

x=3.

A44

Solve $\begin{cases} x+y=5 \\ x-y=1\end{cases}$ .

Intermediate

Add: $2x=6\Rightarrow x=3$ , then $y=2$ :

(3,2).

A45

Solve $\begin{cases} y=2x \\ x+y=9\end{cases}$ .

Intermediate

Substitute: $x+2x=9\Rightarrow x=3$ , so $y=6$ :

(3,6).

A46

Solve the inequality $3x-4>11$ .

Basic

$3x>15$ :

x>5.

A47

Solve $-2x+1\le 7$ .

Intermediate

$-2x\le 6$ ; dividing by $-2$ flips the sign:

x\ge -3.

A48

Find the 10th term of the arithmetic sequence $2,5,8,\dots$

Basic

$a_1=2,\ d=3$ :

a_{10}=2+9(3)=29.

A49

Find the sum of the first 20 terms of $2,5,8,\dots$

Intermediate

$a_{20}=2+19(3)=59$ , then $S_{20}=\tfrac{20}{2}(2+59)$ :

S_{20}=10\cdot 61=610.

A50

Evaluate the infinite sum $16+8+4+\cdots$

Intermediate

Geometric with $a=16,\ r=\tfrac12$ :

S_\infty=\frac{16}{1-\tfrac12}=32.

✎

Test generator

Build a randomised test from this subject’s question bank. Pick the number of questions and a difficulty, generate, then reveal the worked answers when you’re ready — or print a clean copy to hand out.

Questions

Difficulty

Source

↓

Downloads

Take Algebra offline. The PDF packet prints straight from your browser; native Word versions with editable equation-editor math are available on request.

PDF · ready now

Algebra notes & practice

All eleven chapters plus every worked solution, formatted as a clean print document.

DOCX · on request

Native Word packet

Editable .docx with real Word equations (OMML) — ready for a worksheet or quiz.

Official · Past Papers

Past Papers & Official Tests

Officially released exams and mark schemes from the examining body.

ACT Free Practice Tests Official ACT full-length practice tests

Worksheet Library

92 real worksheets with answer keys, organized by unit — free to download and print.

Loading worksheet library…

📋 Mistake Log

📊 Your Progress

Algebra

Linear Equations & Lines

Three forms of a line

Quadratic Functions

Three forms of a quadratic

Exponents & Polynomials

Special products & factoring

Functions & Transformations

Exponential & Logarithmic Functions

Systems & Inequalities

Sequences & Series

Factoring Polynomials

Radicals & Exponents

Rational Expressions

Statistics & Data Analysis

Practice set

Test generator

Downloads

Algebra notes & practice

Native Word packet

Past Papers & Official Tests

Worksheet Library

Privacy Policy

What's stored on your device

Visitor counter

Advertising & cookies

Google Forms integration

Third-party links

Contact

Terms of Service

Use of content

No guarantee of results

AI tutor

Contact

Mock Exam

⚙️ AI API Key