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Algebra I

Common Core · 5 units · high school

Algebra I: every unit as Common Core publishes it, with practice questions per unit and a countdown to the exam.

Set up Algebra I → One set per unit, 5 of them pre-written and answer-checked. Free, no account needed.

Units

Unit 1: Relationships Between Quantities and Reasoning with Equations

Units and quantities, interpreting expressions, creating equations and inequalities in one and two variables, solving linear equations and inequalities, rearranging formulas.

3 of 15 questions in this unit, with the worked answer

A car gets 28 miles per gallon and gas costs \$3.50 per gallon. Using units to guide the calculation, find the cost to drive 420 miles.

Answer. \$52.50

Use units to plan the steps: divide miles by miles-per-gallon to get gallons (420 ÷ 28 = 15 gallons), then multiply gallons by dollars-per-gallon to get dollars (15 × 3.50 = \$52.50). Tracking units shows exactly which operation to perform at each step.

The expression $200 + 15n$ represents the total cost, in dollars, of a gym membership after n months. What does the term $15n$ represent?

Answer. The total amount paid in monthly fees over n months (before adding the initial \$200 fee).

The constant 200 is the one-time fee, and 15 is the cost per month, so multiplying by n gives the running total of monthly charges alone. The full expression adds this to the fixed fee to get the overall cost.

An investment's balance after t years is modeled by $P(1+r)^t$, where P is the initial deposit and r is the annual interest rate. What does the factor $(1+r)$ represent?

Answer. The growth factor applied each year — the balance is multiplied by $(1+r)$ every year because it keeps 100% of its value and gains r more.

Since the whole balance is multiplied by $(1+r)$ once per year (that's what raising to the power t means), $(1+r)$ must be the single-year growth multiplier, combining the original amount (1) with the rate of increase (r).

All 15 questions for Unit 1, with the working →

What people get wrong

  1. Forgetting to reverse the inequality symbol when multiplying or dividing both sides by a negative number. Instead, flip the sign every time you multiply or divide by a negative, and check with a test value if unsure.
  2. Distributing a negative sign incorrectly, for example treating $-3(x+2)$ as $-3x+2$ instead of $-3x-6$. Instead, multiply every term inside the parentheses by the full coefficient, sign included.
  3. When rearranging a formula, dividing only one term on a side instead of the entire side by the divisor, writing $w = P - 2l/2$ instead of $w = (P-2l)/2$. Instead, treat the whole side being divided as one quantity in parentheses.
  4. Accepting any algebraic solution to a constraint without checking it makes sense in context, such as a negative number of tickets or a fractional number of people. Instead, always test the solution against the real-world limits of the variable.

Unit 2: Linear and Exponential Relationships

Systems of linear equations and inequalities, functions and function notation, domain and range, sequences, linear versus exponential growth, graphing and transforming linear and exponential functions.

3 of 13 questions in this unit, with the worked answer

Solve the system: $y = 2x - 3$ and $y = -x + 9$.

Answer. x = 4, y = 5

Since both equations equal y, set them equal: $2x-3=-x+9$. Adding x to both sides and adding 3 gives $3x=12$, so $x=4$. Substituting back gives $y=2(4)-3=5$.

Solve the system by elimination: $3x + 2y = 16$ and $x - 2y = 0$.

Answer. x = 4, y = 2

The y-terms are opposites, so adding the equations eliminates y: $4x = 16$, giving $x=4$. Substituting into $x-2y=0$ gives $4-2y=0$, so $y=2$.

Which point satisfies both $y > x + 1$ and $y \le -2x + 6$?

Answer. (1, 3)

Test each point in both inequalities. For (1,3): $3>1+1=2$ is true, and $3\le -2(1)+6=4$ is true, so it satisfies both. The other points each fail at least one inequality.

All 13 questions for Unit 2, with the working →

What people get wrong

  1. Forgetting to flip the inequality symbol when multiplying or dividing both sides by a negative number while solving a system of inequalities — always check the sign of the coefficient before dividing.
  2. Treating f(-2) as -f(2), i.e. evaluating the function then negating it, instead of substituting -2 directly for every x in the rule — always replace the variable itself with the given input.
  3. Reading a graph's domain or range and including an endpoint that has an open circle — an open circle means that value is not included, so the inequality must be strict ( ), not ≤ or ≥.
  4. Applying the geometric sequence formula (multiplying by r) to a sequence that is actually arithmetic, or vice versa — check whether consecutive terms have a constant difference or a constant ratio before choosing a formula.

Unit 3: Descriptive Statistics

Dot plots, histograms and box plots, centre and spread, two-way tables, scatterplots and linear fits, residuals, correlation versus causation.

3 of 14 questions in this unit, with the worked answer

A data set of six values is 5, 6, 7, 8, 9, 50. Which measure of center best describes a typical value, and why?

Answer. The median (7.5), because the mean (about 14.2) is pulled upward by the outlier 50 and no longer represents a typical value.

Mean = (5+6+7+8+9+50)/6 ≈ 14.17, far above every value except the outlier. Median = (7+8)/2 = 7.5, which sits right in the middle of the ordinary values, so it better represents the typical data point when an outlier is present.

Find the interquartile range of the data set: 3, 7, 8, 10, 12, 13, 14, 18, 20, 22.

Answer. IQR = 10

With 10 values, the lower half is 3,7,8,10,12 (Q1 = median = 8) and the upper half is 13,14,18,20,22 (Q3 = median = 18). IQR = Q3 − Q1 = 18 − 8 = 10.

A data set is 2, 4, 5, 6, 7, 8, 9, 10, 30. Using the 1.5×IQR rule, is 30 an outlier?

Answer. Yes, 30 is an outlier.

Lower half 2,4,5,6 gives Q1 = 4.5; upper half 8,9,10,30 gives Q3 = 9.5; IQR = 5. Upper fence = 9.5 + 1.5(5) = 17. Since 30 > 17, it is classified as an outlier.

All 14 questions for Unit 3, with the working →

What people get wrong

  1. Using the mean and standard deviation to describe a skewed data set or one with an outlier. Instead, check the shape first (a box plot or dot plot showing asymmetry or an extreme value) and switch to median and IQR when the data are not roughly symmetric.
  2. Including the overall median in both halves when finding Q1 and Q3 for an odd-sized data set. Instead, split the data into a strict lower half and upper half, excluding the middle value itself, before taking each half's median.
  3. Computing a residual as predicted minus observed. Instead, always use residual = observed y − predicted y (ŷ), so a positive residual means the actual point lies above the line.
  4. Reading a two-way table by dividing by the grand total when the question asks for a conditional relative frequency. Instead, identify whether the question restricts attention to one row or one column, and divide only by that row's or column's total.

Unit 4: Expressions and Equations

Polynomial arithmetic, factoring, completing the square, the quadratic formula, solving quadratic equations, systems with one linear and one quadratic equation.

3 of 14 questions in this unit, with the worked answer

Multiply: $(2x-3)(x+5)$

Answer. $2x^2+7x-15$

Distribute each term: $2x\cdot x=2x^2$, $2x\cdot5=10x$, $-3\cdot x=-3x$, $-3\cdot5=-15$. Combine like terms $10x-3x=7x$ to get $2x^2+7x-15$.

Subtract: $(4x^2-3x+2)-(x^2+5x-6)$

Answer. $3x^2-8x+8$

Distribute the negative sign to every term of the second polynomial: $4x^2-3x+2-x^2-5x+6$. Combine like terms: $4x^2-x^2=3x^2$, $-3x-5x=-8x$, $2+6=8$.

Expand and simplify: $(x-4)^2$

Answer. $x^2-8x+16$

Use the square-of-a-binomial pattern $(a-b)^2=a^2-2ab+b^2$ with $a=x$, $b=4$: $x^2-2(x)(4)+16=x^2-8x+16$.

All 14 questions for Unit 4, with the working →

What people get wrong

  1. When subtracting one polynomial from another, students distribute the negative sign to only the first term of the second polynomial. Instead, rewrite the subtraction as adding the opposite of every term before combining like terms.
  2. When completing the square, students add $(b/2)^2$ to only one side of the equation, which breaks the equality. Instead, add the same value to both sides, or move the constant to the other side first and add there too.
  3. When factoring a trinomial with a leading coefficient other than 1 (like $2x^2+7x+3$), students apply the simple $(x+p)(x+q)$ pattern that only works when $a=1$. Instead, use the ac-method or check for a common factor first, and always verify by multiplying back.
  4. When applying the quadratic formula, students mishandle the negative sign on $b$, especially when squaring it, writing $b^2$ instead of correctly computing $(-b)^2$ or $b^2-4ac$ with a negative $b$. Instead, substitute all values in parentheses first, then simplify.

Unit 5: Quadratic Functions and Modeling

Graphs of quadratic functions, vertex and intercepts, comparing linear, quadratic and exponential models, transformations, square-root and cube-root functions, piecewise functions.

3 of 14 questions in this unit, with the worked answer

Find the vertex of f(x) = 2x^2 - 8x + 5.

Answer. (2, -3)

Use x=-b/(2a) = 8/4 = 2. Then f(2) = 2(4) - 16 + 5 = 8 - 16 + 5 = -3, so the vertex is (2, -3).

Rewrite g(x) = -3(x+1)^2 - 2 in standard form.

Answer. g(x) = -3x^2 - 6x - 5

Expand (x+1)^2 = x^2+2x+1, multiply by -3 to get -3x^2-6x-3, then subtract 2 more: -3x^2-6x-5.

A parabola has x-intercepts at x=-3 and x=1 and passes through (0,-6). Write its equation in standard form.

Answer. f(x) = 2x^2 + 4x - 6

Start with factored form f(x)=a(x+3)(x-1). Plug in (0,-6): a(3)(-1) = -3a = -6, so a=2. Then f(x)=2(x+3)(x-1)=2x^2+4x-6.

All 14 questions for Unit 5, with the working →

What people get wrong

  1. Dropping the negative sign in x=-b/(2a) — always write the formula with the negative explicitly before substituting values, since a common error is computing b/(2a) instead.
  2. Getting the sign of h backwards in vertex form — in f(x)=a(x-h)^2+k, a term like (x+3)^2 means h=-3 (shift left 3), not h=3; rewrite (x+3) as (x-(-3)) before reading off h.
  3. Assuming a square-root function has domain 'all real numbers' — always set the radicand ≥0 and solve for x before stating the domain.
  4. Misreading which piece of a piecewise function applies at a boundary point — check whether the inequality on each piece is strict or non-strict, since the point can only belong to one piece.

How it works

Choosing this curriculum creates a course in Chalkset with one study set per unit and the exam date set to the start of the next testing window. Each unit is drilled card by card, and the review queue brings back what you miss. Pre-written units come with worked explanations and an independently checked answer key; the rest are written by the agent when you open them.

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