Algebra Lessons

Algebra 1: Every Topic, With Lessons and a Practice Test

Every Algebra 1 topic in order, a lesson for each, and a 12-question practice test with a guide to what to review.

Professor Chacha October 10, 2026 6 min read 1 views
Teenage boy with a backpack standing in front of a chalkboard covered with area formulas

Algebra 1 is the course that every college placement test, high school equivalency test, and entrance exam builds on. If you can solve linear equations, work with lines and slope, handle exponents and polynomials, and factor and solve simple quadratics, you have most of what the ACCUPLACER, TSIA2, GED, HiSET, ASVAB, and the SAT expect. Many states also give an Algebra 1 end-of-course exam in high school.

This page maps out every major Algebra 1 topic, links to a step-by-step lesson for each one, and ends with a 12-question practice test that tells you exactly what to review.

The Algebra 1 topics, in order

Topic What you learn Lesson
Expressions and equations Simplify, distribute, solve linear equations and inequalities Percent word problems (setting up equations from words)
Lines and slope Slope, slope-intercept and point-slope form, graphs How to find slope
Systems of equations Substitution, elimination, word problems Systems of equations word problems
Exponents Product, quotient, power, zero, and negative exponents Exponent rules
Polynomials and factoring Adding, multiplying, and factoring polynomials How to factor trinomials
Quadratics Solving by factoring and with the quadratic formula Quadratic formula, explained

Functions, absolute value, and sequences also appear in most Algebra 1 courses; the practice test below includes a question on each.

Algebra 1 practice test

Twelve questions, roughly one per topic, from the start of the course to the end. Work them without a calculator first.

1.Simplify: \(4(2x - 3) - (x + 5)\)

  1. A\(7x - 17\)
  2. B\(7x - 7\)
  3. C\(9x - 17\)
  4. D\(7x + 2\)
Show solution \[8x - 12 - x - 5 = 7x - 17\]

Common trap: Distributing the minus sign to the \(x\) but not to the 5, which gives \(7x - 7\). The minus sign in front of the parentheses changes both signs inside.

Answer: A (\(7x - 17\))

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

  1. A1
  2. B\(-5\)
  3. C3
  4. D5
Show solution \[\begin{gathered}2x + 6 = 5x - 9 \\ \Rightarrow\; 15 = 3x \\ \Rightarrow\; x = 5\end{gathered}\]

Common trap: Moving terms without changing their signs. When \(2x\) moves to the right side, it is subtracted; when \(-9\) moves to the left, it is added.

Answer: D (5)

3.Solve: \(|x - 4| = 7\)

  1. A\(x = 11\) only
  2. B\(x = 3\) or \(x = 11\)
  3. C\(x = -3\) or \(x = 11\)
  4. D\(x = -11\) or \(x = 3\)
Show solution

The expression inside can be 7 or \(-7\).

\[\begin{gathered}x - 4 = 7 \\ \Rightarrow\; x = 11 \\ x - 4 = -7 \\ \Rightarrow\; x = -3\end{gathered}\]

Common trap: Solving only \(x - 4 = 7\). An absolute value equation usually has two solutions, one for each sign.

Answer: C (\(x = -3\) or \(x = 11\))

4.Solve: \(-2x + 5 \le 11\)

  1. A\(x \le -3\)
  2. B\(x \ge -3\)
  3. C\(x \ge 3\)
  4. D\(x \le 8\)
Show solution \[\begin{gathered}-2x \le 6 \\ \Rightarrow\; x \ge -3\end{gathered}\]

Dividing by a negative number flips the inequality sign.

Common trap: Keeping the \(\le\) sign after dividing by \(-2\). Test a value: \(x = 0\) gives \(5 \le 11\), which is true, and \(0 \ge -3\) agrees.

Answer: B (\(x \ge -3\))

5.What is the slope of the line \(3x - 4y = 8\)?

  1. A\(\tfrac{3}{4}\)
  2. B\(-\tfrac{3}{4}\)
  3. C\(\tfrac{4}{3}\)
  4. D\(-2\)
Show solution \[\begin{gathered}-4y = -3x + 8 \\ \Rightarrow\; y = \tfrac{3}{4}x - 2\end{gathered}\]

Common trap: Reading the slope before solving for \(y\), or dropping a negative when dividing by \(-4\). Both negatives cancel, so the slope is positive.

Answer: A (\(\tfrac{3}{4}\))

6.The table shows a linear function. Which equation fits it?

\(x\) 1 2 3
\(y\) 5 8 11

  1. A\(y = 2x + 3\)
  2. B\(y = 3x + 5\)
  3. C\(y = x + 4\)
  4. D\(y = 3x + 2\)
Show solution

Each time \(x\) goes up by 1, \(y\) goes up by 3, so the slope is 3. Then \(5 = 3(1) + b\) gives \(b = 2\).

Common trap: Using the first \(y\)-value, 5, as the intercept. The intercept is the value of \(y\) when \(x = 0\), not when \(x = 1\).

Answer: D (\(y = 3x + 2\))

7.Solve the system: \(y = 2x + 1\) and \(y = -x + 7\)

  1. A\((1, 3)\)
  2. B\((3, 4)\)
  3. C\((2, 5)\)
  4. D\((6, 1)\)
Show solution

Set the two expressions for \(y\) equal.

\[\begin{gathered}2x + 1 = -x + 7 \\ \Rightarrow\; 3x = 6 \\ \Rightarrow\; x = 2, \; y = 5\end{gathered}\]

Common trap: Finding \(x\) and forgetting \(y\). The solution of a system is a point, so you need both coordinates.

Answer: C (\((2, 5)\))

8.Simplify: \((3x^2)(4x^5)\)

  1. A\(7x^7\)
  2. B\(12x^7\)
  3. C\(12x^{10}\)
  4. D\(7x^{10}\)
Show solution \[3 \cdot 4 \cdot x^{2 + 5} = 12x^7\]

Common trap: Adding the numbers or multiplying the exponents. Multiply the coefficients; add the exponents.

Answer: B (\(12x^7\))

9.Subtract: \((5x^2 - 3x + 2) - (2x^2 + x - 6)\)

  1. A\(3x^2 - 4x + 8\)
  2. B\(3x^2 - 2x - 4\)
  3. C\(3x^2 - 4x - 4\)
  4. D\(7x^2 - 2x - 4\)
Show solution \[\begin{gathered}5x^2 - 3x + 2 - 2x^2 - x + 6 \\ = 3x^2 - 4x + 8\end{gathered}\]

Common trap: Subtracting only the first term of the second polynomial. Change the sign of every term in the second parentheses.

Answer: A (\(3x^2 - 4x + 8\))

10.Factor: \(x^2 - x - 20\)

  1. A\((x - 4)(x + 5)\)
  2. B\((x - 10)(x + 2)\)
  3. C\((x + 5)(x + 4)\)
  4. D\((x - 5)(x + 4)\)
Show solution

Two numbers that multiply to \(-20\) and add to \(-1\): \(-5\) and \(4\).

Common trap: Swapping the signs, \((x - 4)(x + 5)\), which gives \(+x\) in the middle. Multiply back to check.

Answer: D (\((x - 5)(x + 4)\))

11.Solve: \(x^2 - 6x = 0\)

  1. A\(x = 6\) only
  2. B\(x = 0\) or \(x = 6\)
  3. C\(x = -6\) or \(x = 6\)
  4. D\(x = 0\) or \(x = -6\)
Show solution

Factor out \(x\), then set each factor to zero.

\[\begin{gathered}x(x - 6) = 0 \\ \Rightarrow\; x = 0 \text{ or } x = 6\end{gathered}\]

Common trap: Dividing both sides by \(x\), which loses the solution \(x = 0\). Never divide by a variable that could be zero; factor instead.

Answer: B (\(x = 0\) or \(x = 6\))

12.What is the 20th term of the arithmetic sequence 7, 11, 15, 19, ...?

  1. A87
  2. B80
  3. C83
  4. D79
Show solution

The common difference is 4. The \(n\)th term is \(a_1 + (n - 1)d\).

\[7 + (20 - 1)(4) = 7 + 76 = 83\]

Common trap: Multiplying \(20 \times 4\) and adding 7 to get 87. From the 1st term to the 20th, there are only 19 steps.

Answer: C (83)

Answer key and what to review

  1. 1 A
  2. 2 D
  3. 3 C
  4. 4 B
  5. 5 A
  6. 6 D
  7. 7 C
  8. 8 B
  9. 9 A
  10. 10 D
  11. 11 B
  12. 12 C
If you missed Review
1, 2, 3, or 4 Expressions, equations, and inequalities: practice with our percent word problems and the ACCUPLACER QAS practice test
5 or 6 How to find slope
7 Systems of equations
8 Exponent rules
9, 10, or 11 How to factor trinomials

Where Algebra 1 shows up next

Once these topics feel solid, try a full practice set for the test you are facing: ACCUPLACER, TSIA2, GED, HiSET, ASVAB, or SAT.

Teacher's note: Algebra 1 is cumulative: trouble with factoring is often a sign problem from much earlier in the course. If a later topic keeps going wrong, go back two steps rather than forward: redo distributing and combining like terms until it is automatic, and the "hard" topics usually get easier on their own.

Tags: Algebra 1 Equations Factoring Functions Practice test
Professor Chacha
Professor Chacha Math teacher and educational psychologist

Math teacher and educational psychologist with more than 20 years of classroom experience. He writes every practice question on this site from scratch and solves it step by step, the way he explains it to his own students.

Keep practicing

Comments (0)

Be the first to comment.

Leave a comment