CMClearMathAcademy

Algebra Foundations Checkpoint

A free Algebra I lesson from the “Algebra Foundations” unit, with a worked example and practice problems including step-by-step solutions.

This checkpoint samples all seven Algebra Foundations lessons: expressions, multi-step equations, linear inequalities, compound inequalities, absolute value, literal equations, and formulas. Questions are mixed, so the first job is to recognize the structure and choose a method. For inequalities, give the whole solution set: the direction of the symbol and whether each endpoint is included both matter. For formulas, say what a rearrangement assumes, such as a divisor that cannot be zero or a length that must be positive. The 25 review questions and the 25 checkpoint questions were written separately from each other and from the lesson practice and quizzes. Several questions are multiple choice so that complete solution sets and rearranged formulas are graded reliably. A score reports performance on this sample of proposed outcomes A1.1 to A1.3; it is not an external certification. The written reasoning tasks below are reviewed with a rubric and do not change the score.

What you'll learn

Why it matters: Planning problems rarely announce which algebra they need: a capacity limit is an inequality, a tolerance is an absolute value, and a unit rate may need a formula rearranged before any numbers are substituted.

Worked example

Problem. Solve 7 − 2(x + 1) ≥ x − 4 and describe the complete solution set.

  1. Distribute the −2 to both terms: 7 − 2x − 2 ≥ x − 4, which simplifies to 5 − 2x ≥ x − 4.
  2. Collect terms: add 2x and 4 to both sides to get 9 ≥ 3x. Dividing by positive 3 keeps the symbol: 3 ≥ x.
  3. Read 3 ≥ x as x ≤ 3. The boundary 3 is included because the original symbol is ≥.
  4. Check the boundary and one other value: x = 3 gives 7 − 8 = −1 and 3 − 4 = −1 (equal, so included); x = 0 gives 5 ≥ −4 (true).

Answer: x ≤ 3

Lesson preview: this page shows 5 practice items. The interactive activity contains 25 practice items and 25 quiz items (typed response and multiple choice). Counts include repeated prompts and variants.

Practice problems

1. Simplify 7(y − 2) − 3(2y − 5).

Show solution
  1. Distribute each factor: 7(y − 2) = 7y − 14 and −3(2y − 5) = −6y + 15.
  2. The −3 multiplies both terms, so −3 times −5 is +15.
  3. Combine like terms: 7y − 6y = y and −14 + 15 = 1, so the result is y + 1.

Answer: y + 1

2. A kayak rental costs 12 + 3.5h dollars for h hours. What does the term 3.5h represent?

Choices: The part of the cost that depends on the number of hours · The fixed fee paid for any rental · The total cost of a one-hour rental · The number of hours rented

Show solution
  1. The term 3.5h changes when h changes: it is $3.50 for each hour.
  2. The constant term 12 is the fixed fee, and a one-hour rental costs 12 + 3.5 = 15.5 dollars in total.
  3. So 3.5h is the hourly part of the cost, not the fee, the total, or the hours themselves.

Answer: The part of the cost that depends on the number of hours

3. A student rewrote 5 − 2(x − 4) as 3(x − 4) and then as 3x − 12. Which statement is correct?

Choices: Multiplication comes before subtraction; the correct result is 13 − 2x · The work is correct; 3(x − 4) is equivalent to 5 − 2(x − 4) · Only the last sign is wrong; the correct result is 3x + 12 · The 2 multiplies only x; the correct result is 1 − 2x

Show solution
  1. The 2 multiplies (x − 4) before anything is subtracted, so 5 − 2 cannot be combined first.
  2. Distribute −2: 5 − 2x + 8.
  3. Combine constants: 13 − 2x. Check with x = 0: 5 − 2(−4) = 13, while 3(0 − 4) = −12, so the student's expression is not equivalent.

Answer: Multiplication comes before subtraction; the correct result is 13 − 2x

4. Solve (x + 7)/3 = x − 5. What is x?

Show solution
  1. Multiply both sides by 3 to clear the fraction: x + 7 = 3x − 15.
  2. Subtract x and add 15: 22 = 2x, so x = 11.
  3. Check: (11 + 7)/3 = 6 and 11 − 5 = 6.

Answer: 11

5. Which equation is true for every real value of x?

Choices: 3(x − 2) + x = 4x − 6 · 3(x − 2) + x = 4x − 2 · 3(x − 2) + x = 3x − 6 · 3(x − 2) + x = 4(x − 2)

Show solution
  1. Simplify the common left side: 3x − 6 + x = 4x − 6.
  2. 4x − 6 = 4x − 6 is an identity, so every real x works.
  3. 4x − 6 = 4x − 2 and 4x − 6 = 4x − 8 are contradictions with no solution; 4x − 6 = 3x − 6 has only x = 0.

Answer: 3(x − 2) + x = 4x − 6

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Explain "between" and "outside" with distance

Solve |x − 2| ≤ 5 and |x − 2| > 5. For each, write the solution set as an inequality and sketch it on a number line with correctly open or closed endpoints. Use distance from 2 to explain why one solution set lies between two numbers and the other lies outside them. Then decide whether x = 7 and x = −3 belong to each set, and justify your decisions.

Review rubric and solution
  • Solves |x − 2| ≤ 5 as −3 ≤ x ≤ 7 with both endpoints included.
  • Solves |x − 2| > 5 as x < −3 or x > 7 with both endpoints excluded.
  • Draws both number lines with closed circles for the first set, open circles for the second, and shading in the correct regions.
  • Explains that |x − 2| is the distance from x to 2, so at most 5 units gives the numbers between −3 and 7, and more than 5 units gives the numbers beyond them.
  • Places 7 and −3 in the first set only, justified by their distance of exactly 5 from 2.
  1. |x − 2| ≤ 5 means −5 ≤ x − 2 ≤ 5, so −3 ≤ x ≤ 7. Draw closed circles at −3 and 7 and shade between them.
  2. |x − 2| > 5 means x − 2 < −5 or x − 2 > 5, so x < −3 or x > 7. Draw open circles at −3 and 7 and shade outward.
  3. |x − 2| measures the distance between x and 2. Points within 5 units of 2 lie between 2 − 5 = −3 and 2 + 5 = 7; points more than 5 units away lie beyond those numbers.
  4. Both 7 and −3 are exactly 5 units from 2. They satisfy "at most 5" but not "more than 5", so they belong to the first set and not the second.

Build equations with one, none, or infinitely many solutions

Fill in the blank in 2(x − 3) + 4 = ____ with an expression of the form mx + b three different ways, so that the equation has (a) exactly one solution, (b) no solution, and (c) infinitely many solutions. For each, simplify both sides, finish solving, and explain how the final statement shows the number of solutions.

Review rubric and solution
  • Simplifies the left side correctly to 2x − 2.
  • Gives a right side with an x-coefficient other than 2 for (a), finds its single solution, and checks it.
  • Gives a right side 2x + k with k ≠ −2 for (b) and explains that the false statement −2 = k means no value works.
  • Gives a right side equivalent to 2x − 2 for (c) and explains that the resulting true statement holds for every x.
  • Connects each conclusion to the result of solving, not to a guess from the form of the equation.
  1. The left side simplifies to 2x − 6 + 4 = 2x − 2.
  2. (a) Example: 2x − 2 = x + 1 gives x = 3; check: 2(3 − 3) + 4 = 4 and 3 + 1 = 4.
  3. (b) Example: 2x − 2 = 2x + 5. Subtracting 2x leaves −2 = 5, which is false for every x, so there is no solution.
  4. (c) Example: 2x − 2 = 2(x − 1). Subtracting 2x leaves −2 = −2, which is true for every x, so every real number is a solution.

Rearrange a formula and justify the restriction

A course grade is the average of three tests: A = (a + b + c)/3. First, solve for c and say what each step does. Then a student has scores of 78 and 85, and every test is scored from 0 to 100. Write and solve an inequality for the third score c that gives an average of at least 84. State the complete set of third-test scores that meet the goal, and explain which limit comes from the algebra and which comes from the context.

Review rubric and solution
  • Rearranges correctly to c = 3A − a − b, naming each inverse operation.
  • Writes (78 + 85 + c)/3 ≥ 84 or an equivalent inequality.
  • Solves to c ≥ 89 with the boundary included.
  • States the complete set 89 ≤ c ≤ 100 using the scoring limit.
  • Explains that the lower limit comes from solving the inequality and the upper limit of 100 comes from the context, and interprets 89 as the lowest score that meets the goal.
  1. Multiply both sides by 3: 3A = a + b + c. Subtract a and b: c = 3A − a − b.
  2. The goal is (78 + 85 + c)/3 ≥ 84. Multiply by 3: 163 + c ≥ 252, so c ≥ 89.
  3. Scores cannot exceed 100, so the scores that meet the goal are 89 ≤ c ≤ 100.
  4. The algebra gives c ≥ 89; the context adds c ≤ 100. A score of exactly 89 gives an average of exactly 84, which meets "at least 84".

Practice this interactively with instant feedback and an AI tutor.

Practice Algebra Foundations Checkpoint Take the free placement check

More Algebra I lessons