Class 12 is where the arithmetic of the first 11 classes turns into algebra. Students take an algebraic expression — something with letters like x, y, or a — and either evaluate it (plug numbers in and compute a value) or simplify it (combine the parts that go together). Both skills are the foundation of every equation-solving class that follows.
The two big ideas
Evaluating an expression
plug the number in, then compute
substitute → use PEMDAS → number
Combining like terms
add coefficients, keep the variable part
3x + 5x = 8x
What counts as a like term?
Like terms have the exact same variable part — same letters, each raised to the same power. Only the coefficient (the number in front) can be different.
3x and −7x→ LIKE terms (both are “x”)
5x² and 2x→ NOT like (different exponents)
4xy and 9yx→ LIKE terms (order doesn’t matter)
−6 and 11→ LIKE terms (both constants)
7a²b and −3ab²→ NOT like (a²b vs ab²)
The distributive property (your simplifying superpower)
a(b + c) = a·b + a·c. Multiply the outside factor by every term inside the parentheses. Example: 3(2x + 5) = 6x + 15.
A leading minus sign is really −1. So −(x − 4) means −1·(x − 4) = −x + 4. This is the #1 place students lose points — the second sign flips.
Three rules students always forget
1. Write the substitution in parentheses. When you plug a negative number in, put it in parentheses so the sign is safe. Evaluating x² at x = −3 means (−3)² = 9, not −9.
2. Unlike terms cannot be combined. 3x + 5y stays 3x + 5y. There is no simpler form. Only like terms combine.
3. The variable part never changes. When you combine 4x + 3x you get 7x, not 7x². Adding coefficients does not add exponents.
The 30 questions below start with pure substitution (Q1–Q6), move through identifying and combining like terms (Q7–Q15), then distributive property (Q16–Q22), and finish with multi-step simplify-and-evaluate problems and one word problem (Q23–Q30). Every level is tagged so you can jump to what you want to practice.
30 practice questions with hidden answers
Level 1 — Evaluating single-variable expressions (Q1–Q6)
Q1Easy
Evaluate 3x + 5 when x = 4.
17
Why. Substitute 4 for x: 3(4) + 5. Multiplication before addition (PEMDAS): 12 + 5 = 17.
Q2Easy
Evaluate 2n − 7 when n = 10.
13
Why. Substitute 10 for n: 2(10) − 7 = 20 − 7 = 13.
Q3Easy
Evaluate 1⁄2 y + 6 when y = 8.
10
Why. Substitute 8: 1⁄2(8) + 6 = 4 + 6 = 10.
Q4Easy
Evaluate 5x − 12 when x = −3. (Put the substitution in parentheses so the negative is safe.)
−27
Why. Substitute (−3) for x: 5(−3) − 12 = −15 − 12 = −27.
Q5Easy
Evaluate x² + 4 when x = −3.
13
Why. Substitute (−3): (−3)² + 4 = 9 + 4 = 13. The parentheses matter — (−3)² = 9, not −9.
Q6Easy
Evaluate 3(2x + 1) when x = 5.
33
Why. Parentheses first: 2(5) + 1 = 11. Then multiply: 3(11) = 33. (Or distribute first: 6x + 3, then substitute: 6(5) + 3 = 33.)
Level 2 — Identifying and combining like terms (Q7–Q15)
Q7Easy
For each pair, decide if they are LIKE terms or NOT like terms.
(a) 4x and 9x
(b) 3y and 3y²
(c) 5ab and −2ab
(d) 6 and 11
(a) like (b) NOT like (c) like (d) like
Why. Like terms need the exact same variable part. (a) both are x — like. (b) one is y, the other y² — the exponents differ, NOT like. (c) both are ab — like. (d) two constants (no variables at all) are always like.
Q8Easy
Simplify: 3x + 5x
8x
Why. Both terms have the variable x, so they are like terms. Add the coefficients: 3 + 5 = 8. Keep x. Result: 8x.
Q9Easy
Simplify: 9y − 4y
5y
Why. Subtract coefficients: 9 − 4 = 5. Keep y. Result: 5y.
Q10Easy
Simplify: 7a + 3a − 5a
5a
Why. All three terms are like (they share the variable a). Combine coefficients left to right: 7 + 3 = 10, then 10 − 5 = 5. Result: 5a.
Q11Medium
Simplify: 6x + 4 − 2x + 9
4x + 13
Why. Group like terms first. x-terms: 6x − 2x = 4x. Constants: 4 + 9 = 13. Write the sum: 4x + 13. Unlike terms cannot be combined further.
Q12Medium
Simplify: 5x + 3y − 2x + 7y
3x + 10y
Why. Two families of like terms. x-terms: 5x − 2x = 3x. y-terms: 3y + 7y = 10y. Answer: 3x + 10y. (x and y are NOT like, so they stay separate.)
Q13Medium
Simplify: 8m − 3 − 5m + 12 − m
2m + 9
Why. m-terms: 8m − 5m − m = (8 − 5 − 1)m = 2m. Constants: −3 + 12 = 9. Answer: 2m + 9.
Q14Medium
Simplify: 4x² + 3x − x² + 5x
3x² + 8x
Why. Terms with x² are one family; terms with x are another. x²-terms: 4x² − x² = 3x². x-terms: 3x + 5x = 8x. Answer: 3x² + 8x. Never combine x² with x.
Q15Medium
Simplify: 6ab + 4a − 2ab + 3b − 5a
4ab − a + 3b
Why. Three families of terms. ab-terms: 6ab − 2ab = 4ab. a-terms: 4a − 5a = −a. b-terms: 3b (nothing to combine). Answer: 4ab − a + 3b. Note ab, a, and b are all different variable parts.
Level 3 — The distributive property (Q16–Q22)
Q16Medium
Expand: 3(x + 4)
3x + 12
Why. Distribute 3 to every term inside: 3·x + 3·4 = 3x + 12.
Q17Medium
Expand: 5(2y − 3)
10y − 15
Why. 5·(2y) + 5·(−3) = 10y − 15. Distribute to both terms, including the subtraction sign.
Q18Medium
Expand: −4(x + 6)
−4x − 24
Why. The outside factor is −4, so distribute the negative too. (−4)·x + (−4)·6 = −4x − 24. Both signs come out negative.
Q19Medium
Expand: −2(3x − 5)
−6x + 10
Why. (−2)·(3x) + (−2)·(−5) = −6x + 10. The negative outside times the negative inside makes the second term POSITIVE. This sign-flip is a top source of errors.
Q20Medium
Expand: −(x − 7)
−x + 7
Why. A leading minus sign means “multiply by −1.” −1·(x − 7) = −1·x + (−1)·(−7) = −x + 7. Every sign inside the parentheses flips.
Q21Hard
Expand and simplify: 3(x + 2) + 5(x − 4)
8x − 14
Why. Distribute each parenthesis first. 3(x + 2) = 3x + 6. 5(x − 4) = 5x − 20. Now the expression is 3x + 6 + 5x − 20. Combine like terms: x-terms 3x + 5x = 8x; constants 6 − 20 = −14. Answer: 8x − 14.
Q22Hard
Expand and simplify: 4(2x + 3) − 2(x − 5)
6x + 22
Why. Watch the second distribution — the −2 is negative. 4(2x + 3) = 8x + 12. −2(x − 5) = −2x + 10 (the second sign flipped because negative × negative = positive). Combine: (8x − 2x) + (12 + 10) = 6x + 22.
Level 4 — Simplify then evaluate; multi-step (Q23–Q30)
Q23Hard
Simplify then evaluate at x = 2: 2(x + 3) + 4x
18
Why. Simplify first: 2(x + 3) + 4x = 2x + 6 + 4x = 6x + 6. Substitute x = 2: 6(2) + 6 = 12 + 6 = 18. (Sanity check without simplifying: 2(2+3) + 4(2) = 2·5 + 8 = 10 + 8 = 18 ✓.)
Q24Hard
Evaluate 3x² − 2x + 7 when x = −2.
23
Why. Substitute (−2) in parentheses. 3(−2)² − 2(−2) + 7. Exponent first: (−2)² = 4, so 3·4 = 12. Middle term: −2(−2) = +4. Constant: 7. Sum: 12 + 4 + 7 = 23.
Q25Hard
Evaluate 2a − 3b + 5ab when a = 4 and b = −1.
−9
Why. Substitute both values in parentheses. 2(4) − 3(−1) + 5(4)(−1) = 8 − (−3) + (−20) = 8 + 3 − 20 = −9.
Q26Hard
Simplify: 3(2x − 1) − (x + 4) + 5
5x − 2
Why. Distribute both parentheses. 3(2x − 1) = 6x − 3. −(x + 4) = −x − 4 (leading minus flips both signs). Now: 6x − 3 − x − 4 + 5. Combine like terms: x-terms 6x − x = 5x; constants −3 − 4 + 5 = −2. Answer: 5x − 2.
Q27Hard
Simplify: 4x² − 3(x² − 2x) + 6x
x² + 12x
Why. Distribute −3 first: −3(x² − 2x) = −3x² + 6x (both signs flip). Now the expression is 4x² − 3x² + 6x + 6x. Combine x²-terms: 4x² − 3x² = x². Combine x-terms: 6x + 6x = 12x. Answer: x² + 12x.
Q28Challenge
Simplify: 2(3x + 4) − 5(2x − 1) + 3(x + 6)
−x + 31
Why. Distribute each group. 2(3x + 4) = 6x + 8. −5(2x − 1) = −10x + 5 (negative × negative = positive on the second term). 3(x + 6) = 3x + 18. Sum: 6x + 8 − 10x + 5 + 3x + 18. x-terms: 6x − 10x + 3x = −x. Constants: 8 + 5 + 18 = 31. Answer: −x + 31.
Q29Challenge
Simplify then evaluate at x = −3 and y = 4: 2(x + 3y) − (4x − y)
Simplified: −2x + 7y · Value: 34
Why. Distribute first. 2(x + 3y) = 2x + 6y. −(4x − y) = −4x + y (leading minus flips both signs). Combine: x-terms 2x − 4x = −2x; y-terms 6y + y = 7y. So the expression is −2x + 7y. Substitute (−3) and (4): −2(−3) + 7(4) = 6 + 28 = 34.
Q30Challenge
A rectangle has length (2x + 5) and width (x + 3). Write a simplified expression for its perimeter, then evaluate at x = 4.
Perimeter: 6x + 16 · At x = 4: 40 units
Why. Perimeter of a rectangle = 2·length + 2·width. P = 2(2x + 5) + 2(x + 3). Distribute: 4x + 10 + 2x + 6. Combine like terms: x-terms 4x + 2x = 6x; constants 10 + 6 = 16. So P = 6x + 16. Substitute x = 4: P = 6(4) + 16 = 24 + 16 = 40 units. (Check by finding length and width first: length = 2(4)+5 = 13, width = 4+3 = 7, P = 2(13) + 2(7) = 26 + 14 = 40 ✓.)
About the Young Fermats Pre-Algebra course
Class 12 is one of 48 rolling classes in our Young Fermats Pre-Algebra program for grade 5–6 students (ages 10–12). The full arc runs from integers on the number line through systems of linear equations and probability — the direct on-ramp to Algebra 1 in 7th or 8th grade. Classes are small-group (max 6 students), 120 minutes per week, and students can start any Monday because the syllabus is rolling.
Every new family starts with a free 30-minute evaluation in our 226 W 79th Street classroom plus a written diagnostic delivered within 48 hours. The diagnostic maps your child's current level against the readiness signals for Pre-Algebra and Algebra 1 — and gives you a clear next step whether you enroll or not.
Ready to see the whole 48-class arc?
Class 12 is a taste — Young Fermats runs 48 rolling classes, students can join any Monday, and the syllabus takes them from integers through systems of equations. Small-group (max 6), 120 minutes a week, at 226 W 79th Street. First class is free. Cancel any time with 15 days’ notice.
See the course → Book a free evaluation or call (646) 668-6151
Related posts