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11 multiple-choice questions and 15 flashcards on Systems of Equations and Inequalities, about 7% of the High School Algebra 1 bank. Every one carries a written rationale.
Systems of Equations and Inequalities is one of 8 chapters in CoStudy's High School Algebra 1 bank, and it holds 11 of the bank's 150 multiple-choice questions — roughly 7% of the total. That proportion is not arbitrary: chapters follow the certifying body's published exam outline, and the number of questions in each is set by that domain's published weight, so the share of your practice time this chapter takes matches the share of the real exam it accounts for.
Studying by chapter is worth doing once you have a diagnostic score. A single overall percentage tells you whether you are close; it does not tell you which domain is dragging. Working a weak chapter in isolation, and re-testing it in isolation, is the fastest way to move a score that has stalled — and it is why the mock exams in CoStudy report by domain rather than as one number.
8 questions drawn from this chapter, with the full rationale shown — the controlling principle behind the right answer, and why each wrong option tempts and fails.
Factor: x² - 9
Answer: D — (x - 3)(x + 3) (difference of squares: a² - b² = (a-b)(a+b))
Difference of squares pattern: a² - b² = (a - b)(a + b). Here a = x, b = 3. So x² - 9 = (x - 3)(x + 3). Verify by FOIL: (x-3)(x+3) = x² + 3x - 3x - 9 = x² - 9 ✓.
Solve by factoring: x² + 5x + 6 = 0
Answer: A — x = -2, -3 (factor as (x+2)(x+3) = 0; zero-product gives x = -2 or x = -3)
Find factors of 6 that sum to 5: 2 and 3. So x² + 5x + 6 = (x+2)(x+3). Zero-product property: x + 2 = 0 or x + 3 = 0 → x = -2 or x = -3.
Solve the system: y = 2x and x + y = 3.
Answer: C — (1, 2)
A) Used y = -2x in substitution. B) Swapped x and y at the end. C) Substitute y: x + 2x = 3 → x = 1, then y = 2(1) = 2. D) Used x - y = 3 instead of x + y = 3. Substitution is the cleanest method when one variable is already isolated.
The equation y = 3x - 5 in slope-intercept form has y-intercept:
Answer: E — -5 (in y = mx + b, b is y-intercept; here b = -5, so line crosses y-axis at (0, -5))
y = mx + b: m = slope, b = y-intercept. y = 3x - 5 means slope = 3, y-intercept = -5. Setting x = 0 confirms: y = 3(0) - 5 = -5. Point (0, -5).
Solve the system: x + y = 10, x - y = 4.
Answer: C — (7, 3) — add equations: 2x = 14, x = 7; then y = 3
Elimination method: add equations to eliminate y. (x+y) + (x-y) = 10+4 → 2x = 14 → x = 7. Substitute: 7 + y = 10 → y = 3. Verify: 7+3=10 ✓ and 7-3=4 ✓.
Simplify: 3(x - 4) - 2(x + 1)
Answer: A — x - 14 (distribute: 3x - 12 - 2x - 2 = x - 14)
Distribute first: 3(x-4) = 3x - 12, -2(x+1) = -2x - 2. Combine: 3x - 12 - 2x - 2 = (3x-2x) + (-12-2) = x - 14. Common error: sign on -2 distribution.
Solve the inequality: -3x + 5 ≥ 11.
Answer: B — x ≤ -2
A) Forgot to flip the inequality when dividing by a negative. B) Subtract 5: -3x ≥ 6; divide by -3 and flip direction: x ≤ -2. C) Did not flip + sign error on 6. D) Sign error on 6 only. Multiplying or dividing by a negative reverses ≥ to ≤.
Multiply: (x + 4)(x - 3)
Answer: A — x² + x - 12 (FOIL: x² - 3x + 4x - 12 = x² + x - 12)
FOIL: First x·x = x²; Outer x·(-3) = -3x; Inner 4·x = 4x; Last 4·(-3) = -12. Combine: x² - 3x + 4x - 12 = x² + x - 12.
4 cards from the 15 in this chapter.
Solve by elimination: 2x + 3y = 12, 4x − 3y = 6.
Add: 6x = 18 → x = 3. Then 6 + 3y = 12 → y = 2.
What does it mean graphically when a system has one solution?
The lines intersect at exactly one point.
Solve by substitution: y = 2x + 1, 3x + y = 16.
3x + (2x + 1) = 16 → 5x = 15 → x = 3, y = 7.
What does it mean when a system has infinitely many solutions?
The two equations represent the same line.
These are a sample. The full Systems of Equations and Inequalities chapter runs 26 items with per-chapter progress tracking, on the web and in the iOS app.
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