Calculus Prep · Linear Equations & Inequalities · Grades 10–12
Calculus Prep: 33 Linear Equations & Inequalities Problems, Solved
Every one of the 33 problems from Chapter 1 of The Humongous Book of Calculus Problems — the pre-calculus warm-up chapter that every AP Calculus AB, AP Calculus BC, and Calculus 1 student should be able to blow through in under 45 minutes. Read the problem, click the + to reveal the answer, and check your reasoning against the full worked solution. Written by the SOMATH team at 226 W 79th Street on the Upper West Side.
Grab the original PDF (Chapter 1, 12 pages)
The full chapter with all 33 problems, the author's worked solutions, and the classic hand-written margin notes.
Source: The Humongous Book of Calculus Problems by W. Michael Kelley, Chapter One — Linear Equations and Inequalities (pages 2–13). Shared for study use by SOMATH.
What's in this walkthrough
- Linear Geometry (Problems 1.1 – 1.10) — slope, forms of a line, parallelograms, midpoints, distances
- Linear Inequalities & Interval Notation (Problems 1.11 – 1.20) — interval notation, graphing on a number line and coordinate plane
- Absolute Value Equations & Inequalities (Problems 1.21 – 1.27) — solve two things for the price of one
- Systems of Equations & Inequalities (Problems 1.28 – 1.33) — substitution, elimination, indeterminate systems, three-variable systems
- Why this chapter matters for calculus
- FAQ
How to use this page: Each problem is exactly as it appears in the book. Click the gold + to reveal the answer and the full worked solution. If a problem takes you more than about two minutes without the reveal, that's the skill to drill before starting the calculus chapters. If you want a real diagnostic, come in for a free 30-minute evaluation at 226 W 79th St or call (646) 668-6151.
Linear Geometry (Problems 1.1 – 1.10)
Creating, graphing, and measuring lines and line segments.
Fast-refresh reminders. Slope-intercept form: y = mx + b (m is slope, b is y-intercept). Standard form: Ax + By = C with slope −A/B. Slope between two points: m = (y₂ − y₁)/(x₂ − x₁). Point-slope: y − y₁ = m(x − x₁). Distance: √[(x₂ − x₁)² + (y₂ − y₁)²]. Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2). Parallel lines: equal slopes. Perpendicular lines: slopes are negative reciprocals.
Click a problem to reveal its solution
All 33 problems from Chapter 1 in order, with full worked steps from the book.
Problem 1.1Solve the equation: 3x − (x − 7) = 4x − 5.
Answer: x = 6.
Distribute the −1 through the parentheses and combine like terms.
3x − x + 7 = 4x − 5 2x + 7 = 4x − 5
Subtract 4x and 7 from both sides to separate the variable and constant terms.
2x + 7 = 4x − 5 −4x − 7 −4x − 7 ───────────────── −2x = −12
Divide both sides by −2: x = 6.
Problem 1.2Calculate the slope, m, of the line 4x − 3y = 9.
Answer: m = 4/3.
Solve for y to rewrite it in slope-intercept form (y = mx + b).
4x − 3y = 9
−3y = −4x + 9
y = (4/3)x − 3
The slope of the line is the coefficient of x: m = 4/3.
Problem 1.3Prove that the slope of a line in standard form, Ax + By = C, is −A/B.
Answer: m = −A/B.
Write the equation in slope-intercept form by solving for y.
Ax + By = C
By = −Ax + C
y = −(A/B) x + C/B
The coefficient of x is the slope of the line: m = −A/B. This is the "slope shortcut" used throughout the chapter.
Problem 1.4Rewrite the linear equation 3x − 4(x − (2/3)y) = (4/5)x − (7y + 3) in standard form.
Answer: 27x − 145y = 45.
Distribute the constants and combine like terms.
3x − 4x + (8/3)y = (4/5)x − 7y − 3
−x + (8/3)y = (4/5)x − 7y − 3
Multiply through by 15 (the least common denominator) to clear fractions.
−15x + 40y = 12x − 105y − 45
Separate variables and constants.
−27x + 145y = −45
Multiply the entire equation by −1 so the x-coefficient is positive (required by standard form): 27x − 145y = 45.
Problem 1.5Write the equation of the line passing through the points (−3,−8) and (−6,2) in slope-intercept form.
Answer: y = −(10/3)x − 18.
Calculate the slope using m = (y₂ − y₁)/(x₂ − x₁).
m = (2 − (−8)) / (−6 − (−3)) = 10 / −3 = −10/3
Substitute the slope into y = mx + b and use one of the coordinate pairs to solve for b.
−8 = −(10/3)(−3) + b −8 = 10 + b b = −18
Final equation: y = −(10/3)x − 18.
Problem 1.6Calculate the x- and y-intercepts of 3x − 4y = −6 and use them to graph the line.
Answer: x-intercept (−2, 0); y-intercept (0, 3/2).
For the y-intercept, substitute x = 0 and solve for y.
3(0) − 4y = −6
−4y = −6
y = 3/2
For the x-intercept, substitute y = 0 and solve for x.
3x − 4(0) = −6
3x = −6
x = −2
The graph of 3x − 4y = −6 crosses the x-axis at (−2, 0) and the y-axis at (0, 3/2).
See Figure 1-1 in the original PDF for the graph.
Problem 1.7Assume that line p contains the point (−3, 1) and is parallel to x − 4y = 1. Write the equation of p in slope-intercept form.
Answer: y = (1/4)x + 7/4.
Calculate the slope of x − 4y = 1 using the shortcut m = −A/B from Problem 1.3.
m = −A/B = −1/(−4) = 1/4
Parallel lines share the same slope, so line p also has slope 1/4. Plug that slope and the point (−3, 1) into point-slope form.
y − y₁ = m(x − x₁)
y − 1 = (1/4)(x − (−3))
y − 1 = (1/4)x + 3/4
y = (1/4)x + 7/4
Final answer: y = (1/4)x + 7/4.
Problem 1.8According to a basic Euclidean geometry theorem, the diagonals of a parallelogram bisect each other. Demonstrate this theorem for parallelogram ABCD with A = (2,1), B = (7,1), C = (11,4), D = (6,4).
Answer: Both midpoints equal (13/2, 5/2), confirming the theorem.
Calculate the midpoints of the two diagonals, AC and BD.
Midpoint of AC = ((2 + 11)/2, (1 + 4)/2) = (13/2, 5/2) Midpoint of BD = ((7 + 6)/2, (1 + 4)/2) = (13/2, 5/2)
The midpoints are equal, so the diagonals bisect each other. Theorem verified.
See Figure 1-2 in the original PDF for parallelogram ABCD.
Problem 1.9Prove that ABCD is a rhombus by verifying that its sides are congruent.
Answer: AB = BC = CD = AD = 5, so all four sides are congruent. ABCD is a rhombus.
Apply the distance formula four times, one for each side.
AB = √[(7−2)² + (1−1)²] = √25 = 5 BC = √[(11−7)² + (4−1)²] = √(16 + 9) = √25 = 5 CD = √[(6−11)² + (4−4)²] = √25 = 5 AD = √[(6−2)² + (4−1)²] = √(16 + 9) = √25 = 5
All four sides equal 5, so ABCD is a rhombus.
Problem 1.10Prove that ABCD is a rhombus by verifying that its diagonals are perpendicular to one another.
Answer: Slope of AC = 1/3 and slope of BD = −3. These are negative reciprocals, so the diagonals are perpendicular.
Compute the slopes of the two diagonals.
Slope of AC = (4 − 1)/(11 − 2) = 3/9 = 1/3 Slope of BD = (4 − 1)/(6 − 7) = 3/(−1) = −3
(1/3) · (−3) = −1, so the slopes are negative reciprocals. The diagonals are perpendicular, which is a defining property of a rhombus (parallelogram with perpendicular diagonals).
Linear Inequalities & Interval Notation (Problems 1.11 – 1.20)
Goodbye equal sign, hello parentheses and brackets.
Interval notation quick reference. A bracket [ or ] means the endpoint is included; a parenthesis ( or ) means it is excluded. Always use parentheses next to ∞ or −∞. When you multiply or divide an inequality by a negative number, flip the inequality sign. On a number line, a closed dot = included endpoint, open dot = excluded endpoint.
Click a problem to reveal its solution
Interval notation and graphing linear inequalities.
Problem 1.11Write the expression x ≥ −4 using interval notation.
Answer: [−4, ∞).
The lower bound is −4 and it is included (≥), so use a bracket. The upper bound is infinity, which always takes a parenthesis. Result: [−4, ∞).
Problem 1.12Write the expression x < 10 using interval notation.
Answer: (−∞, 10).
The upper bound is 10 and it is excluded (strict <), so use a parenthesis. The lower bound is −∞, which always takes a parenthesis. Result: (−∞, 10).
Problem 1.13Write the expression 6 ≥ x > −1 using interval notation.
Answer: (−1, 6].
The lower bound must always precede the upper bound. Here x is between −1 (excluded, strict >) and 6 (included, ≥). Result: (−1, 6].
Problem 1.14Write the solution to the inequality using interval notation: 4x − 2 > x + 13.
Answer: (5, ∞).
Separate variables and constants, then divide by the coefficient of x.
4x − x > 13 + 2
3x > 15
x > 5
Interval notation: (5, ∞).
Problem 1.15Write the solution to the inequality using interval notation: 3(2x − 1) − 5 ≤ 10x + 19.
Answer: [−27/4, ∞).
Distribute, combine like terms, and isolate x on the left.
6x − 3 − 5 ≤ 10x + 19
6x − 8 ≤ 10x + 19
−4x ≤ 27
Divide by −4. Dividing by a negative reverses the inequality sign.
x ≥ −27/4
Interval notation: [−27/4, ∞).
Problem 1.16Graph the inequality: −2 ≤ x < 3.
Answer: Interval [−2, 3). Closed dot at −2, open dot at 3, shaded between.
Rewrite the inequality as an interval: [−2, 3). On a number line, place a closed dot at x = −2 (included), an open dot at x = 3 (excluded), and shade the segment between them.
See Figure 1-3 in the original PDF for the number-line graph.
Problem 1.17Graph the inequality: x > −1.
Answer: Interval (−1, ∞). Open dot at −1, shaded to the right with an arrow.
All values greater than −1 satisfy the inequality. Place an open dot at −1 (excluded) and shade the number line to the right toward infinity.
See Figure 1-4 in the original PDF.
Problem 1.18Solve and graph the inequality: −7 ≤ 1 − 2x < 11.
Answer: (−5, 4], graphed on the number line with an open dot at −5 and a closed dot at 4.
Subtract 1 from each of the three expressions to isolate −2x in the middle.
−7 − 1 ≤ −2x < 11 − 1 −8 ≤ −2x < 10
Divide each expression by −2, reversing both inequality signs.
4 ≥ x > −5
Interval notation: (−5, 4].
See Figure 1-5 in the original PDF for the number-line graph.
Problem 1.19Graph the inequality: y < −(1/3)x + 2.
Answer: Dotted (dashed) line with y-intercept (0, 2) and slope −1/3; shade the region below the line.
The inequality is already solved for y. Graph the boundary line y = −(1/3)x + 2. Because the inequality uses a strict < (not ≤), draw it dotted to show the line itself is excluded.
Starting at the y-intercept (0, 2), go down 1 and right 3 to plot another point. Connect the dots with a dashed line.
Since y < ..., shade the region below the line.
See Figure 1-6 in the original PDF for the graph.
Problem 1.20Solve the equation: 2x − y ≤ 4.
Answer: Solid line y = 2x − 4; shade the region above the line (y ≥ 2x − 4).
Solve the inequality for y.
2x − y ≤ 4
−y ≤ −2x + 4
y ≥ 2x − 4 (flip sign when dividing by −1)
Graph the boundary line y = 2x − 4 as a solid line (because ≥ includes the boundary). Shade the region above the line (all (x, y) with y ≥ 2x − 4).
See Figure 1-7 in the original PDF for the shaded region.
Absolute Value Equations & Inequalities (Problems 1.21 – 1.27)
Solve two things for the price of one.
Absolute value rule. If |A| = b (with b ≥ 0), then A = b or A = −b — two equations for the price of one. If |A| < b, rewrite as the compound inequality −b < A < b. If |A| > b, rewrite as A > b or A < −b. Always isolate the absolute value first before splitting. Absolute values are never negative, so |A| = negative has no solution.
Click a problem to reveal its solution
Absolute value equations and inequalities.
Problem 1.21Solve the equation: |3x − 7| = 8.
Answer: x = 5 or x = −1/3.
The expression inside the absolute value must equal either 8 or −8.
3x − 7 = 8 3x − 7 = −8
3x = 15 3x = −1
x = 5 x = −1/3
Solutions: x = 5 or x = −1/3.
Problem 1.22Solve the equation: 1 − 2|x + 6| = −4.
Answer: x = −7/2 or x = −17/2.
Isolate the absolute value expression.
1 − 2|x + 6| = −4
−2|x + 6| = −5
|x + 6| = 5/2
Split into two equations.
x + 6 = 5/2 x + 6 = −5/2
x = 5/2 − 12/2 x = −5/2 − 12/2
x = −7/2 x = −17/2
Solutions: x = −7/2 or x = −17/2.
Problem 1.23Solve the equation: 9 − 3|x + 2| = 15.
Answer: No solution.
Isolate the absolute value expression.
9 − 3|x + 2| = 15
−3|x + 2| = 6
|x + 2| = −2
Absolute values always produce a non-negative number, so |x + 2| = −2 is impossible. No solution.
Problem 1.24Solve the inequality: |x − 5| < 1.
Answer: (4, 6).
An absolute value inequality of form |x + a| < b is equivalent to the compound inequality −b < x + a < b.
−1 < x − 5 < 1
Add 5 to each expression.
−1 + 5 < x < 1 + 5
4 < x < 6
Interval notation: (4, 6).
Problem 1.25Graph the solution to the inequality: 2|x − 7| − 5 ≤ −1.
Answer: [5, 9]. Closed dots at both endpoints, shaded between.
Isolate the absolute value expression.
2|x − 7| − 5 ≤ −1
2|x − 7| ≤ 4
|x − 7| ≤ 2
Create a compound inequality (as in Problem 1.24) and solve.
−2 ≤ x − 7 ≤ 2
−2 + 7 ≤ x ≤ 2 + 7
5 ≤ x ≤ 9
Interval notation: [5, 9].
See Figure 1-8 in the original PDF for the number-line graph (closed interval, both endpoints included).
Problem 1.26Solve the inequality: |2x + 5| ≥ 3.
Answer: (−∞, −4] ∪ [−1, ∞).
An inequality of form |ax + b| ≥ c splits into two inequalities: ax + b ≥ c OR ax + b ≤ −c.
2x + 5 ≥ 3 2x + 5 ≤ −3
2x ≥ −2 2x ≤ −8
x ≥ −1 x ≤ −4
The union of the two solutions: (−∞, −4] ∪ [−1, ∞).
Problem 1.27Solve the inequality and graph the solution: 2 − 3|x + 1| < −5.
Answer: (−∞, −10/3) ∪ (4/3, ∞).
Isolate the absolute value expression.
2 − 3|x + 1| < −5
−3|x + 1| < −7
|x + 1| > 7/3 (dividing by −3 flips the inequality)
Split into two inequalities (as in Problem 1.26).
x + 1 > 7/3 x + 1 < −7/3
x > 4/3 x < −10/3
Interval notation: (−∞, −10/3) ∪ (4/3, ∞).
See Figure 1-9 in the original PDF. On the number line, open dots at −10/3 and 4/3, shaded outward.
Systems of Equations & Inequalities (Problems 1.28 – 1.33)
Find a common solution shared by multiple equations or inequalities.
Systems methods. Substitution: solve one equation for a variable and plug it into the other. Elimination: scale the equations so one variable cancels when they're added. A dependent (indeterminate) system reduces to a true statement like 8 = 8 and has infinitely many solutions (the two lines are the same). An inconsistent system reduces to a false statement (e.g. 0 = 5) and has no solution (parallel lines). For three-variable systems: eliminate one variable to reduce to a 2×2 system, then back-substitute.
Click a problem to reveal its solution
Systems of equations and inequalities.
Problem 1.28Solve the system using substitution: { −8x + 2y = −5 ; 2x − y = 1 }.
Answer: (3/4, 1/2).
Solve the second equation for y: y = 2x − 1. Substitute into the first equation.
−8x + 2(2x − 1) = −5
−8x + 4x − 2 = −5
−4x = −3
x = 3/4
Substitute x = 3/4 back into y = 2x − 1.
y = 2(3/4) − 1 = 6/4 − 1 = 1/2
Solution: (3/4, 1/2).
Problem 1.29Solve the system using elimination: { 2x − 5y = −11 ; 3x + 13y = 4 }.
Answer: (−3, 1).
To eliminate x, multiply the first equation by −3 and the second by 2, then add.
−6x + 15y = 33
6x + 26y = 8
─────────────────
41y = 41
y = 1
Substitute y = 1 into either original equation.
2x − 5(1) = −11
2x = −6
x = −3
Solution: (−3, 1).
Problem 1.30Solve the system: { x − 6y = 24 ; (1/3)x − 2y = 8 }.
Answer: Infinitely many solutions (dependent system).
Solve the first equation for x: x = 6y + 24. Substitute into the second equation.
(1/3)(6y + 24) − 2y = 8
2y + 8 − 2y = 8
8 = 8
The variables vanish and the result is a true statement (8 = 8). This means the two equations describe the same line (the first equation divided by 3 gives the second), so the system is dependent and has infinitely many solutions.
Problem 1.31Determine the real number value of k in the system below that makes the system indeterminate (no solution): { x − 6y = −13 ; 4x − ky = 1 }.
Answer: k = 24.
An "indeterminate" (inconsistent) system has no solution when its two lines are parallel — they must share the same slope but have different intercepts. Compute the slopes with the shortcut from Problem 1.3.
Slope₁ = −(1)/(−6) = 1/6 Slope₂ = −(4)/(−k) = 4/k
Set the slopes equal and solve for k.
4/k = 1/6 4·6 = k·1 24 = k
So k = 24. (You can check that at k = 24, the two lines are parallel but not identical, so the system has no solution.)
Problem 1.32Graph the solution to the system of inequalities: { y < 3 ; x ≥ −4 ; y > (2/3)x − 1 }.
Answer: The two-dimensional region shaded by all three inequalities simultaneously (below y = 3, right of x = −4, above y = (2/3)x − 1).
Graph each inequality on the same coordinate plane. Use a dashed line for strict inequalities (<, >) and a solid line for ≥ or ≤. The solution to the system is the region where all three shaded regions overlap.
y = 3 is a horizontal line 3 units above the x-axis; shade below it. x = −4 is a vertical line four units left of the y-axis; shade to its right (including the line). y = (2/3)x − 1 is a line with y-intercept (0, −1) and slope 2/3; shade above it (dashed).
See Figure 1-10 in the original PDF for the overlapping shaded region.
Problem 1.33Solve the system: { 3x + 2y − z = 0 ; 5x − y − 8z = 9 ; x + 4y − 3z = −22 }.
Answer: (x, y, z) = (4, −5, 2).
Solve the first equation for z: z = 3x + 2y. Substitute this expression for z into the other two equations to reduce the system to two equations in x and y.
5x − y − 8(3x + 2y) = 9
5x − y − 24x − 16y = 9
−19x − 17y = 9
x + 4y − 3(3x + 2y) = −22
x + 4y − 9x − 6y = −22
−8x − 2y = −22
−4x − y = −11
Now solve the reduced 2×2 system by substitution. Solve the second reduced equation for y: y = −4x + 11. Plug into the first reduced equation.
−19x − 17(−4x + 11) = 9
−19x + 68x − 187 = 9
49x = 196
x = 4
Back-substitute x = 4 into y = −4x + 11.
y = −4(4) + 11 = −5
Back-substitute into z = 3x + 2y.
z = 3(4) + 2(−5) = 12 − 10 = 2
Solution: (x, y, z) = (4, −5, 2).
Why this chapter matters for calculus
Chapter 1 is short on purpose. Every problem in this walkthrough is algebra a calculus student is already expected to own — and every problem is a skill that will show up buried inside a limit, derivative, or integral. A short map of where each block reappears:
- Slopes and forms of a line (1.1 – 1.10) — The derivative is a slope. Tangent-line problems, linear approximations, and the definition of the derivative all live here. Point-slope form is the tangent-line equation you'll write on every AP Calculus free response.
- Distance, midpoint, perpendicular (1.8 – 1.10) — Perpendicular slopes come back as normal lines to curves. Distance shows up in arc length and in Newton's method.
- Interval notation (1.11 – 1.20) — This is the language calculus uses to describe domains, ranges, intervals of continuity, intervals of increase/decrease, and intervals of concavity. Every calculus textbook uses interval notation from Chapter 2 onward.
- Absolute value (1.21 – 1.27) — The formal definition of a limit uses |x − a| < δ and |f(x) − L| < ε. If absolute value inequalities aren't automatic, epsilon-delta is unnecessarily hard.
- Systems of equations (1.28 – 1.33) — Every intersection problem, every "find the tangent line to two curves at their intersection" problem, and every 3-variable optimization problem starts with a system.
The point of a good calculus prep is not "did you get the answer" — it's "did you get the answer in 30 seconds without a calculator." If any of the 33 problems above took you more than about two minutes, that's the topic to tune up before opening the derivative chapter.
Want a real diagnostic? Every SOMATH student who signs up for AP Calculus AB, BC, or Calculus 1 tutoring starts with a free 30-minute evaluation at our Upper West Side classroom (226 W 79th St, 1st Floor). We diagnose which of the 33 topics is the weak link, then build the shortest path to fluency. Call (646) 668-6151 to book.
FAQ
Why is a chapter of linear equations and inequalities in a calculus book?
Because every calculus problem contains an algebra problem inside it. Slopes are the geometric meaning of the derivative. Solving for a variable is the last step of almost every optimization and related-rates problem. Interval notation is how domains, ranges, and continuity are written. Absolute value shows up in convergence and epsilon-delta arguments. Systems of equations appear in every intersection, tangent, and equilibrium problem. Chapter 1 is short on purpose: it is the algebra a calculus student is expected to already own.
What is The Humongous Book of Calculus Problems?
It is a well-known calculus problem book by W. Michael Kelley — over 1000 worked calculus problems with hand-written margin notes explaining every step. Chapter 1 (Linear Equations and Inequalities) is the pre-calculus warm-up chapter — 33 problems on lines, slopes, interval notation, absolute value, and systems. It's a favorite of AP Calculus AB, AP Calculus BC, and Calculus 1 students who want a quick algebra tune-up before starting derivatives.
Should I do these problems before or after starting calculus?
Before, if you're not certain you can solve every one without a calculator in under three minutes. The algebra in these 33 problems is the same algebra that shows up buried inside limits, derivatives, and integrals — but once you're inside a derivative chapter, you don't want to be slowed down by "wait, how do I write this in slope-intercept form?" or "how do I solve an absolute value inequality?" Chapter 1 is the tune-up.
How does this connect to SOMATH's calculus prep courses?
SOMATH runs in-person calculus prep and full AP Calculus AB, BC, and Calculus 1 tutoring on the Upper West Side. Every calculus student starts by demonstrating fluency on exactly the skills in this chapter — slopes, forms of a line, interval notation, absolute value, and systems — before we start derivatives. If a student is not fluent, we spend a session or two here first. The classroom is at 226 W 79th Street, 1st Floor. Book a free 30-minute evaluation at (646) 668-6151.
Are these problems representative of what an AP Calculus student sees on the exam?
The problems themselves are pre-calculus by design (Chapter 1 is the algebra chapter). But the SKILLS tested are the exact algebra skills that show up embedded inside AP Calculus AB and BC free response questions: slope from two points, point-slope and slope-intercept form, parallel and perpendicular lines, interval notation, absolute value, and systems of equations. If any of the 33 problems takes you more than a minute or two, it's a signal to tune that skill up before moving on.
Where can I get the full book?
The Humongous Book of Calculus Problems is widely available at bookstores and online retailers. The chapter shared here is the first chapter of the book. If you like the format — question in a gray box, hand-written margin notes, and a full worked solution — the full book has 1000+ more calculus problems in the same style.
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