Chapter 3 — Linear Equations
Topics in this chapter:
3.1 What is a Linear Equation?
Key Definitions:
- Equation: Any expression with an equal sign.
- Variable: A quantity in an equation whose value is unknown.
- Constant: A quantity that can take only one fixed value.
- Coefficient: A constant multiplied by a variable term. In $3x + y = 2$, the coefficient of $x$ is $3$ and the coefficient of $y$ is $1$.
- Degree of a term: The sum of the powers of the variables in that term. E.g. the degree of $x$ is $1$; the degree of $xy^2$ is $3$.
- Degree of an equation: The highest degree among all terms.
- Linear equation: An equation whose degree is one, with no variables raised to fractional powers.
Example 3-1: Which of these equations are linear?
i. $x^2 + y = 4$ — Degree two, not linear.
ii. $xy = 4$ — Degree two (the term $xy$ has degree $1 + 1 = 2$), not linear.
iii. $x + y + z + w = 0$ — Degree one, linear ✓.
iv. $3^x + y = 0$ — The term $3^x$ is exponential; we cannot define a polynomial degree for it. Not linear. (Same applies to equations with trig functions or logarithms.)
3.2 One Equation, One Variable
Linear equations in one variable (like $x + 4 = 5$ or $3y + 4 = -5$) are solved in two steps:
- Move all constants to one side and the variable to the other (using addition/subtraction).
- Divide both sides by the coefficient of the variable.
Example 3-2: Solve $3x + 5 = 11 + x$ for $x$.
Solution:
\[3x + 5 = 11 + x\]Subtract $5$: $\;3x = 6 + x$
Subtract $x$: $\;2x = 6$
Divide by $2$: $\;x = 3$
Check: $3(3) + 5 = 14 = 11 + 3$ ✓
Example 3-3: Solve $ax + b = c$ for $x$, where $a, b, c$ are constants.
Solution: Subtract $b$: $\;ax = c - b$. Divide by $a$:
\[x = \frac{c - b}{a}\]Example 3-4: Solve for $y$: $\;\dfrac{1}{y} + 1 = 3$.
Solution: This isn’t linear as written, but multiply both sides by $y$:
\[1 + y = 3y\]Collecting: $2y = 1$, so $y = \dfrac{1}{2}$.
Exercise 3-1: Solve for $y$: $\;3y + 2 = y - 3 + 4y$.
Exercise 3-2: Solve for $y$: $\;\dfrac{1}{y} - 3 = y$.
3.3 Two Equations, Two Variables
A single equation in two variables (e.g. $x + y = 3$) has infinitely many solutions. We need a system of two equations to (usually) pin down a unique solution.
Substitution Method
Solve one equation for one variable, then substitute into the other.
Strategy: Pick the equation and variable that is easiest to isolate. “Solve for $x$” means put the equation in the form $x = \text{(something not involving } x\text{)}$.
Example: Given $x + y = 3$ and $x - y = 1$.
From the first equation: $x = 3 - y$.
Substitute into the second: $(3 - y) - y = 1$, so $2y = 2$, giving $y = 1$ and $x = 2$.
Elimination Method
Add (or subtract) multiples of the two equations to cancel one variable.
Example (same system): Add the equations:
\[(x + y) + (x - y) = 3 + 1 \implies 2x = 4 \implies x = 2, \quad y = 1.\]General elimination recipe: Given $a_1 x + a_2 y = a_3$ and $b_1 x + b_2 y = b_3$:
- Multiply the first equation by $-b_1$ and the second by $a_1$.
- Add the results — the $x$ terms cancel, leaving an equation in $y$ alone.
Example 3-5: Solve $\;2x + 3y = -1,\quad 3x - 4y = 7$.
Elimination: Multiply the first by $-3$ and the second by $2$:
\[-6x - 9y = 3, \qquad 6x - 8y = 14.\]Add: $-17y = 17 \implies y = -1$, then $x = 1$.
Substitution: From the first equation $x = \dfrac{-1 - 3y}{2}$. Substituting into the second:
\[3\!\left(\frac{-1 - 3y}{2}\right) - 4y = 7\]Solving yields $y = -1$, $x = 1$. Solution: $(x,y) = (1,-1)$. ✓
Special Cases
Example 3-6 (No solution — parallel lines): Solve $\;2x - 4y = 7,\quad x - 2y = 2$.
Attempting elimination gives $0 = 3$, a contradiction. The lines are parallel — no intersection, no solution.
Example 3-7 (Infinitely many solutions — identical lines): Solve $\;2x - 2y = 6,\quad x - y = 3$.
Elimination gives $0 = 0$. The first equation is just twice the second — they describe the same line. There are infinitely many solutions.
Warning: Always substitute your solution back into the original equations to verify correctness.
Example 3-8: Find $(x,y)$ such that $2x = 2y - 4$ and $2y = 2 + x$.
Solution: Rewrite as $2x - 2y = -4$ and $-x + 2y = 2$. Adding eliminates $y$: $x = -2$, so $y = 0$.
$(x,y) = (-2, 0)$.
Example 3-9: Find all $(x,y)$ such that $2\sqrt{x} + 4\sqrt{y} = 10$ and $2\sqrt{x} - 3\sqrt{y} = 3$.
Solution: These are not linear in $x,y$, but they are linear in $\sqrt{x}$ and $\sqrt{y}$. Subtracting the second from the first: $7\sqrt{y} = 7$, so $\sqrt{y} = 1$. Then $2\sqrt{x} + 4 = 10$ gives $\sqrt{x} = 3$. Squaring: $(x,y) = (9,1)$.
Exercise 3-3: Solve for $(x,y)$:
i. $\;3x = 5 + 2y,\quad 2x - 2y = 7$
ii. $\;2x + 3y = 4,\quad x + 6y = 9$
iii. $\;0.1x + y = 3,\quad 0.5x - 3y = 7$
iv. $\;x - y = 2x + 3,\quad x - 2y = 5 - 3y$
3.4 Word Problems
Word problems are equations written in words. The three-step procedure:
- Define all variables.
- Translate the words into equations.
- Solve the equations.
Age Problems
Example 3-10: Johnny is twice as old as Gina. Johnny is five years older than Gina. How old is Johnny?
Solution: Let $J$ = Johnny’s age, $G$ = Gina’s age.
\[J = 2G, \qquad J = G + 5.\]Substituting: $2G = G + 5 \implies G = 5$, so $J = 10$. Johnny is 10 years old.
Example 3-11: The units digit of a two-digit number is three times the tens digit. If the digits are reversed, the resulting number is 54 more than the original. Find the original number. (MAΘ 1990)
Solution: Let $t$ = tens digit, $u$ = units digit. The number’s value is $10t + u$.
From the problem: $u = 3t$ and $10u + t - (10t + u) = 54$.
The second equation simplifies: $9u - 9t = 54 \implies u - t = 6$.
Since $u = 3t$: $3t - t = 2t = 6$, so $t = 3$, $u = 9$.
The original number is 39.
Exercise 3-4: Adult tickets to a football game were $3.25 and student tickets were $1.75. If 1350 fans paid a total of $2700 to attend the game, how many adults attended? (MAΘ 1992)
Rate / Distance / Time Problems
Fundamental formula:
\[\text{rate} \times \text{time} = \text{distance}\]Example 3-12: Jim drives to his mother’s house (40 miles away) at 40 mph, then drives back at 20 mph. What is his average speed for the whole trip?
Solution: The answer is not 30 mph!
- Time there: $40/40 = 1$ hour.
- Time back: $40/20 = 2$ hours.
- Total: $80$ miles in $3$ hours.
Exercise 3-5: Jim drives 40 mph to his mother’s house and 20 mph on the way back. Show that his average speed for the round trip is $26\frac{2}{3}$ mph regardless of the distance to his mother’s house.
Example 3-13: A frog swims 8 miles downstream in 2 hours. She returns upstream in 14 hours. How fast does the frog swim in still water? (MAΘ 1987)
Solution: Let $x$ = frog’s rate in still water, $y$ = current rate.
- Downstream: $(x + y)(2) = 8$
- Upstream: $(x - y)(14) = 8$
From the first: $x + y = 4$. From the second: $x - y = \frac{4}{7}$.
Adding: $2x = \frac{32}{7}$, so $x = \frac{16}{7}$ mph.
Work Problems
Work formula:
\[(\text{time at work}) \times (\text{fraction of work done per unit time}) = (\text{fraction of job done})\]If a person can do a job in $T$ hours, they complete $\frac{1}{T}$ of the job per hour.
Example 3-14: Pipe A fills a pool in 5 hours; pipe B fills it in 4 hours. How long for both together?
Solution: Together they fill $\frac{1}{5} + \frac{1}{4} = \frac{9}{20}$ of the pool per hour.
\[x \cdot \frac{9}{20} = 1 \implies x = \frac{20}{9} = 2\tfrac{2}{9} \text{ hours}\]Example 3-15: Tom and Huck paint a fence for 4 hours, then Jim helps and they finish 2 hours later. Without Jim, it would have taken 5 more hours. How long would Jim take alone?
Solution: Let $T$, $H$, $J$ be the solo times for Tom, Huck, and Jim.
Tom + Huck can do the whole job in 9 hours, so $\frac{1}{T} + \frac{1}{H} = \frac{1}{9}$.
In the first 4 hours they complete $\frac{4}{9}$ of the fence. The remaining $\frac{5}{9}$ is done in 2 hours by all three:
\[2\!\left(\frac{1}{T} + \frac{1}{H} + \frac{1}{J}\right) = \frac{5}{9}\]Since $\frac{1}{T} + \frac{1}{H} = \frac{1}{9}$:
\[2\!\left(\frac{1}{9} + \frac{1}{J}\right) = \frac{5}{9} \implies \frac{1}{9} + \frac{1}{J} = \frac{5}{18} \implies \frac{1}{J} = \frac{5}{18} - \frac{2}{18} = \frac{3}{18} = \frac{1}{6}\]Jim can paint the fence alone in 6 hours.
Exercise 3-6: A canoeist paddled upstream for 2 hours, then downstream for 3 hours. The current was 2 mph. She ended up 20 miles downstream from her start. How many hours will it take her to paddle back? (MAΘ 1992)
Exercise 3-7: One knight can storm a castle in 15 days. He and his partner together can do it in 10 days. How long does the partner take alone? (MAΘ 1987)
Exercises & Problems
End-of-Chapter Problems
Problem 27. At a dance party, one boy dances with 5 girls, a second with 6 girls, and so on, the last boy dancing with all girls. If $b$ = number of boys and $g$ = number of girls, find $b$ in terms of $g$. (AHSME 1958)
Problem 28. The tens digit of a two-digit number exceeds its units digit by 4. The number exceeds twice the reversed number by 10. Find the original number. (MAΘ 1987)
Problem 29. There are 16 coins in a bank, all nickels and dimes, totaling $1.05. How many nickels are there? (MATHCOUNTS 1990)
Problem 30. George Washington was born 11 years before Thomas Jefferson. In 1770, Washington’s age was 3 more than 7 times Jefferson’s age in 1748. What was the sum of the two men’s ages in 1750? (MAΘ 1991)
Problem 31. The number 66 is divided into two smaller numbers. One number is 3 more than twice the other. Find the larger number. (MATHCOUNTS 1990)
Problem 32. Four pounds of onions costs the same as 2 pounds of string beans. One pound of string beans costs 3 times as much as a pound of potatoes, and 1 pound of onions costs 4 cents less than 2 pounds of potatoes. What is the total cost of 1 pound of each vegetable? (MAΘ 1991)
Problem 33. Find two consecutive odd integers such that $\frac{1}{3}$ of the smaller plus twice the larger equals 7 more than their sum. (MAΘ 1990)
Problem 34. In a basketball game, the US has four times as many points as Croatia. A Croatian makes a 3-point basket, at which point the US has only three times as many points. How many points does the US have? (Mandelbrot #3)
Problem 35. Mike and Joey bought identical loaves of bread and packages of bologna. Mike made sandwiches with 5 slices of bologna and had 4 slices of bread left when he ran out of meat. Joey made sandwiches with 4 slices of bologna and had 4 slices of meat left when he ran out of bread. How many slices of bread were in each loaf? (MAΘ 1992)
Problem 36. Sue has $3.08 in pennies, nickels, and quarters. She has four more pennies than quarters and one more nickel than pennies. How many nickels does she have? (MAΘ 1990)
Problem 37. Solve for $c$ in terms of $a$ and $b$ given that
\[\frac{a}{b+c} = \frac{b}{a+c}.\](AHSME 1955)
Problem 38. K takes 30 minutes less than M to travel 30 miles. K travels 5 mph faster than M. If $x$ is K’s speed in mph, find K’s time for the distance in terms of $x$. (AHSME 1952)
Problem 39. What is the value of $x$ if $1$ minus the reciprocal of $(1 - x)$ equals the reciprocal of $(1 - x)$? (MAΘ 1992)
Problem 40. A train traveling from Aytown to Beetown meets with an accident after 1 hour. The train is stopped for 30 minutes, after which it proceeds at $\frac{4}{5}$ of its usual rate, arriving 2 hours late. If the train had covered 80 miles more before the accident, it would have been just 1 hour late. What is the usual rate of the train? (AHSME 1955)
Problem 41. Adam can do a job in 10 days; Brenda takes 15 days. After Brenda works alone for 3 days, they work together to finish. How many days did Adam work? (MAΘ 1990)
Problem 42. A car travels 120 miles from A to B at 30 mph but returns at 40 mph. What is the average speed for the round trip? (AHSME 1950)
Problem 43. One car left a city at 2:00 PM traveling at 40 mph. A second car left at 4:00 PM on the same route and overtook the first at 9:00 PM. What was the second car’s average speed? (MATHCOUNTS 1991)
Problem 44. A man can do a job in 9 days and his son in 16 days. They start together. After 4 days the son leaves. How many more days does the man take to finish? (MAΘ 1991)
Problem 45. Twenty-five women did $\frac{1}{5}$ of a job in 8 days. How many additional women must be added to complete the rest in the next 20 days? (MATHCOUNTS 1989)
Problem 46. Two bicyclists are $\frac{7}{8}$ of the way through a mile-long tunnel when a train approaches the closer end at 40 mph. They ride in opposite directions at the same speed and each escapes as the train passes them. How fast did they ride? (Mandelbrot #3)
Problem 47. A train $x$ meters long, traveling at constant speed, takes 20 seconds from entering a 300-meter tunnel until completely emerging. A ceiling light is directly above the train for 10 seconds. Find $x$. (MAΘ 1992)
Problem 48. Two men start at a point on a circular 1-mile track and walk in opposite directions with uniform speeds, meeting in 6 minutes. Walking in the same direction, the faster gains a lap in 1 hour. Find the rate of the slower walker. (MAΘ 1991)
Problem 49. A crew of 30 can build a road in 60 days. After 10 days, the company wants it done in 30 total days. How many more people must be hired? (MAΘ 1992)
Problem 50. Jack and Jill went up the hill at 8 units/min and came tumbling down at 8 units/sec. What was their average rate in units/min for the round trip? (MAΘ 1992)
Problem 51. Two dogs run toward each other from 500 ft apart, each at 10 ft/s. A flea flies between them at 25 ft/s until crushed when they collide. How far did the flea fly? (MAΘ 1992)
Problem 52. Find the ordered pair $(x,y)$ that solves (MAΘ 1990):
\[\frac{x + 2y}{xy} = \frac{11}{12}, \qquad \frac{2x - 3y}{xy} = \frac{2}{3}.\]Problem 53. Find $x/y$ if $\;\dfrac{3}{\sqrt{y}} - \dfrac{1}{\sqrt{x}} = \dfrac{2}{\sqrt{x}+\sqrt{y}}$. (MAΘ 1992)
Problem 54. When three numbers are added two at a time, the sums are 29, 46, and 53. What is the sum of all three? (MATHCOUNTS 1991)
Problem 55. Each valve A, B, C releases water into a tank at its own constant rate. All three open: fills in 1 hr. A and C only: 1.5 hr. B and C only: 2 hr. How long for A and B only? (AHSME 1973)