Unit 18 · Algebra I: Linear Equations, Functions and Systems

Look at three sentences: 3 + 4 = 7, 3 + 4 = 8, and x + 4 = 7. The first is true. The second is false. The third is neither, because it has a blank in it. A sentence with a variable is called an open sentence: it becomes true or false only after you fill in a number. Fill in 3 and you get 3 + 4 = 7, true. Fill in 5 and you get 5 + 4 = 7, false. So an equation is not a command to "do something." It is a claim that two sides are equal, and your job is to find out when that claim holds.

A number that makes the equation true is a solution. To test a number, you substitute it for the variable and simplify both sides. Is 5 a solution of 2x − 3 = 7? Substitute: 2(5) − 3 = 10 − 3 = 7. The left side is 7, the right side is 7, so yes. Is 4 a solution? 2(4) − 3 = 8 − 3 = 5, and 5 is not 7, so no. This testing step is the most important habit in algebra, because it works no matter how you found the number.

Students sometimes decide a number is a solution because it "looks right" or because it appeared in the problem. The test does not care about looks. For 3x + 2 = 17, the number 17 is in the problem, but 3(17) + 2 = 53, not 17. The number 5 is not in the problem, but 3(5) + 2 = 17, so 5 is the solution. Always substitute and compare the two sides before you trust an answer.

Picture a balance scale that is level. Add 2 pounds to the left pan only and it tips. Add 2 pounds to both pans and it stays level. An equation works the same way. The properties of equality say you can add, subtract, multiply or divide both sides by the same number and the new equation has the same solution as the old one. That is the whole toolkit for solving. You use it to isolate the variable, which means getting it alone on one side.

Solve 3x + 7 = 22. The variable is wrapped in two operations: multiplied by 3, then 7 is added. Undo them in reverse order using inverse operations. First subtract 7 from both sides: 3x + 7 − 7 = 22 − 7, so 3x = 15. Then divide both sides by 3: x = 5. Check by substituting: 3(5) + 7 = 15 + 7 = 22. True. Notice the coefficient 3 was handled last, because it was applied to x first.

One more: x/4 − 2 = 3. Add 2 to both sides: x/4 = 5. Multiply both sides by 4: x = 20. Check: 20/4 − 2 = 5 − 2 = 3. The most common mistake is doing the step on one side only, writing 3x = 22 − 7 but then forgetting the 7 was also on the left. Another is dividing only one term, turning 3x + 7 = 22 into x + 7 = 22/3. Every operation touches the whole of both sides.

Some equations have parentheses, several terms, and variables on both sides. Solve 4(x − 3) + 2x = 3x + 6. Start by cleaning up each side. The distributive property turns 4(x − 3) into 4x − 12, because the 4 multiplies both the x and the 3. Now the left side is 4x − 12 + 2x. Combine the like terms 4x and 2x to get 6x − 12. The equation is now 6x − 12 = 3x + 6.

Next, get all the x terms on one side. Subtract 3x from both sides: 3x − 12 = 6. Add 12 to both sides: 3x = 18. Divide both sides by 3: x = 6. Check in the original equation: 4(6 − 3) + 2(6) = 4(3) + 12 = 24, and 3(6) + 6 = 24. Both sides are 24, so x = 6 is correct. Writing the reason next to each step is called justifying it: distributive property, combine like terms, subtraction property of equality, addition property, division property.

Two mistakes cause most wrong answers here. The first is distributing to only the first term, writing 4(x − 3) as 4x − 3. Try it with numbers: 4(5 − 3) = 8, but 4(5) − 3 = 17, so those are not the same. The second is moving a term across the equals sign without changing its sign. Subtracting 3x from the right side means subtracting 3x from the left side too. Justifying each step forces you to notice both errors.

The perimeter of a rectangle is P = 2l + 2w. Usually you know the length and width and want P. But a fence builder often knows the total fencing and the length, and needs the width. A formula with more than one letter is a literal equation, and you can solve for any letter in it. Treat every other letter as if it were a known number.

Try it with numbers first. If P = 50 and l = 15, then 50 = 2(15) + 2w, so 50 = 30 + 2w. Subtract 30 from both sides: 20 = 2w. Divide by 2: w = 10. Now do the identical steps with letters. Start with P = 2l + 2w. Subtract 2l from both sides: P − 2l = 2w. Divide both sides by 2: w = (P − 2l)/2. Plug in P = 50 and l = 15 to see it give the same 10. The letter version is a new formula you can reuse for any fence.

Jasmine’s fertilizer rule was pounds = 0.003 × A. To solve for A, divide both sides by 0.003: A = pounds/0.003. Six pounds covers 6/0.003 = 2,000 square feet. The classic error in rearranging is dividing only part of one side. From P − 2l = 2w, some students write w = P − 2l/2, which divides only the 2l. Parentheses matter: the whole expression P − 2l is divided by 2.

Weather in Chicago is reported in degrees Fahrenheit, but science class and most of the world use Celsius. The two scales are linked by F = (9/5)C + 32. Water freezes at 0 °C, and the formula gives (9/5)(0) + 32 = 32 °F. Water boils at 100 °C: (9/5)(100) + 32 = 180 + 32 = 212 °F. A warm 25 °C day becomes (9/5)(25) + 32 = 45 + 32 = 77 °F.

To go the other way, solve for C. Start with F = (9/5)C + 32. Subtract 32 from both sides: F − 32 = (9/5)C. Multiply both sides by 5/9, which undoes multiplying by 9/5: C = (5/9)(F − 32). Test it on a hot summer afternoon of 86 °F: 86 − 32 = 54, and (5/9)(54) = 30, so 86 °F is 30 °C. A cold January morning of −4 °F: −4 − 32 = −36, and (5/9)(−36) = −20, so −4 °F is −20 °C. Both formulas are equivalent; they describe the same relationship.

The order of operations trap here is subtracting 32 after multiplying. (5/9)(86) − 32 = 47.8 − 32 = 15.8, which is wrong. The 32 is added last in the forward formula, so it must be removed first when going backward. Whenever you rearrange a formula, test your new version with one pair of values you already trust, such as 0 °C and 32 °F.

The absolute value of a number is its distance from zero on the number line, ignoring direction. So |5| = 5 and |−5| = 5. That means the equation |x| = 5 has two solutions, 5 and −5, because two numbers sit five units from zero. The equation |x − 3| = 5 asks which numbers are 5 units from 3. Split it into two ordinary equations: x − 3 = 5 gives x = 8, and x − 3 = −5 gives x = −2. Check: |8 − 3| = 5 and |−2 − 3| = |−5| = 5.

If the absolute value is not alone, isolate it first. Solve 3|x| + 2 = 14. Subtract 2: 3|x| = 12. Divide by 3: |x| = 4, so x = 4 or x = −4. Now try |2x + 1| = 7. Case one: 2x + 1 = 7, so 2x = 6 and x = 3. Case two: 2x + 1 = −7, so 2x = −8 and x = −4. Check the second: |2(−4) + 1| = |−7| = 7. Both work.

A distance can never be negative, so |x| + 5 = 2 has no solution at all: it would need |x| = −3. Absolute value shows up in manufacturing as tolerance. A part must measure 50 millimeters, give or take 0.2 millimeters. The acceptable extremes solve |m − 50| = 0.2, giving m = 50.2 and m = 49.8. The common mistake is forgetting the second case and reporting only the positive answer.

An inequality compares two sides with <, >, ≤ or ≥ instead of an equals sign. The statement x + 4 > 9 is true for x = 6, x = 10, x = 5.5 and infinitely many other values. All of those values together form the solution set. To find it, use the same balancing moves as with equations. Subtract 4 from both sides: x > 5. Every number greater than 5 works, and 5 itself does not.

Solve 2x − 3 ≤ 11. Add 3 to both sides: 2x ≤ 14. Divide by 2: x ≤ 7. Check with one number inside the range, say x = 0: 2(0) − 3 = −3, and −3 ≤ 11 is true. Check one number outside, say x = 10: 2(10) − 3 = 17, and 17 ≤ 11 is false. Checking both an inside and an outside value catches a wrong direction.

Draw the solution on a number line. The boundary number, 7, gets a closed dot because ≤ includes it, and an arrow points left toward all smaller numbers. For x > 5 the boundary 5 gets an open circle, because 5 is not included, and the arrow points right. The picture is the answer; a single number is not.

Start with a true statement: 2 < 5. Multiply both sides by −1. You get −2 and −5, and on the number line −2 is to the right of −5, so −2 > −5. The direction changed. Multiplying or dividing by a negative coefficient flips every number to the other side of zero and swaps the order. So the rule: whenever you multiply or divide both sides of an inequality by a negative number, reverse the sign.

Solve −3x + 4 ≥ 19. Subtract 4 from both sides: −3x ≥ 15. Now divide both sides by −3 and reverse the sign: x ≤ −5. Check with x = −6: −3(−6) + 4 = 18 + 4 = 22, and 22 ≥ 19 is true. Check with x = 0: 0 + 4 = 4, and 4 ≥ 19 is false. The inside value works and the outside value fails, so x ≤ −5 is right.

You can avoid the flip entirely by moving the variable to the other side. From −3x + 4 ≥ 19, add 3x to both sides: 4 ≥ 19 + 3x. Subtract 19: −15 ≥ 3x. Divide by positive 3: −5 ≥ x, which is the same equivalent inequality x ≤ −5. The common mistake is flipping at the wrong moment, such as when subtracting a negative number. Adding and subtracting never flip the sign; only multiplying or dividing by a negative does.

A refrigerator should stay between 34 °F and 40 °F. That is two conditions at once: T ≥ 34 and T ≤ 40. Written as one compound inequality, it is 34 ≤ T ≤ 40. The solution is the intersection, the numbers that satisfy both parts, which forms an interval on the number line from 34 to 40 with closed dots at both ends.

Solve −2 < x + 3 ≤ 6. There are three parts, and whatever you do must be done to all three. Subtract 3 from every part: −5 < x ≤ 3. The solution is every number greater than −5 and at most 3. Check x = 0, which is inside: −2 < 3 ≤ 6, true. Check x = 5, outside: −2 < 8 ≤ 6 is false because 8 is not ≤ 6. A frequent error is subtracting 3 from only the middle and the right, leaving −2 unchanged.

An ’or’ inequality is different. Solve 2x − 1 < −7 or 3x + 2 > 11. Handle each separately: 2x < −6 gives x < −3, and 3x > 9 gives x > 3. The solution is the union: x < −3 or x > 3, two arrows pointing away from each other with a gap between −3 and 3. Numbers like 0 are not solutions. With ’and’ you keep only what both parts share; with ’or’ you keep everything either part allows.

A gym charges $25 to join and $15 per month. Someone has paid $130 in total. How many months have they been a member? A model is an equation that captures the situation. First define the variable: let m be the number of months. The join fee is paid once and the monthly fee is paid m times, so 25 + 15m = 130. Subtract 25: 15m = 105. Divide by 15: m = 7. Finally interpret: they have been a member for 7 months. Check: 25 + 15(7) = 25 + 105 = 130.

Number puzzles use the same steps. Three consecutive integers add up to 48. Let n be the smallest. The next two are n + 1 and n + 2. The equation is n + (n + 1) + (n + 2) = 48, which simplifies to 3n + 3 = 48. Subtract 3: 3n = 45. Divide: n = 15. The integers are 15, 16 and 17, and 15 + 16 + 17 = 48.

Watch the key phrases. ’Twice a number decreased by 7’ is 2n − 7, not 2(n − 7); the decrease happens after the doubling. ’Seven less than a number’ is n − 7, not 7 − n. If the wording is unclear, plug in an easy number and see which expression matches what the sentence says. Then, after solving, always read your answer back into the story to see whether it makes sense.

A freight elevator is rated for 2,000 pounds. A cart weighing 180 pounds is already inside, and each worker weighs about 150 pounds. How many workers can ride? The limit is a constraint, and ’rated for’ means the total must be at most 2,000. Let p be the number of workers: 180 + 150p ≤ 2000. Subtract 180: 150p ≤ 1820. Divide by 150: p ≤ 12.13. You cannot have 0.13 of a person, so at most 12 workers can ride. Rounding up to 13 would break the limit: 180 + 150(13) = 2,130.

A club needs to raise at least $500 and has $124 so far. It sells candles for $8 each. Let c be the number of candles: 124 + 8c ≥ 500. Subtract 124: 8c ≥ 376. Divide by 8: c ≥ 47. They must sell 47 or more candles. Check: 124 + 8(47) = 124 + 376 = 500 exactly, so 47 just reaches the goal and 46 falls short at 492.

Translate the phrases carefully. ’At most 30’ and ’no more than 30’ both mean ≤ 30. ’At least 30’ and ’no fewer than 30’ both mean ≥ 30. ’More than 30’ is strictly > 30. After solving, decide which numbers are actually feasible: whole numbers for people or candles, and never negative. A model with a solution like p ≤ 12.13 is telling you 12, not 12.13, and not 13.

Back to the fertilizer bag: 3 pounds for every 1,000 square feet. A bed 40 feet by 60 feet has area 40 ft × 60 ft = 2,400 square feet. The rate is 3 lb per 1,000 sq ft, so pounds = 2,400 sq ft × (3 lb / 1,000 sq ft) = 7.2 lb. Notice the square feet cancel and only pounds remain. This tracking of units is called unit analysis. If a student multiplied by 3 without dividing by 1,000, the answer would be 7,200, and the leftover units would be pound-square-feet, which is meaningless. The units flag the error.

An estimate catches errors too. Before computing, ask: is 2,400 square feet more or less than 1,000? It is a bit more than double, so the answer should be a bit more than double 3 pounds, roughly 6 to 8. An answer of 7.2 is reasonable; an answer of 720 or 0.72 is not. Estimating first takes ten seconds and saves a whole wrong page.

Time is the unit that trips people up most. A trip of 60 miles at 40 miles per hour takes 60 ÷ 40 = 1.5 hours. That is 1 hour and 30 minutes, not 1 hour and 50 minutes, because 0.5 hour × 60 minutes per hour = 30 minutes. Likewise 0.6 hour is 36 minutes. And some answers are impossible on their face: −3 tickets, 4.5 people, or a 90-year-old kindergartner. When the number is impossible, the model or the arithmetic is wrong, not the world.

A taxi meter is a machine: you feed in a number of miles and it gives back a fare. Feed in 1 mile, get $5. Feed in 2 miles, get $7. Feed in 3 miles, get $9. No matter how many times you feed in 3 miles, the meter gives back $9, never $8 and never $10. That is the key property of a function: every input is matched with exactly one output. The set of miles is the input, the set of fares is the output, and the rule connecting them is the function.

Any set of matched pairs is called a relation, but not every relation is a function. Suppose a table listed 2 miles → $5 and also 2 miles → $7. The same input has two different outputs, so the table is not a function; the meter would be broken. On the other hand, two different inputs can share an output. A rule that charges $10 for any ride under a mile, so that 0.3 miles → $10 and 0.8 miles → $10, is still a function, because each input still has only one answer.

On a graph, the input is on the horizontal axis and the output on the vertical axis. If any vertical line crosses the graph twice, then one input has two outputs and the graph is not a function. This is the vertical line test. A straight slanted line always passes the test. A circle fails, because a vertical line through its middle hits the top and the bottom.

Function notation gives a rule a name. The taxi rule is f(x) = 2x + 3. Read it as "f of x equals 2x plus 3." The letter f is the name, x is the input value, and 2x + 3 is what the rule does. To evaluate f(3), replace every x with 3: f(3) = 2(3) + 3 = 9. In the story, that means a 3-mile ride costs $9. The parentheses do not mean multiplication here. f(3) is not f × 3.

The notation runs backward too. If someone asks, "How far can I ride for $15?" you are solving f(x) = 15. Write 2x + 3 = 15, subtract 3 to get 2x = 12, and divide to get x = 6. A $15 ride is 6 miles. Notice the difference between the two questions: f(6) asks for the output when the input is 6 (answer: 15), while f(x) = 15 asks which input gives an output of 15 (answer: 6).

Be careful with expressions inside and outside the parentheses. f(x + 1) means put x + 1 into the rule: f(3 + 1) = f(4) = 2(4) + 3 = 11. But f(x) + 1 means find f(x) first and then add 1: f(3) + 1 = 9 + 1 = 10. Different rules get different names so they do not get mixed up: the cell data rule is d(g) = 8g + 20, and d(3) = 44 while f(3) = 9.

The cell plan rule d(g) = 8g + 20 works fine on paper for g = 1,000, but the plan caps data at 10 gigabytes. So the inputs that actually make sense run from 0 to 10. That set of allowed inputs is the domain: 0 ≤ g ≤ 10. Feed the smallest and largest inputs through the rule to find the outputs: d(0) = 20 and d(10) = 8(10) + 20 = 100. Since the rule rises steadily, every cost from $20 to $100 is possible. That set of outputs is the range: 20 ≤ d ≤ 100.

Some functions accept only certain kinds of numbers. Movie tickets cost $12 each, so c(n) = 12n, but n must be a whole number: you cannot buy 2.5 tickets. If the website allows up to 6 tickets, the domain is {0, 1, 2, 3, 4, 5, 6} and the range is {0, 12, 24, 36, 48, 60, 72}. A domain made of separate values like this is discrete, and its graph is dots. The taxi domain, miles ≥ 0, is continuous, because a ride can be 2.37 miles, and its graph is a solid line.

Domain and range often get swapped, so anchor them to the axes: domain is the set of x-values (horizontal), range is the set of y-values (vertical). When a problem gives a context, use it. Time cannot be negative, tickets come in whole numbers, and a plan has a cap. The context shrinks the domain, and the rule then decides the range.

On the taxi graph, the point (1, 5) means 1 mile costs $5 and (4, 11) means 4 miles cost $11. Between them the fare rose by 11 − 5 = 6 dollars while the miles rose by 4 − 1 = 3. The slope is the vertical change divided by the horizontal change: 6/3 = 2. In the story that is $2 per mile, the rate of change. The vertical change is often called the rise and the horizontal change the run, so slope = rise/run.

The formula for two points (x1, y1) and (x2, y2) is slope = (y2 − y1)/(x2 − x1). For (2, 7) and (5, 16): (16 − 7)/(5 − 2) = 9/3 = 3. Order does not matter as long as you are consistent. Going the other way, (7 − 16)/(2 − 5) = −9/−3 = 3, the same. The mistake to avoid is mixing the orders: (16 − 7)/(2 − 5) = 9/−3 = −3, which is wrong.

Slope can be negative. A tank starts with 500 gallons and holds 400 gallons after 4 minutes: (400 − 500)/(4 − 0) = −100/4 = −25. The tank loses 25 gallons per minute. For a linear function the slope is the same between any two points, which is exactly what makes it linear. For a curve, the same calculation gives the average rate of change over that interval, which is why the formula is worth memorizing.

Every linear function can be written in slope-intercept form, y = mx + b. The number m is the slope, and b is the y-intercept, the output when the input is 0. In the cell plan d = 8g + 20, m = 8 means $8 per gigabyte and b = 20 means the bill is $20 before any data is used. That b is often called the initial value because it is where the story starts.

You can build the equation from two points. Say a bill was $28 for 1 GB and $52 for 4 GB, so the points are (1, 28) and (4, 52). Slope: (52 − 28)/(4 − 1) = 24/3 = 8. Now find b by substituting one point into y = 8x + b: 28 = 8(1) + b, so b = 20. The function is d = 8g + 20. Check with the other point: 8(4) + 20 = 52. It matches.

From a graph, read b where the line crosses the vertical axis, then count rise over run to a second clear point. The common mistake is swapping m and b: reading y = 3x + 5 as "starts at 3, goes up 5 each time." Remember that the number attached to x is the one that repeats with every step, and the lone number is paid once.

A stadium’s first row has 20 seats, and each row behind it has 2 more: 20, 22, 24, 26, … A list where you add the same amount each time is an arithmetic sequence, and the amount added, here 2, is the common difference. Each number in the list is a term. Row 1 is term 1, row 2 is term 2. To find any term, notice that row n has taken n − 1 steps of size 2 from the first row: a(n) = 20 + 2(n − 1), which simplifies to a(n) = 2n + 18. Row 15 has a(15) = 30 + 18 = 48 seats.

Look at the sequence 5, 9, 13, 17, … The common difference is 4. The explicit rule is a(n) = 5 + 4(n − 1) = 4n + 1. Check: a(1) = 5, a(2) = 9, a(3) = 13. Then a(10) = 41. The same sequence has a recursive rule: a(1) = 5 and a(n) = a(n − 1) + 4, which says "start at 5 and keep adding 4." The recursive rule is easy to state but slow to use for a(100); the explicit rule gets there in one step.

The explicit rule a(n) = 4n + 1 is a linear function with slope 4 and intercept 1, but its domain is only the positive integers, so its graph is a row of dots on a line, not the line itself. The common difference is the slope. The usual error is writing a(n) = 5 + 4n, which starts at 9 instead of 5. Always test your rule on the first term before trusting it.

A tank holds 500 gallons and drains at 25 gallons per minute, so y = 500 − 25x. Its graph crosses the vertical axis at 500: that is the y-intercept, the amount at time 0. It crosses the horizontal axis where y = 0. Solve 500 − 25x = 0: 25x = 500, so x = 20. That is the x-intercept, also called the zero of a function, and it means the tank is empty after 20 minutes.

The line falls from left to right, so the function is decreasing: every extra minute gives fewer gallons. A savings function like s(w) = 30w + 120 is increasing, because each week adds $30. Its y-intercept, 120, is the starting balance. Its zero would be at 30w + 120 = 0, so w = −4, but negative weeks are outside the domain, so the graph never reaches zero in the story.

A graph is also positive where it sits above the horizontal axis and negative where it sits below. For the tank, y is positive for 0 ≤ x < 20 and would be negative after that, which is impossible, so the sensible domain is 0 ≤ x ≤ 20. When you read any graph, ask four questions: where does it start (y-intercept), where does it hit zero (x-intercept), does it rise or fall (increasing or decreasing), and which inputs actually make sense (domain).

A bike rental posts these prices: 1 hour $14, 2 hours $20, 3 hours $26, 4 hours $32. Each extra hour adds 20 − 14 = 6, then 26 − 20 = 6, then 32 − 26 = 6. Constant differences mean a linear model fits. The slope is 6 dollars per hour. To find the intercept, step back one hour from (1, 14): 14 − 6 = 8. The model is c = 6h + 8, meaning an $8 flat fee plus $6 per hour. Check: 6(4) + 8 = 32.

A model earns its keep by making a prediction. For 7 hours, c = 6(7) + 8 = 50. That prediction lies just outside the table’s range of 1 to 4 hours, which is called extrapolating. Predicting a value inside the table’s range, such as 2.5 hours giving 6(2.5) + 8 = 23, is interpolating and is usually safer. Far extrapolation is risky: for 40 hours the model gives $248, but the shop probably switches to a daily rate long before that.

Real data rarely lines up perfectly. If the differences were 6, 7, 5 and 6, the pattern is still roughly linear and a line drawn through the middle of the points is a reasonable model. State the slope with its units, state what the intercept means, and say how far you trust the prediction. A model is a tool for the situation it came from, not a law of nature.

A downtown garage charges $5 for the first hour and $3 for each additional hour or part of an hour. One hour costs $5. Two hours cost $8. Two and a half hours also cost $11, because the half hour counts as a full extra hour. The graph is flat at $5 up to 1 hour, jumps to $8, stays flat, jumps to $11, and so on. A graph made of flat pieces with jumps is a step function. It is still a function: every parking time has exactly one price.

More generally, a piecewise function has a different rule on each interval of its domain. Overtime pay is a classic case: a worker earns $15 an hour for the first 40 hours and $22.50 an hour after that. So p(h) = 15h when h ≤ 40, and p(h) = 600 + 22.5(h − 40) when h > 40, where 600 is the pay for the first 40 hours. For a 44-hour week, use the second rule: 600 + 22.5(4) = 600 + 90 = 690.

To evaluate a piecewise function, first find which interval the input belongs to, then apply that rule only. Let f(x) = 2x for x < 3 and f(x) = x + 3 for x ≥ 3. Then f(2) = 4 because 2 < 3, and f(5) = 8 because 5 ≥ 3. The boundary matters: f(3) uses the second rule since 3 ≥ 3, so f(3) = 6. The common mistake is applying both rules or the wrong one at the boundary.

Start with f(x) = 2x, the taxi rate without a starting fee. Now the company adds a $3 flag fee: g(x) = f(x) + 3 = 2x + 3. Every output is 3 more than before, so the whole line slides up 3 units. This is a vertical shift, one kind of transformation. At x = 0, 1 and 2, f gives 0, 2 and 4, while g gives 3, 5 and 7. Same slope, higher line.

Adding inside the parentheses changes the input instead. h(x) = f(x − 1) = 2(x − 1) = 2x − 2. To get the output f used to give at x = 1, you now must put in x = 2. The graph slides right 1 unit, a horizontal shift. Notice the direction: x − 1 moves right and x + 1 moves left, the opposite of what many people first guess. Check with a point: f(1) = 2, and h(2) = 2(2) − 2 = 2, so the point moved from (1, 2) to (2, 2).

Multiplying the output makes the line steeper or flatter. k(x) = 3·f(x) = 6x has slope 6 instead of 2, a vertical stretch. The intercept stays at 0, but each output is tripled: at x = 2, f gives 4 and k gives 12. Fastest way to see a transformation is a small table of three inputs. If every output changed by the same amount, it was a shift; if the outputs were multiplied, it was a stretch.

Two cell plans: Plan A charges $20 plus $8 per gigabyte, so A(g) = 8g + 20. Plan B charges $35 plus $5 per gigabyte, so B(g) = 5g + 35. Which is cheaper? Try 3 GB: A(3) = 44 and B(3) = 50, so A wins. Try 8 GB: A(8) = 84 and B(8) = 75, so B wins. Somewhere between 3 and 8 the plans must cost the same. That crossing is the break-even point.

To find it exactly, set the outputs equal: 8g + 20 = 5g + 35. Subtract 5g: 3g + 20 = 35. Subtract 20: 3g = 15. Divide: g = 5. Check: A(5) = 60 and B(5) = 60. Both plans cost $60 at 5 GB. On a graph, this is the intersection point (5, 60) where the two lines cross. Below 5 GB the lower-intercept plan A is cheaper; above 5 GB the lower-slope plan B is cheaper.

This is a good comparison habit: the plan with the smaller starting cost wins for small inputs, and the plan with the smaller rate wins for large inputs. The crossing point tells you where the advantage flips. Setting two rules equal and solving is the door into the next chapter, where two equations are solved together as a system.

A single equation with two variables, like c + b = 120, has endless solutions: 0 and 120, 1 and 119, 60 and 60. It cannot tell you how many cookies were sold. A second fact, c + 2b = 170, also has endless solutions on its own. But only one pair fits both. Two equations that must hold at the same time form a system of equations, and the pair that makes both true is the solution of the system.

Test a candidate pair the same way you test a solution of one equation, but check every equation. Is (70, 50), meaning c = 70 and b = 50, the solution? First equation: 70 + 50 = 120, true. Second: 70 + 2(50) = 170, true. So yes. Is (60, 60) the solution? First: 60 + 60 = 120, true. Second: 60 + 120 = 180, not 170. It fails one equation, so it is not the solution. A pair must pass every test.

You can find a solution by guess and check in a table. Try b = 40: then c = 80 and the money is 80 + 80 = 160, too low. Try b = 50: c = 70 and the money is 70 + 100 = 170, right on. Each extra brownie adds $1, so the table climbs by 1 each row, and you can see exactly where it lands. The rest of this chapter replaces the table with faster methods, but the meaning stays the same: find the pair that satisfies both.

Each equation in a system is a line, and every point on a line satisfies that equation. The solution of the system satisfies both, so it must lie on both lines. That is the intersection, the point where they cross. Take y = 2x + 1 and y = −x + 7. The first has y-intercept 1 and rises 2 for each step right. The second starts at 7 and falls 1 per step. Plot each line from its intercept and slope, and they cross at (2, 5).

Confirm by substituting: 2(2) + 1 = 5 and −2 + 7 = 5. Both equations give y = 5 at x = 2, so (2, 5) is the solution. When an equation is not in slope-intercept form, rewrite it first. For c + b = 120, solve for b: b = −c + 120. For c + 2b = 170, solve for b: 2b = −c + 170, so b = −0.5c + 85. The lines cross at c = 70, b = 50, the bake-sale answer.

Graphing shows the answer, but a hand-drawn graph is an estimate. If the lines cross at (2.4, 5.8), reading that from grid paper is hard. Use the graph to see roughly where the answer is and whether it makes sense, then use algebra to get it exactly. If two lines look parallel, expect no solution; if they lie on top of each other, every point is a solution.

Solve y = 2x + 1 and 3x + y = 16. The first equation already says what y is. Replace y in the second equation with 2x + 1: 3x + (2x + 1) = 16. Now there is one variable. Combine: 5x + 1 = 16. Subtract 1: 5x = 15. Divide: x = 3. Then find y from the first equation: y = 2(3) + 1 = 7. The solution is (3, 7). This method is substitution, and it works best when one equation is already solved for a variable or easily can be.

Check in both equations: 7 = 2(3) + 1 is true, and 3(3) + 7 = 16 is true. For the bake sale, c + b = 120 gives c = 120 − b. Substitute into c + 2b = 170: (120 − b) + 2b = 170, so 120 + b = 170 and b = 50. Then c = 120 − 50 = 70. The algebra is Priya’s ’extra $50’ reasoning written out.

Two errors are common. The first is substituting into the same equation you took the expression from, which gives a useless 0 = 0. Always substitute into the other one. The second is forgetting parentheses: in 4x − (x + 3) = 9, the minus must apply to the whole x + 3, giving 4x − x − 3 = 9, so 3x = 12 and x = 4. Writing 4x − x + 3 = 9 by mistake gives x = 2, and the check fails.

Take x + y = 10 and x − y = 4. The y terms are opposites: +y and −y. Add the two equations, left side to left side and right side to right side: (x + y) + (x − y) = 10 + 4, which is 2x = 14, so x = 7. The y vanished. Substitute back: 7 + y = 10, so y = 3. Solution (7, 3). Check the second: 7 − 3 = 4. This method is elimination.

Why is adding equations allowed? The second equation says x − y and 4 are the same number. Adding the same number to both sides of the first equation keeps it true; you just wrote that number two different ways. This is the addition property of equality from the first chapter, applied to a whole equation. If the matching terms are equal rather than opposite, subtract instead. From 3x + 2y = 19 and x + 2y = 9, subtracting gives 2x = 10, so x = 5 and then 2y = 4, y = 2.

Sign errors are the usual trouble. When subtracting, subtract every term of the second equation, including its right side. And after finding one variable, find the other by substituting into an original equation, not into the combined one. Finally, check the pair in both originals: 3(5) + 2(2) = 19 and 5 + 4 = 9. Both true, so (5, 2) is right.

Solve 2x + 3y = 12 and 4x − y = 10. Nothing cancels yet: adding gives 6x + 2y = 22, still two variables. But multiplying the whole second equation by 3 gives 12x − 3y = 30, and now the y terms are opposites of the 3y in the first equation. Multiplying both sides by 3 is the multiplication property of equality, so the new equation is equivalent to the old one. Add: (2x + 3y) + (12x − 3y) = 12 + 30, so 14x = 42 and x = 3.

Substitute into an original: 4(3) − y = 10, so y = 2. Check the first: 2(3) + 3(2) = 6 + 6 = 12. Solution (3, 2). Sometimes both equations need multiplying. For 3x + 2y = 7 and 2x + 5y = 12, multiply the first by 2 and the second by −3 to get 6x + 4y = 14 and −6x − 15y = −36. Add: −11y = −22, so y = 2. Then 3x + 4 = 7 gives x = 1. Check the second: 2 + 10 = 12.

Choose the multiplier by finding a common multiple of the two coefficients you want to cancel, and give one of them a negative sign. The classic slip is multiplying only the left side of an equation. Every term, including the constant on the right, gets multiplied. Another is forgetting the sign change, which makes the terms add instead of cancel. When the combined equation still has two variables, look back for one of those two mistakes.

Consider y = 2x + 1 and y = 2x − 3. Both have slope 2, so they are parallel lines that never cross. Try substitution anyway: 2x + 1 = 2x − 3. Subtract 2x from both sides: 1 = −3. That is false no matter what x is. When the variables vanish and leave a false statement, the system has no solution. No pair of numbers can be on both lines.

Now consider 2x + 4y = 8 and x + 2y = 4. Multiply the second by 2: 2x + 4y = 8, identical to the first. They are the same line written two ways, so every point on the line works: (4, 0), (0, 2), (2, 1) and infinitely many more. Elimination shows it: subtract the doubled second equation from the first and you get 0 = 0, a statement that is always true. That signals infinitely many solutions.

So there are three outcomes. If solving gives a specific x, the lines cross once and there is one solution. If the variables cancel and leave something false like 1 = −3, there is no solution. If they cancel and leave something true like 0 = 0, there are infinitely many. In a story, no solution might mean two plans never cost the same, and infinitely many might mean two ’different’ facts were really the same fact. A quick slope check predicts the outcome before you solve.

The equation y = 2x + 1 is a line. The inequality y ≤ 2x + 1 is a whole region: every point on the line or below it. To graph it, first draw the boundary line y = 2x + 1. Because the inequality includes ’equal to’, draw it solid. Then pick a test point not on the line, usually (0, 0): is 0 ≤ 2(0) + 1? Yes, 0 ≤ 1 is true. So the side containing (0, 0), the region below the line, gets shaded. That shaded area is the half-plane of solutions.

For y > −x + 3, the boundary y = −x + 3 is drawn dashed, because points on the line make the inequality false; only ’greater than’ counts. Test (0, 0): 0 > 3 is false, so shade the other side, above the line. Any point in the shaded area satisfies the inequality. Check one: (4, 2) gives 2 > −4 + 3, that is 2 > −1, true.

Rewrite first if needed. For 2x + y ≥ 6, solve for y: y ≥ −2x + 6, solid line, and (0, 0) gives 0 ≥ 6, false, so shade above. Two habits prevent most errors: always test a point rather than guessing ’above for greater,’ because the rule flips if you divide by a negative when solving for y, and always match the line style to the symbol, dashed for < and >, solid for ≤ and ≥.

Priya’s bake sale has limits. The table holds at most 150 items: c + b ≤ 150. Ingredients cost $0.30 per cookie and $0.60 per brownie, with $60 to spend: 0.30c + 0.60b ≤ 60. And you cannot bake a negative number: c ≥ 0 and b ≥ 0. Together these form a system of inequalities. A pair (c, b) is allowed only if it satisfies all four. Graphing each one as a shaded half-plane and keeping the overlap gives the feasible region.

Find the region’s edges by turning each ≤ into =. The line c + b = 150 runs from (150, 0) to (0, 150). The budget line 0.30c + 0.60b = 60 runs from (200, 0) to (0, 100). Where do they cross? Multiply the budget equation by 10: 3c + 6b = 600. Multiply the table equation by 3: 3c + 3b = 450. Subtract: 3b = 150, so b = 50 and c = 100. Check the budget: 0.30(100) + 0.60(50) = 30 + 30 = 60.

The feasible region is the four-sided shape with vertices at (0, 0), (150, 0), (100, 50) and (0, 100). Any point inside works: (60, 40) has 100 items and costs 18 + 24 = $42. Any point outside breaks a rule: (120, 60) has 180 items, too many for the table. The vertices are where two limits are both used up at once, and, as the next lesson shows, that is where the best plans live.

A juice stand has a 10% juice drink and a 40% juice drink. It wants 30 liters of a 20% mix. How much of each? Let x be liters of the 10% drink and y be liters of the 40% drink. The amount equation counts liters: x + y = 30. The concentration equation counts pure juice: 0.10x + 0.40y is the juice in the mix, and it must equal 20% of 30 liters, which is 0.20 × 30 = 6. So 0.10x + 0.40y = 6.

Solve by substitution. From the first, y = 30 − x. Substitute: 0.10x + 0.40(30 − x) = 6. Distribute: 0.10x + 12 − 0.40x = 6. Combine: −0.30x + 12 = 6. Subtract 12: −0.30x = −6. Divide: x = 20. Then y = 10. Check: 20 + 10 = 30 liters, and 0.10(20) + 0.40(10) = 2 + 4 = 6 liters of juice, which is 20% of 30. Twenty liters of weak and ten of strong.

Coin and ticket problems are mixture problems too. A jar holds 40 coins, all nickels and dimes, worth $3.10. Count: n + d = 40. Value: 0.05n + 0.10d = 3.10. Substitute n = 40 − d: 0.05(40 − d) + 0.10d = 3.10, so 2 − 0.05d + 0.10d = 3.10, so 0.05d = 1.10 and d = 22, n = 18. Check: 18(0.05) + 22(0.10) = 0.90 + 2.20 = 3.10. The common mistake is writing the value equation with the counts instead of the values per item.

A student startup prints T-shirts. Equipment costs $200 and each shirt costs $3 to make, so cost is C = 200 + 3x for x shirts. Each shirt sells for $8, so revenue is R = 8x. The break-even point is where R = C: 8x = 200 + 3x. Subtract 3x: 5x = 200. So x = 40. At 40 shirts, both cost and revenue are $320. Selling fewer than 40 loses money; selling more earns profit, which is R − C = 5x − 200. At 100 shirts, profit is 500 − 200 = $300.

Ticket sales are a two-equation problem. A school play sold 200 tickets for $1,520. Adult tickets were $10 and student tickets $6. Let a and s be the counts. Count: a + s = 200. Money: 10a + 6s = 1520. Substitute s = 200 − a: 10a + 6(200 − a) = 1520, so 10a + 1200 − 6a = 1520, so 4a = 320 and a = 80. Then s = 120. Check: 80 + 120 = 200 and 800 + 720 = 1,520.

Elimination works too: multiply the count equation by 6 to get 6a + 6s = 1200 and subtract it from the money equation to get 4a = 320. Same answer. Whatever method you use, label the answer in words: 80 adult tickets and 120 student tickets. A bare pair like (80, 120) is easy to reverse, and reversed, it fails the money check: 800 + 480 is only $1,280.

Priya wants the most profit: $0.70 per cookie and $1.50 per brownie, so the goal is P = 0.70c + 1.50b. This is her objective. The constraints are the table, c + b ≤ 150; the budget, 0.30c + 0.60b ≤ 60; the advisor’s rule, c ≥ 40; and b ≥ 0. Choosing the best point in a feasible region for a linear goal is called linear programming, and its key fact is simple: the best value always occurs at a vertex of the region.

So list the vertices. Two of them come from c = 40. On the c-axis it is (40, 0). On the budget line, 0.30(40) + 0.60b = 60 gives 12 + 0.60b = 60, so b = 80: the point (40, 80). The other corners are (150, 0), where the table is full of cookies, and (100, 50), where the table and budget lines cross. Now evaluate the objective at each: (40, 0) gives 28; (150, 0) gives 105; (100, 50) gives 70 + 75 = 145; (40, 80) gives 28 + 120 = 148.

The optimal plan is 40 cookies and 80 brownies, for $148 profit. Check it against every constraint: 120 items fits the table, 12 + 48 = $60 uses the whole budget, and 40 cookies satisfies the advisor. Notice that the winner is not the corner with the most items; it uses the budget fully on the higher-profit item while meeting the cookie rule. If the constraints change, the region changes, the vertices move, and the best plan can change with them. That is why businesses redo this calculation whenever prices shift.