Unit 15 · Functions and Linear Equations

Picture one step of a staircase. The step goes up a certain height and forward a certain distance. The height it goes up is the rise. The distance it goes forward is the run. The slope is the rise divided by the run. A step that rises 7 inches and runs 11 inches has a slope of 7/11. If you write that as a decimal, 7 ÷ 11 is about 0.64.

Try a wheelchair ramp. A ramp rises 2 feet from the sidewalk to a door, and it stretches 24 feet along the ground. Slope = rise ÷ run = 2/24. Simplify by dividing the top and bottom by 2: 2/24 = 1/12. That is the gentlest slope most building rules allow for a ramp, so this ramp passes. A ramp that rose 2 feet over only 12 feet would have slope 2/12 = 1/6, twice as steep, and it would be hard to push a chair up.

Slope lets you compare steepness even when the numbers look different. Hill A rises 30 feet over 200 feet: 30/200 = 0.15. Hill B rises 12 feet over 60 feet: 12/60 = 0.2. Hill B is steeper, even though it rises fewer feet, because it does its rising in a shorter distance. To compare slopes, turn each fraction into a decimal or give them the same denominator.

The most common mistake is flipping the fraction and putting run over rise. A ramp with rise 2 and run 24 has slope 2/24, not 24/2 = 12. Remember that a gentle ramp should have a small slope number. If your answer for a gentle ramp is a big number, you flipped it. A second mistake is mixing units, like rise in inches and run in feet. Change both to the same unit before you divide.

On a graph, every point is an ordered pair (x, y). The x tells how far right, and the y tells how far up. To find the slope of the line through two points, find the rise by subtracting the y-values and the run by subtracting the x-values. Take the points (2, 3) and (6, 11). Rise = 11 − 3 = 8. Run = 6 − 2 = 4. Slope = 8/4 = 2. The line goes up 2 for every 1 it goes right.

The order matters, but only in one way: subtract in the same order on top and bottom. With (2, 3) and (6, 11), you could also do rise = 3 − 11 = −8 and run = 2 − 6 = −4. Then slope = −8/−4 = 2. Same answer. The mistake is mixing the orders, like 11 − 3 on top and 2 − 6 on the bottom, which gives 8/−4 = −2 and the wrong sign.

Lines can go downhill too. Take (1, 9) and (4, 3). Rise = 3 − 9 = −6. Run = 4 − 1 = 3. Slope = −6/3 = −2. A negative slope means the line drops as you move right: down 2 for every 1 to the right. On a graph of a car’s gas tank over time, the slope is negative because the gas goes down as time goes up.

Here is the rule in general. For points (x1, y1) and (x2, y2), slope = (y2 − y1) ÷ (x2 − x1). Say it as change in y over change in x. One more check: the run is always the x part. A student who writes (x2 − x1) on top has flipped the fraction and will get the wrong slope.

Draw a line through (0, 0), (2, 3), (4, 6) and (6, 9). Now draw a right triangle under the line from (0, 0) to (2, 3): its legs are the run 2 and the rise 3. This is a slope triangle. Draw a bigger one from (0, 0) to (6, 9): run 6 and rise 9. The big triangle is just the small one enlarged by 3. Its slope is 9/6, and 9/6 simplifies to 3/2, the same as the small triangle’s 3/2.

Triangles that are enlargements of each other are similar triangles. They have the same angles, and their matching sides are in the same ratio. Every slope triangle on one line has the same angle at the bottom, because the line always tilts the same way. So all the slope triangles are similar, and rise ÷ run comes out the same for every one of them. This is why a straight line has exactly one slope.

You can use this fact to test whether points are on one line. Take (1, 1), (3, 5) and (6, 11). From the first to the second: rise 4, run 2, slope 2. From the second to the third: rise 6, run 3, slope 2. Same slope, so the three points sit on one line. Now try (0, 1), (2, 5) and (5, 10). First pair: 4/2 = 2. Second pair: 5/3. Different slopes, so these three points do not line up.

A common mistake is to think a bigger slope triangle means a bigger slope. The triangle from (0, 0) to (6, 9) is three times as large as the one to (2, 3), but its slope is still 3/2. Slope is a ratio, and a ratio does not change when you scale both parts. Another mistake is checking only the y-values. Points whose y-values grow by the same amount are on a line only if their x-values also grow by the same amount.

Dev works at a car wash. After 2 hours he has earned $30. After 5 hours he has earned $75. How much does he earn per hour? The pay went up 75 − 30 = 45 dollars while the time went up 5 − 2 = 3 hours. So the rate of change is 45 ÷ 3 = 15 dollars per hour. That is exactly the slope between the points (2, 30) and (5, 75): rise 45, run 3, slope 15.

The rate of change always carries units. Here it is dollars per hour. Try a water tank. After 3 minutes it holds 40 gallons. After 8 minutes it holds 90 gallons. Change in gallons: 90 − 40 = 50. Change in minutes: 8 − 3 = 5. Rate = 50 ÷ 5 = 10 gallons per minute. On a graph with minutes on the x-axis and gallons on the y-axis, the slope of the line is 10.

Notice what the rate does not tell you. It does not say how much water was in the tank at the start. Work backward: at 3 minutes there were 40 gallons, and the tank gains 10 per minute, so at 0 minutes there were 40 − 30 = 10 gallons. The rate tells the speed of change, and the starting amount is a separate fact. You will need both to write the equation of a line in the next lesson.

Watch for the mistake of dividing the wrong way. A student who computes 3 ÷ 45 for Dev’s pay gets 0.067 hours per dollar, which is a real number but not the question asked. Always divide the change in the output (pay, gallons) by the change in the input (hours, minutes). Another mistake is dividing the totals, like 75 ÷ 5 = 15. That happens to work here only because Dev started at $0. It fails for the tank: 90 ÷ 8 is 11.25, not 10.

A taxi charges $3 just to get in, then $2 for every mile. The cost y for x miles is y = 2x + 3. Try 4 miles: y = 2 × 4 + 3 = 8 + 3 = 11 dollars. Try 0 miles: y = 2 × 0 + 3 = 3 dollars, the starting fee. The number multiplied by x, here 2, is the slope: cost goes up $2 for each mile. The number added at the end, here 3, is the y-intercept: the value of y when x is 0, where the line crosses the y-axis.

The general form is y = mx + b. It is called slope-intercept form because m is the slope and b is the y-intercept. The m is a coefficient, a number multiplied by a variable. So y = 5x + 2 has slope 5 and y-intercept 2. A negative slope works the same way: y = −3x + 10 has slope −3 and y-intercept 10. Put in x = 2: y = −3 × 2 + 10 = −6 + 10 = 4.

The order of the terms can trick you. The equation y = 7 + 4x has slope 4, not 7, because 4 is the number attached to x. The 7 is added alone, so it is the y-intercept. Rewrite it as y = 4x + 7 if that helps. Likewise y = x − 6 has slope 1, because x means 1x, and its y-intercept is −6.

A gym charges $25 to join and $10 each month. Write the equation: the rate is 10 per month, and the starting amount is 25, so y = 10x + 25, where x is the number of months. After 6 months: y = 10 × 6 + 25 = 60 + 25 = 85 dollars. The mistake to avoid is swapping m and b: y = 25x + 10 would mean $25 every month, which is a very different gym.

To graph y = (2/3)x + 1, start at the y-intercept. The y-intercept is 1, so plot the point (0, 1) on the y-axis. Next use the slope 2/3 as directions: rise 2, run 3. From (0, 1), go right 3 and up 2 to reach (3, 3). Do it again to reach (6, 5). Draw a straight line through the points. Check one: at x = 3, y = (2/3) × 3 + 1 = 2 + 1 = 3. Correct.

A whole-number slope is a fraction with 1 on the bottom. For y = 4x − 2, the slope is 4/1: rise 4, run 1. Start at (0, −2), go right 1 and up 4 to (1, 2), then to (2, 6). A negative slope means go down instead of up. For y = −x + 4, the slope is −1, so rise −1, run 1. Start at (0, 4), go right 1 and down 1 to (1, 3), then (2, 2), then (3, 1), and at (4, 0) the line crosses the x-axis.

Now try y = −(1/2)x + 5. Start at (0, 5). The slope −1/2 means right 2, down 1. That gives (2, 4), then (4, 3), (6, 2), (8, 1) and (10, 0). Check with the equation: at x = 10, y = −(1/2) × 10 + 5 = −5 + 5 = 0. So the line hits the x-axis at (10, 0).

Two mistakes come up often. First, starting on the x-axis: the y-intercept 1 is the point (0, 1), not (1, 0). Second, reading the slope upside down: for 2/3, go up 2 and right 3, not up 3 and right 2. If you go the wrong way, the line will be steeper or flatter than it should be. Always test one of your points in the equation before you draw the line.

A line is drawn on a grid. To write its equation, find two things: where it crosses the y-axis and how steep it is. Say the line crosses the y-axis at 2 and also passes through (4, 5). The y-intercept is b = 2. For the slope, draw a slope triangle from (0, 2) to (4, 5): rise 5 − 2 = 3, run 4 − 0 = 4, slope 3/4. The equation is y = (3/4)x + 2.

Choose points where the line passes exactly through a grid corner. Such a point is called a lattice point, and it gives whole-number coordinates you can trust. A point where the line passes between grid corners is a guess, and a guess makes the slope wrong. Two lattice points far apart give the most accurate slope.

Now a line that falls. It crosses the y-axis at 6 and crosses the x-axis at 3. The place it crosses the x-axis is the x-intercept, the point (3, 0). Slope from (0, 6) to (3, 0): rise 0 − 6 = −6, run 3 − 0 = 3, slope −2. Equation: y = −2x + 6. Check with the x-intercept: y = −2 × 3 + 6 = 0. Correct.

The most common error is using the x-intercept as b. In the example above, a student might write y = −2x + 3 because the line hits the x-axis at 3. Test it: at x = 0 that equation gives y = 3, but the line is at 6 when x = 0. The b in y = mx + b is always where the line crosses the y-axis. Always test your equation with a point you can see on the graph.

A table shows x = 1, 2, 3 and y = 7, 10, 13. Each time x goes up by 1, y goes up by 3, so the slope is 3. But the table does not show x = 0, so you cannot read b directly. Take any row and substitute it into y = 3x + b. Using (1, 7): 7 = 3 × 1 + b, so 7 = 3 + b and b = 4. The equation is y = 3x + 4. Check another row: 3 × 3 + 4 = 13. Correct.

Two points work the same way. Take (2, 1) and (6, 9). Slope = (9 − 1) ÷ (6 − 2) = 8/4 = 2. Now solve for b using (2, 1): 1 = 2 × 2 + b, so 1 = 4 + b and b = 1 − 4 = −3. The equation is y = 2x − 3. Check with the other point: 2 × 6 − 3 = 12 − 3 = 9. It matches, so the equation is right.

Here is the whole method in three steps. One: find the slope from two points. Two: put the slope and one point into y = mx + b and solve for b. Three: check with the other point. If the check fails, the slope or b is wrong, and it is worth redoing both.

The trap in a table is grabbing the first y-value as b. In the table above, the first row is (1, 7), and 7 is not the y-intercept, because x is 1 there, not 0. The y-intercept is 4. The only time you can read b straight from a table is when a row has x = 0. Otherwise, substitute and solve.

Two jobs pay $15 an hour. The first pays only that: y = 15x. The second also gives a $20 bonus on day one: y = 15x + 20. Both graphs are lines with slope 15. But only the first goes through the origin, the point (0, 0). At 0 hours the first job pays $0 and the second pays $20. The first is a proportional relationship; the second is linear but not proportional.

In a proportional relationship, y is always the same multiple of x. The equation is y = kx, and k is the constant of proportionality. It is also the slope, and the y-intercept is 0. To test a table, divide each y by its x. Try x = 2, 4, 6 with y = 10, 20, 30: 10/2 = 5, 20/4 = 5, 30/6 = 5. Same every time, so it is proportional with k = 5, and y = 5x.

Now test x = 2, 4 with y = 12, 18. Divide: 12/2 = 6 and 18/4 = 4.5. Not the same, so it is not proportional. Is it still a line? From (2, 12) to (4, 18) the slope is 6/2 = 3. Solve for b: 12 = 3 × 2 + b, so b = 6. The equation is y = 3x + 6. It is a line, but it does not pass through the origin, so it is not proportional.

Students often say a table is proportional because y goes up by the same amount each time. That only shows the relationship is linear. Proportional needs more: y ÷ x must be the same for every row, which is the same as saying the line passes through (0, 0). Every proportional relationship is linear, but most linear relationships are not proportional.

A thermostat holds a room at 68°F for five hours. Graph the temperature against time: every point has y = 68, so the graph is a flat, horizontal line. Its equation is simply y = 68. Find its slope from (1, 68) and (4, 68): rise 68 − 68 = 0, run 4 − 1 = 3, slope 0/3 = 0. A horizontal line has slope 0 because the rise is always 0.

Now think of a wall standing straight up. Every point on it has the same x-value. A vertical line through (2, 1) and (2, 7) has the equation x = 2. Try the slope: rise 7 − 1 = 6, run 2 − 2 = 0, slope 6/0. You cannot divide by 0, so the slope is undefined. That is not the same as slope 0. A slope of 0 means flat, and an undefined slope means straight up and down.

Any line through (1, 5) and (4, 5) is horizontal, because the y-values match. Its equation is y = 5. Any line through (−3, 2) and (−3, 9) is vertical, because the x-values match. Its equation is x = −3. Look at which coordinate stays the same. If y stays the same, the line is horizontal. If x stays the same, the line is vertical.

The common mix-up is writing y = 2 for a vertical line at x = 2. Ask which axis the line is parallel to. A horizontal line runs parallel to the x-axis, but its equation names y, because y is the value that never changes. A vertical line runs parallel to the y-axis, and its equation names x. Also note that y = 68 fits y = mx + b with m = 0: y = 0x + 68. A vertical line does not fit that form at all, because there is no y in its equation.

Press B4 on a vending machine and one item drops out, the same item every time. That machine is working like a function. A function is a rule that takes an input and gives back exactly one output. The input is what you put in, like the button. The output is what comes out, like the snack. If B4 sometimes gave chips and sometimes gave pretzels, the machine would not be a function.

Most functions in math are number rules. Take the rule output = 2 × input + 1. Put in 3: 2 × 3 + 1 = 7. Put in 10: 2 × 10 + 1 = 21. Put in 0: 2 × 0 + 1 = 1. Each input gives one and only one output. We usually call the input x and the output y, so this rule is y = 2x + 1. A table of inputs and outputs is one way to show it.

Try a rule with subtraction: output = 3 × input − 4. Input 1: 3 − 4 = −1. Input 2: 6 − 4 = 2. Input 5: 15 − 4 = 11. Input 10: 30 − 4 = 26. You can also run a function backward with one known output. If the output is 20, then 3 × input − 4 = 20, so 3 × input = 24 and the input is 8. Check: 3 × 8 − 4 = 24 − 4 = 20.

Functions do not have to use numbers. The rule that sends each student to their birthday is a function, because every student has exactly one birthday. The rule that sends each birthday to the students born on that day is not, because one date can belong to two students, and then the input has more than one output. The test is always the same: does each input have exactly one output?

A list of pairs shows inputs and outputs: (1, 3), (2, 5), (1, 7). Look for an input that repeats, one that appears more than once. The input 1 appears twice, once with output 3 and once with output 7. One input, two outputs: this is not a function. Now look at (1, 4), (2, 4), (3, 4). The output 4 repeats, but that is fine. Each input still has exactly one output, so this is a function.

That is the key difference. Repeated outputs are allowed. Repeated inputs with different outputs are not. A list like (2, 5), (3, 5), (4, 5) is a function even though the output never changes. A list like (2, 5), (2, 6) is not, because the input 2 cannot decide between 5 and 6.

On a graph, use the vertical line test. Imagine sliding a vertical line across the graph from left to right. Each vertical line marks one input, the x-value. If the line ever crosses the graph at two points, that input has two outputs, and the graph is not a function. A straight line that is not vertical always passes the test. A circle fails: a vertical line through its middle crosses it at the top and at the bottom.

Students sometimes run the test the wrong way, with a horizontal line. A horizontal line crossing twice only means two inputs share an output, which is allowed. Keep the line vertical. Students also sometimes think a table with a repeated output is not a function. Check the inputs, not the outputs.

A pool starts with 500 gallons and loses 3 gallons every hour through a leak. That sentence is the function in words. Turn it into an equation: let x be the hours and y the gallons. Each hour takes away 3, so y = 500 − 3x. Now make a table of values. At x = 0, y = 500. At x = 1, y = 497. At x = 2, y = 494. At x = 10, y = 500 − 30 = 470.

Plot those table rows as points, (0, 500), (1, 497), (2, 494), (10, 470), and connect them: that is the graph. It is a straight line that starts at 500 on the y-axis and falls 3 for every hour. Words, equation, table and graph are four views of one function. Each view is good for something. The equation gives any value fast. The table shows a few exact values. The graph shows the shape at a glance.

You can go in any direction. From a table with x = 0, 1, 2 and y = 9, 7, 5, find the rule. The y-value at x = 0 is 9, so the starting amount is 9. Each step down is 2, so the rule is y = 9 − 2x, which is the same as y = −2x + 9. From words like start with 12 and add 5 each week, write y = 5x + 12, where x counts the weeks.

A common slip is reading the first row of a table as the starting amount when that row is not x = 0. If a table begins at x = 1, the starting amount is one step earlier. Another slip is mixing the two numbers in the words: start with 12 and add 5 each week is y = 5x + 12, not y = 12x + 5. The number that repeats every week is the one that multiplies x.

A table shows x = 0, 1, 2, 3 and y = 4, 7, 10, 13. Look at how y changes as x goes up by 1: 7 − 4 = 3, 10 − 7 = 3, 13 − 10 = 3. The change is always 3. A function that changes by the same amount for each equal step in x has a constant rate, and it is a linear function. Its graph is a straight line, and its equation can be written as y = mx + b.

For this table, m is the constant rate, 3, and b is the value at x = 0, which is 4. So y = 3x + 4. Check: at x = 3, 3 × 3 + 4 = 9 + 4 = 13. It matches. At x = 10, the equation gives 3 × 10 + 4 = 34, which the table would reach if it kept going.

You can spot a linear function in any form. In words: a fixed amount added or taken away every unit of time, like 3 gallons every hour. In a table: equal steps in x give equal steps in y. In an equation: x appears only multiplied by a number and maybe added to a number, like y = 6x − 1 or y = 500 − 3x. In a graph: a straight line.

The classic mistake is deciding a function is linear because y keeps going up. Going up is not enough. It must go up by the same amount every step. The table x = 0, 1, 2, 3 and y = 1, 2, 4, 8 goes up every time, but the steps are 1, 2 and 4. That is not a constant rate, so it is not linear. Always compute the differences before you decide.

The area of a square with side x is y = x^2, which means x times x. Make a table: x = 0, 1, 2, 3, 4 gives y = 0, 1, 4, 9, 16. Now find the changes: 1 − 0 = 1, 4 − 1 = 3, 9 − 4 = 5, 16 − 9 = 7. The steps are 1, 3, 5, 7, growing each time. The rate is not constant, so this is a nonlinear function. Plot the points and you get a curve that bends upward, not a straight line.

Doubling is another nonlinear pattern. A rule where each output is twice the one before gives x = 1, 2, 3, 4 and y = 2, 4, 8, 16. The changes are 2, 4 and 8. Again the steps grow. The equation is y = 2^x, 2 to the power x. It grows faster and faster, which is why doubling patterns get huge so quickly.

Nonlinear functions are still functions. Each input has exactly one output: the square of 6 is 36 and only 36. What makes them different from linear functions is only the rate. A linear function adds the same amount each step. A nonlinear function does not. In a graph, that shows up as bending. In an equation, it shows up as x being squared, used as an exponent, or placed under a fraction bar.

A common mistake is calling y = x^2 linear because its table goes up steadily and its equation looks simple. Compute the differences and the bending shows: 1, 3, 5, 7. Another mistake is using the linear method on a nonlinear table. If someone takes the first step of y = x^2, which is 1, and predicts y = 1x, they will get 6 for x = 6 instead of 36.

A candle is 20 cm tall and burns down 2 cm every hour. Two numbers describe this function. The initial value is the height at the start, when 0 hours have passed: 20 cm. The rate of change is −2 cm per hour, negative because the height goes down. The equation is y = 20 − 2x, or y = −2x + 20. After 4 hours: 20 − 8 = 12 cm. The candle is gone when y = 0, which is 20 ÷ 2 = 10 hours.

From an equation, read the two numbers straight off. In y = 12x + 30, the rate of change is 12 and the initial value is 30. That could be a gym that costs $30 to join and $12 a month. From a graph, the initial value is where the line crosses the y-axis, and the rate is the slope.

From a table without x = 0, do a little work. Say x = 2 gives y = 19 and x = 5 gives y = 31. Rate = (31 − 19) ÷ (5 − 2) = 12 ÷ 3 = 4 per unit. To find the initial value, go back from x = 2 to x = 0, which is 2 steps of 4, or 8: 19 − 8 = 11. So the initial value is 11 and the equation is y = 4x + 11. Check: 4 × 5 + 11 = 31.

Keep the two numbers separate in your head. The rate answers how fast it changes. The initial value answers where it starts. A common slip is reporting the first number in the table as the initial value, like 19 above, when x is not 0 in that row. Another is dropping the negative sign on a decreasing rate: the candle’s rate is −2, and writing +2 would make it grow.

Two text plans are on offer. Plan A comes as an equation: y = 0.10x + 20, where x is the number of texts and y the monthly cost in dollars. Plan B comes as a table: 0 texts cost $15, 100 texts cost $27, 200 texts cost $39. To make a fair comparison, get both plans into the same terms: a rate and an initial value.

Plan A’s rate is 0.10 dollars per text and its initial value is $20. For Plan B, the rate is (27 − 15) ÷ (100 − 0) = 12 ÷ 100 = 0.12 dollars per text, and the initial value is $15, since the table shows x = 0. So B starts cheaper but grows faster. Its graph is steeper. At 100 texts, A costs 10 + 20 = 30 and B costs 27, so B is cheaper. At 300 texts, A costs 30 + 20 = 50 and B costs 15 + 36 = 51, so A is cheaper.

The same idea works for any pair. Function A: y = 5x + 2. Function B is a table: x = 0, 1, 2 and y = 10, 13, 16. B’s rate is 3 and its initial value is 10. A’s rate is 5 and its initial value is 2. A grows faster because 5 is greater than 3, even though B starts higher. At x = 4, A gives 5 × 4 + 2 = 22 and B gives 10 + 3 × 4 = 22. That is the point where they meet.

Do not compare the wrong numbers. B starts at 10 and A starts at 2, but that says nothing about which grows faster. Growth is the rate. Also, do not read a rate from a table without dividing by the change in x. Plan B’s cost rose 12 dollars, but over 100 texts, so the rate is 0.12 per text, not 12.

Ana walks to the store, shops, and jogs home. Graph her distance from home against time. She walks 0.5 mile in 10 minutes: the first segment rises from (0, 0) to (10, 0.5). She shops for 15 minutes: the distance stays at 0.5, so the segment is flat from (10, 0.5) to (25, 0.5). She jogs home in 5 minutes: the last segment falls from (25, 0.5) to (30, 0).

The slopes tell the speeds. Walking: 0.5 mile ÷ 10 minutes = 0.05 mile per minute. Jogging home: 0.5 ÷ 5 = 0.1 mile per minute, twice as fast. So the jog-home segment is twice as steep as the walk-there segment, and it goes down because Ana is getting closer to home. The flat segment has slope 0: no motion.

A bathtub works the same way. Fill it for 8 minutes: the water level rises. Sit in it for 20 minutes: the level stays put, a flat segment. Pull the plug and it drains in 4 minutes: the level falls, and since 4 minutes is shorter than 8, the draining segment is steeper than the filling one. Three parts of the story, three segments: rise, flat, fall.

When sketching, ask three questions for each part of the story. Is the amount going up, going down or staying the same? Is it changing fast or slowly? How long does that part last? Going up means the segment rises, fast means steep, and the time sets how wide the segment is. A common mistake is drawing the return trip going up again. In a distance-from-home graph, coming home means the distance shrinks, so the segment falls.

Return to the bike ride graph. From 0 to 20 minutes the distance rises from 0 to 3 miles. From 20 to 30 minutes it rises from 3 to 5.5 miles. From 30 to 35 minutes it stays at 5.5. From 35 to 60 minutes it falls from 5.5 back to 0. Four segments, four parts of the story: a slow climb, a fast stretch, a pause, and the ride home.

Turn each segment into a speed. First: 3 miles in 20 minutes. There are 3 twenty-minute pieces in an hour, so that is 3 × 3 = 9 miles per hour. Second: 2.5 miles in 10 minutes, and there are 6 ten-minute pieces in an hour, so 2.5 × 6 = 15 miles per hour. Third: 0 miles in 5 minutes, speed 0. Fourth: 5.5 miles in 25 minutes. Divide 5.5 by 25 to get 0.22 mile per minute, and 0.22 × 60 = 13.2 miles per hour.

Now compare. The steepest rising segment is the second, at 15 miles per hour, so that was the fastest part of the ride out. The flat segment is the water stop. The last segment is decreasing, which means the riders were getting closer to the start. It was fairly steep too, 13.2 miles per hour, so they rode home almost as fast as their fastest stretch.

Two habits keep the story honest. First, read the axes before anything else. If the y-axis were speed instead of distance, a flat segment would mean steady speed, not a stop. Second, compare slopes, not heights. A segment that ends at a high number is not automatically the fast one. The 25-minute ride home covered the most miles, but the 10-minute stretch was faster.

Sometimes you do not need numbers, just the shape. Throw a ball straight up. Its height is increasing while it rises, then decreasing while it falls. The graph of height against time is a curve, so the function is nonlinear. The whole description is: nonlinear, increasing at first, then decreasing. Those few words tell most of what happens.

A savings account starts with $50 and gets $10 every week. The amount is always going up, and it goes up by the same $10 each week. So the function is linear and increasing. The graph is a straight line rising to the right. A phone battery that loses 1% every 4 minutes during a video is linear and decreasing: a straight line falling to the right.

The temperature on a summer day in Chicago might be 65°F at 6 am, rise to 84°F by 3 pm, and fall to 70°F by midnight. The function is increasing on the interval from 6 am to 3 pm and decreasing after 3 pm. It is nonlinear, because the temperature does not rise by the same amount every hour. It climbs quickly at midmorning and slowly near the peak.

Two things to watch. First, increasing and decreasing describe parts of a graph, so name the interval: increasing from 6 am to 3 pm, not just increasing. Second, linear and nonlinear describe the whole shape. A function can be nonlinear and still increasing everywhere, like y = 2^x. A function can be linear and decreasing, like the battery. Mix the words as the graph requires.

Two numbers add up to 10, and their difference is 4. Call the larger number x and the smaller y. The first sentence says x + y = 10. The second says x − y = 4. Together the two equations form a system of equations. Many pairs fit the first equation alone: (5, 5), (8, 2), (9, 1). Many fit the second alone: (6, 2), (8, 4), (10, 6). The solution of a system is a pair that fits both at the same time.

Test (8, 2). First equation: 8 + 2 = 10, true. Second: 8 − 2 = 6, not 4. So (8, 2) is not a solution. Test (7, 3). First: 7 + 3 = 10, true. Second: 7 − 3 = 4, true. Both equations are satisfied, so (7, 3) is the solution. The two numbers are 7 and 3.

On a graph, each equation is a line, and every point on that line makes its equation true. The solution of the system is the point where the two lines cross, because that is the only point on both lines. For x + y = 10 and x − y = 4, the lines cross at (7, 3). There is no other point on both lines, so there is no other solution.

To check a proposed solution, substitute it into both equations, not just one. Take y = x − 4 and y = 2x − 9 with the point (5, 1). First: 5 − 4 = 1, true. Second: 2 × 5 − 9 = 10 − 9 = 1, true. So (5, 1) is a solution. A point that passes one equation and fails the other is not a solution. Both checks must pass.

Solve the system y = 2x + 1 and y = −x + 7 by graphing. Graph the first line: start at (0, 1) and use slope 2 to reach (1, 3), (2, 5), (3, 7). Graph the second: start at (0, 7) and use slope −1 to reach (1, 6), (2, 5), (3, 4). The point (2, 5) is on both lists. That is the intersection, where the two lines cross, so the solution is (2, 5).

Always check the point of intersection in both equations. First: 2 × 2 + 1 = 5. Second: −2 + 7 = 5. Both give 5, so (2, 5) is exact. If a check fails, either the graph was drawn wrong or the crossing was read wrong. Redraw with care and look again.

Try y = x + 3 and y = 3x − 1. First line: (0, 3), (1, 4), (2, 5), (3, 6). Second line: (0, −1), (1, 2), (2, 5), (3, 8). They share (2, 5). Check: 2 + 3 = 5 and 3 × 2 − 1 = 5. Notice that the second line starts lower but climbs faster, so it catches the first, just like the hare catching the tortoise.

Graphing is quick and it shows the picture, but it has limits. It works best when the crossing lands on a lattice point with whole-number coordinates. If the lines cross at something like (1.4, 5.2), reading it off a grid gives only an estimate. The next lesson deals with that. Also watch for careless slopes: one line drawn with the wrong slope will cross in the wrong place, and the check will catch it.

Graph y = 3x + 1 and y = −2x + 8. The first line rises steeply from (0, 1). The second falls from (0, 8). They cross somewhere between x = 1 and x = 2, a little above y = 5. From the picture, a reasonable estimate is (1.5, 5). But is that exact? Check by substitution: 3 × 1.5 + 1 = 5.5, and −2 × 1.5 + 8 = 5. The two equations give different values, so (1.5, 5) is close but not the true crossing.

To get the exact point, set the two expressions for y equal, since y is the same at the crossing: 3x + 1 = −2x + 8. Add 2x to both sides: 5x + 1 = 8. Subtract 1: 5x = 7. Divide: x = 7/5 = 1.4. Then y = 3 × 1.4 + 1 = 4.2 + 1 = 5.2. Check in the second equation: −2 × 1.4 + 8 = −2.8 + 8 = 5.2. Both agree, so the exact solution is (1.4, 5.2).

The estimate was not wasted. It told us the answer should be near x = 1.5 and y = 5, so the algebra result of (1.4, 5.2) makes sense. If the algebra had given (14, 52), the graph would show that something went wrong. Use the graph to predict the size of the answer, and use algebra to nail it down.

Two mistakes are common here. One is trusting a graph reading as exact just because it looks right. Always substitute into both equations. The other is checking only one equation. The point (2, 5) fits y = 2x + 1 perfectly, but for y = −x + 6 it gives −2 + 6 = 4, not 5. A point must pass both checks to be a solution.

Solve y = 2x − 3 and 3x + y = 12. The first equation already has y by itself: y equals 2x − 3. So wherever y appears in the second equation, put 2x − 3 in its place. This is substitution. The second equation becomes 3x + (2x − 3) = 12. Combine: 5x − 3 = 12. Add 3: 5x = 15. Divide: x = 3. Then go back to the first equation: y = 2 × 3 − 3 = 3. The solution is (3, 3). Check: 3 × 3 + 3 = 12, true.

Substitution also works when x is the variable alone. Solve x = y + 4 and 2x + 3y = 23. Replace x in the second equation with (y + 4): 2(y + 4) + 3y = 23. Distribute the 2: 2y + 8 + 3y = 23. Combine: 5y + 8 = 23. Subtract 8: 5y = 15, so y = 3. Then x = 3 + 4 = 7. Check: 2 × 7 + 3 × 3 = 14 + 9 = 23, true. The solution is (7, 3).

If neither equation has a variable alone, isolate one first. In x + y = 20 and y = 3x, the second already gives y. Substitute: x + 3x = 20, so 4x = 20 and x = 5. Then y = 3 × 5 = 15. Check the first: 5 + 15 = 20. The solution is (5, 15). Substitution is the natural choice whenever one equation starts with y = or x =.

The most common error is forgetting the parentheses. In 2x + 3y = 23 with x = y + 4, writing 2y + 4 + 3y instead of 2(y + 4) + 3y loses the 8 and gives the wrong answer. Put the whole expression in parentheses, then distribute. Another error is stopping after finding x. A solution is a pair, so find y too, and check the pair in the equation you did not use to find it.

Solve x + y = 12 and x − y = 2. Look at the y terms: +y in the first and −y in the second. They are opposites. Add the two equations, left side to left side and right side to right side: (x + y) + (x − y) = 12 + 2. The y terms cancel, leaving 2x = 14, so x = 7. Put x = 7 into the first equation: 7 + y = 12, so y = 5. The solution is (7, 5). Check the second: 7 − 5 = 2, true. This method is elimination.

Adding works whenever a variable’s coefficients are opposites. Solve 3x + 2y = 16 and 5x − 2y = 0. The y coefficients are +2 and −2. Add: 8x = 16, so x = 2. Then 3 × 2 + 2y = 16, so 6 + 2y = 16, 2y = 10, y = 5. Check the second: 5 × 2 − 2 × 5 = 10 − 10 = 0, true. The solution is (2, 5).

When the coefficients are the same instead of opposite, subtract one equation from the other. Solve 4x + y = 14 and 2x + y = 8. Both have +y. Subtract the second from the first: (4x − 2x) + (y − y) = 14 − 8, which is 2x = 6, so x = 3. Then 2 × 3 + y = 8 gives y = 2. Check the first: 4 × 3 + 2 = 14, true. The solution is (3, 2).

Two slips to avoid. First, adding the left sides but not the right sides: if you add x + y and x − y, you must also add 12 and 2. Second, subtracting carelessly when the signs are already opposite. With +y and −y, subtracting would give 2y, not 0, and nothing is eliminated. Choose add when the signs differ and subtract when they match.

Solve 2x + 3y = 12 and x + y = 5. No coefficients match yet. But if you multiply through the second equation by −2, it becomes −2x − 2y = −10, and now the x coefficients are +2 and −2. Add: (2x − 2x) + (3y − 2y) = 12 − 10, so y = 2. Then x + 2 = 5, so x = 3. Check the first equation: 2 × 3 + 3 × 2 = 6 + 6 = 12, true. The solution is (3, 2).

Multiplying an equation by a number keeps it true, as long as you multiply every term, including the number on the right. That is why the method works. Sometimes both equations need multiplying. Solve 3x + 2y = 7 and 2x + 5y = 12. To eliminate x, make both x coefficients 6, the least common multiple of 3 and 2. Multiply the first by 2: 6x + 4y = 14. Multiply the second by 3: 6x + 15y = 36.

Now subtract the first new equation from the second: (6x − 6x) + (15y − 4y) = 36 − 14, so 11y = 22 and y = 2. Substitute into 3x + 2y = 7: 3x + 4 = 7, so 3x = 3 and x = 1. Check the other original equation: 2 × 1 + 5 × 2 = 2 + 10 = 12, true. The solution is (1, 2).

The usual mistake is multiplying only the left side. If 2x + y = 7 becomes 4x + 2y = 7 instead of 4x + 2y = 14, everything after is wrong. Multiply every term. Also, choose which variable to eliminate by looking for the smaller multiplier. In 2x + y = 7 and 3x + 2y = 12, doubling the first equation to 4x + 2y = 14 lines up the y terms right away, and subtracting the second gives x = 2, then y = 3.

Most systems have exactly one solution, because two lines with different slopes cross exactly once. But two other things can happen. Take y = 2x + 1 and y = 2x − 4. Both have slope 2, so they are parallel lines: they run side by side and never meet. There is no point on both lines, so the system has no solution.

Algebra says the same thing. Set the expressions equal: 2x + 1 = 2x − 4. Subtract 2x from both sides: 1 = −4. That is false no matter what x is. A false statement with the variables gone means no solution. Compare that with y = 3x + 2 and 2y = 6x + 4. Divide the second by 2: y = 3x + 2. It is the very same line. Set them equal: 3x + 2 = 3x + 2, which gives 2 = 2, true for every x.

When both variables disappear and what is left is true, the two equations describe one line, and every point on it is a solution. That is infinitely many solutions. So there are three cases. Different slopes: one solution. Same slope, different y-intercepts: no solution. Same slope and same y-intercept: infinitely many.

Do not confuse the false case with a solution of 0. If substitution leads to 0 = 7, the answer is not x = 0 or x = 7. It means the lines never meet. And do not read 0 = 0 as no solution. It means the opposite, that every point on the line works. Say the words out loud: 0 = 7 is false, so no solution; 0 = 0 is true, so infinitely many.

Equations do not always arrive in slope-intercept form. Solve 2x + y = 6 and 4x + 2y = 8 for y. Rearrange the first: y = −2x + 6. Rearrange the second: 2y = −4x + 8, so y = −2x + 4. Both have slope −2, but the y-intercepts are 6 and 4. Same slope, different intercept: parallel lines, no solution.

Now x − y = 1 and 3x − 3y = 3. From the first: −y = −x + 1, so y = x − 1. From the second: −3y = −3x + 3, so y = x − 1. Same slope and same intercept. The two lines coincide, meaning they lie exactly on top of each other. Infinitely many solutions: any point on y = x − 1, such as (1, 0) or (5, 4).

One more: y = 5x + 1 and y = −x + 7. The slopes are 5 and −1, different, so there is exactly one solution. Find it: 5x + 1 = −x + 7, so 6x = 6, x = 1, and y = 5 × 1 + 1 = 6. Check the second: −1 + 7 = 6, true. Whenever slopes differ, the lines must cross somewhere, even if the crossing is far off the page.

The trap is comparing equations before rearranging them. The pair 2x + y = 6 and 4x + 2y = 8 does not look parallel, because the coefficients are different. But every term in the second is double the first, except the 8, which would need to be 12. Rearranging shows the shared slope. The other trap is comparing only slopes and forgetting intercepts. Same slope tells you parallel or identical; the intercept decides which.

Plan A costs $25 a month plus $2 for each gigabyte of data. Plan B costs $10 a month plus $5 per gigabyte. Each plan has a fixed cost, the part you pay no matter what, and a rate. Let g be the gigabytes and c the cost. Plan A: c = 2g + 25. Plan B: c = 5g + 10. Which plan is cheaper depends on how much data you use.

Find where the plans cost the same. Set the costs equal: 2g + 25 = 5g + 10. Subtract 2g: 25 = 3g + 10. Subtract 10: 15 = 3g, so g = 5. At 5 GB, Plan A costs 2 × 5 + 25 = 35 and Plan B costs 5 × 5 + 10 = 35. That point, (5, 35), is the break-even point. Below 5 GB, Plan B is cheaper because its fixed cost is lower. Above 5 GB, Plan A is cheaper because its rate is lower.

Test it. At 2 GB: A is 4 + 25 = 29, B is 10 + 10 = 20, so B wins by $9. At 8 GB: A is 16 + 25 = 41, B is 40 + 10 = 50, so A wins by $9. On a graph, Plan B starts lower but climbs faster, and it crosses above Plan A at 5 GB. The steeper line always ends up on top eventually.

The setup is where mistakes happen. The number that multiplies g is the per-gigabyte rate, and the number added alone is the fixed cost. Writing c = 25g + 2 for Plan A would mean $25 per gigabyte, which is a very different plan. Read the words carefully: per means it multiplies, and a month means it is added once.

A school play sells adult tickets for $8 and child tickets for $5. One night 30 tickets sold for $195 in all. How many of each? Let a be the adult tickets and c the child tickets. The count gives one equation: a + c = 30. The money gives the other: 8a + 5c = 195. This is a mixture problem: two kinds of things are mixed, and you know the total count and the total value.

Solve by elimination. Multiply the count equation by 5: 5a + 5c = 150. Subtract it from the money equation: (8a − 5a) + (5c − 5c) = 195 − 150, so 3a = 45 and a = 15. Then c = 30 − 15 = 15. Check the money: 8 × 15 + 5 × 15 = 120 + 75 = 195, true. Fifteen adult tickets and fifteen child tickets.

Coins work the same way. A jar holds 20 coins, all dimes and quarters, worth $3.05. Work in cents to avoid decimals: d + q = 20 and 10d + 25q = 305. Multiply the first by 10: 10d + 10q = 200. Subtract from the second: 15q = 105, so q = 7. Then d = 20 − 7 = 13. Check: 10 × 13 + 25 × 7 = 130 + 175 = 305 cents, which is $3.05. Seven quarters and thirteen dimes.

To set up any problem of this kind, ask two questions. What is being counted, and what is the total count? What is each one worth, and what is the total value? The first answer gives the count equation with plain variables. The second gives the value equation with each variable multiplied by its worth. The common slip is putting the total value into the count equation, like d + q = 305. The count of coins is 20, and the value in cents is 305. Keep them in separate equations.