Create and solve equations and inequalities in one variable from linear, quadratic, rational and exponential situations

Learning goal: Create equations and inequalities in one variable that describe a situation, use them to solve problems, and check the answer against the situation. The situations give rise to linear, quadratic, simple rational and exponential functions, and for the more complicated ones technology may be used to solve.

From a situation to an equation. Creating an equation is a translation. First name the unknown with a variable and say what it measures, including units: let m be the number of minutes. Second, write each quantity in the situation as an expression in that variable. Third, find the sentence in the situation that says two quantities are equal, or that one is at most or at least another, and write it with =, ≤, <, ≥ or >. Only then solve. Students who start solving before writing the equation usually solve the wrong problem. Finally, interpret the solution: a length cannot be negative, a number of people is a whole number, and an inequality may need its endpoint checked.

Linear situations. A phone plan charges $25 a month plus $0.10 per minute of calls. For what number of minutes is the bill exactly $40? Let m be the minutes. The bill is 25 + 0.10m dollars. The equation is 25 + 0.10m = 40, so 0.10m = 15 and m = 150 minutes. Check: 25 + 15 = 40. The equation is linear because m appears to the first power only. An inequality arises when the sentence says "at most" or "no more than": with $60 to spend on an entry fee of $15 plus $4 per ride, the number of rides n satisfies 15 + 4n ≤ 60, so 4n ≤ 45 and n ≤ 11.25. Since n counts whole rides, the answer is at most 11 rides. The solution set of the inequality is every n up to 11.25, drawn on the number line in the diagram, and the whole-number answer is read from it.

Quadratic situations: a fully worked model. A rectangular garden has a length 4 meters more than its width, and its area is 60 square meters. Let w be the width in meters; the length is w + 4. Area is width times length, so the equation is w(w + 4) = 60. Expand and rearrange to standard form: w² + 4w − 60 = 0. Factor: two numbers with product −60 and sum 4 are 10 and −6, so (w + 10)(w − 6) = 0, giving w = −10 or w = 6. A width cannot be negative, so w = 6 meters and the length is 10 meters. Check: 6 × 10 = 60. The negative solution is a genuine solution of the equation but not of the problem; stating why it is rejected is part of the answer. If the quadratic had not factored, the quadratic formula would have finished it.

Simple rational situations. A 300-kilometer trip takes one hour less if the average speed is increased by 10 km/h. What is the original speed? Let x be the original speed in km/h. Time equals distance over speed, so the original time is 300/x hours and the faster time is 300/(x + 10) hours. The sentence "one hour less" gives the equation 300/x = 300/(x + 10) + 1. Multiply every term by x(x + 10), which is allowed since x and x + 10 are positive speeds: 300(x + 10) = 300x + x(x + 10), so 300x + 3000 = 300x + x² + 10x, hence x² + 10x − 3000 = 0, and (x + 60)(x − 50) = 0. Reject x = −60; the original speed is 50 km/h. Check: 300/50 = 6 hours, 300/60 = 5 hours, one hour less. The equation is rational because the variable appears in a denominator; clearing the denominators turned it into a quadratic.

Exponential situations. A culture of 500 bacteria doubles every hour. When will it reach 8000? Let t be the number of hours. The population is 500·2ᵗ, so the equation is 500·2ᵗ = 8000. Divide by 500: 2ᵗ = 16 = 2⁴, so t = 4 hours. Check: 500 → 1000 → 2000 → 4000 → 8000 after four doublings. When the target is not an exact power, for instance 500·2ᵗ = 10000 so that 2ᵗ = 20, the answer is between 4 and 5 hours and a logarithm or a graphing tool gives t ≈ 4.32; this is the sort of more complicated case where technology is a reasonable option. The same model handles other periods and decay. If a quantity doubles every 6 years instead of every hour, then after y years it has doubled y/6 times, so it is multiplied by 2^(y/6); and a quantity that halves every hour is multiplied by 1/2 each hour, so 80 g becomes 80·(1/2)ᵗ g after t hours: 40 g after 1 hour, 20 g after 2.

Inequalities and their solution sets. An inequality's solution is usually a whole interval, and the words of the situation decide which endpoint is included. Saving $35 a week from $200 to reach at least $500 gives 200 + 35w ≥ 500, so 35w ≥ 300 and w ≥ 8.57…; since deposits happen weekly, 9 weeks is the first week the target is met. Multiplying or dividing an inequality by a negative number reverses its direction; that rule is the one students forget, and it comes from the fact that −3 < −2 while 3 > 2.

The misconception to remove. Students often write the equation from the numbers rather than from the sentence: seeing 300, 10 and 1, they might write 300/10 = 1 or add the numbers. The sentence "one hour less" is what produced the equation, and the expressions 300/x and 300/(x + 10) came from the relationship time = distance ÷ speed. Writing the meaning of each expression in words before writing the equation prevents this. Rejecting a solution is also part of creating the equation well: the equation knows nothing about widths being positive, so the solver must apply the situation's constraints.

About the numbers. Every price, fee, amount and business in this lesson's examples, tasks and questions is made up for practice; none describes a real company, product or offer.

Worked linear inequality: Ana has $200 saved and adds $35 each week. After how many weeks will she have at least $500? Let w be the number of weeks. Savings: 200 + 35w. The sentence "at least $500" gives 200 + 35w ≥ 500, so 35w ≥ 300 and w ≥ 300/35 ≈ 8.57. Weeks are whole, so the first week the goal is met is week 9: 200 + 35 × 9 = 515 ≥ 500, while week 8 gives 480, which is short.

Worked quadratic from a falling object: A ball is thrown upward from a 48-foot platform so that its height in feet after t seconds is h = −16t² + 32t + 48, an idealized model that ignores air resistance (−16 is half of gravity's downward acceleration of about 32 feet per second per second, rounded) and applies only while the ball is in the air. When does it hit the ground? Ground level is h = 0, so −16t² + 32t + 48 = 0. Divide by −16: t² − 2t − 3 = 0, which factors as (t − 3)(t + 1) = 0, so t = 3 or t = −1. Time cannot be negative here, so the ball lands after 3 seconds. Check: −16(9) + 96 + 48 = −144 + 144 = 0.

Worked rational from a shared cost: A group plans to split a $240 rental equally. If 2 more people join, each person pays $4 less. How many were in the original group? Let n be the original number. Each pays 240/n originally and 240/(n + 2) afterward, so 240/n − 240/(n + 2) = 4. Multiply by n(n + 2): 240(n + 2) − 240n = 4n(n + 2), so 480 = 4n² + 8n, hence n² + 2n − 120 = 0 and (n + 12)(n − 10) = 0. Reject −12; n = 10 people. Check: 240/10 = 24 and 240/12 = 20, a difference of 4.

Worked exponential: A population of 4 insects triples every day. After how many days is it 972? Let t be the number of days; the population is 4·3ᵗ. Equation: 4·3ᵗ = 972, so 3ᵗ = 243 = 3⁵ and t = 5 days. Check: 4, 12, 36, 108, 324, 972. Had the target been 1000, then 3ᵗ = 250, which is not a power of 3; t is between 5 and 6, and a graphing tool or logarithm gives t ≈ 5.03.

Worked quadratic inequality: For the ball above, h = −16t² + 32t + 48, during what times is it at least 48 feet high? The sentence "at least 48 feet" gives −16t² + 32t + 48 ≥ 48, so −16t² + 32t ≥ 0. Divide by −16 and reverse the direction: t² − 2t ≤ 0, that is t(t − 2) ≤ 0. A product of two factors is negative or zero only when the factors have opposite signs or one is zero, which happens for 0 ≤ t ≤ 2. So the ball is at least 48 feet high from launch until 2 seconds. Check t = 1: −16 + 32 + 48 = 64 ≥ 48; check t = 2.5: −100 + 80 + 48 = 28 < 48, outside the interval as expected.

Worked rational inequality: A 300-kilometer trip must take at most 4 hours. Which average speeds x (km/h) work? Time is 300/x, so 300/x ≤ 4 with x > 0. Because x is positive, multiplying both sides by x keeps the direction: 300 ≤ 4x, so x ≥ 75 km/h. Check x = 100: 3 hours, within the limit; x = 60: 5 hours, too long. Multiplying by x was safe only because the situation makes x positive; if its sign were unknown, the direction could flip.

Worked exponential inequality: When does the insect population 4·3ᵗ exceed 1000? The inequality is 4·3ᵗ > 1000, so 3ᵗ > 250. Since 3ᵗ grows as t grows and 3⁵ = 243 < 250 < 729 = 3⁶, the solution is every t greater than log₃ 250 ≈ 5.03 days; counting whole days, the first count above 1000 is on day 6 (4·3⁶ = 2916), because day 5 gives 972.

Representation task: For the rides situation, 15 + 4n ≤ 60, draw a number line from 0 to 14, shade the solution set n ≤ 11.25, mark the endpoint 11.25 and circle the whole numbers that are actual possible numbers of rides. Underneath, write a three-row table headed "quantity in words", "expression", "unit" for the entry fee, the cost of the rides and the total, then copy the inequality and state which word in the situation produced the ≤ sign.

Independent performance: Write four original situations, one each of linear, quadratic, simple rational and exponential type, each with a question that has a numerical answer. For each: define the variable with its unit, write every quantity as an expression, create the equation or inequality, solve it, reject any solution the situation does not allow with a stated reason, and check the answer in the original words. At least two of the four must be inequalities, and at least one of those must come from a quadratic, rational or exponential situation. Success criteria: each equation follows from a sentence in the situation, not from a guess; the quadratic and rational equations are cleared and factored or solved by formula with steps shown; every rejected solution has a reason; and each check substitutes into the situation, not only into the equation.