Creating/Solving One Variable Equations/Inequalities - Algebra 2

A savings account starts with $1200 and earns 6% interest compounded annually. Write and solve an exponential equation to find how many years it will take for the balance to reach $2000. Let $t$ = the number of years.
Answer: Model $1200(1.06)^t=2000$ . Then $t=\log_{1.06}\left(\frac{2000}{1200}\right)\approx 8.79$ . It takes about 8.79 years. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'grows 8% yearly' means multiply by 1.08 each year, giving formula amount = initial × $(1.08)^t$ . To find when it reaches a specific value, set up equation like $500(1.08)^t = 1000$ and solve using logarithms: $(1.08)^t = 2$ , so $t = \ln(2)/\ln(1.08) \approx 9$ years. The logarithm unlocks the exponent! Starting with $1200$ at 6% annual growth, the balance is $1200(1.06)^t = 2000$ , so $(1.06)^t = 2000/1200 = 5/3$ , $t = \log_{1.06}(5/3) = \ln(5/3)/\ln(1.06) \approx 8.79$ years, interpreting as the time for the investment to grow to the target. Choice A correctly sets up the exponential growth model and solves using logarithms to get approximately 8.79 years. A distractor like Choice C uses a linear model instead of exponential, leading to an unrealistically long time of 755 years, but remember interest compounds multiplicatively, not additively. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A number and its reciprocal have a sum of $\frac{10}{3}$ . Write and solve an equation to find the number. Let $x$ = the number (assume $x\ne 0$ ).
Answer: Set up $x+\frac{1}{x}=\frac{10}{3}$ . Solving gives $x=3$ or $x=\frac{1}{3}$ . Both satisfy the context. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Reciprocal relationships lead to rational equations that become quadratic: for x + 1/x = sum, multiply by x to get x² - sum x + 1 = 0, solve with quadratic formula, both roots often valid if positive. Here, x + 1/x = 10/3, multiply by x: x² + 1 = (10/3)x, x² - (10/3)x + 1 = 0, multiply by 3: 3x² - 10x + 3 = 0, roots [10 ± √(100-36)]/6 = [10 ± 8]/6, x=3 or x=1/3, both satisfy as reciprocals sum to 10/3. Choice A correctly sets up the sum and solves for both valid solutions. A distractor like Choice D discards one root unnecessarily, but both fit the context since the problem doesn't specify which is the 'number' vs. reciprocal. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A concert venue charges a one-time rental fee of $150 plus $12 per ticket printed. The budget for fees and printing is at most $450. Create an inequality and find the maximum number of tickets that can be printed. Let $n$ = number of tickets.
Answer: Model $12n+150\le 450$ . Then $n\le 25$ . Maximum is 25 tickets. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Budget constraints lead to linear inequalities: fixed cost + rate × quantity ≤ total limit, solve for quantity, and round down for maximum integers if discrete like tickets. Here, 150 + 12n ≤ 450, 12n ≤ 300, n ≤ 25, so maximum 25 tickets, as 150 + 12×25 = 450 exactly fits the 'at most' budget. Choice B correctly models the inequality with ≤ for 'at most' and solves to find the maximum integer n=25. A distractor like Choice A uses ≥ instead of ≤, which would model a minimum rather than a maximum, but budgets cap spending, so inequality direction matters. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A streaming service charges a one-time setup fee of $25 plus $8 per month. Create an inequality for a budget of at most $105 and find the maximum number of months you can pay for. Let $m$ = the number of months.
Answer: Set up $8m+25\le 105$ . Then $8m\le 80$ , so $m\le 10$ . Maximum is 10 months. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Budget constraint problems typically lead to linear inequalities: total cost = fixed fee + (rate × quantity). Here, the total cost is $25 + $8m, and we need this to be at most $105, giving the inequality 25 + 8m ≤ 105. Solving: 8m ≤ 80, so m ≤ 10. Since m represents months (a discrete quantity), the maximum is 10 months. The total cost for m months is: setup fee + monthly charges = $25 + $8m. For a budget of at most $105: 25 + 8m ≤ 105. Subtracting 25: 8m ≤ 80. Dividing by 8: m ≤ 10. Since m must be a whole number of months, the maximum is 10 months. We can verify: 25 + 8(10) = 105 ✓. Choice A correctly sets up the inequality 8m + 25 ≤ 105 and solves to find m ≤ 10, giving a maximum of 10 months. Choice B incorrectly uses ≥ instead of ≤, which would mean we want to spend at least $105 rather than at most $105 - this reverses the problem's meaning! Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A rectangular garden has an area of 150 m $^2$ . Its length is 5 meters more than its width. Write and solve an equation to find the garden's dimensions. Let $w$ = width (in meters).
Answer: Set up $w(w+5)=150$ . Then $w=10$ (reject $w=-15$ ). Dimensions are $10$ m by $15$ m. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Area problems with rectangular dimensions lead to quadratic equations: if width is w and length is w+5, then area = width × length gives w(w+5) = 150, which expands to w² + 5w = 150 or w² + 5w - 150 = 0. Factoring gives (w+15)(w-10) = 0, so w = -15 or w = 10, but negative width is impossible, leaving w = 10 meters and length = 15 meters. Choice B correctly sets up the equation as width × length = area (w(w+5) = 150) and solves to get w = 10, rejecting the negative solution, giving dimensions 10m by 15m. Choice A incorrectly uses w-5 for length instead of w+5, C treats it as a perimeter problem using 2w+5 = 150, and D sets up w² + 5 = 150 which doesn't represent the area formula. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A rectangular patio has an area of $96\text{ m}^2$ . Its length is 4 meters more than its width. Set up a quadratic equation and solve to find the patio's width and length. Let $w$ = the width in meters.
Answer: Set up $w(w+4)=96$ . Then $w^2+4w-96=0$ , so $w=8$ (reject $-12$ ). The patio is $8\text{ m}$ by $12\text{ m}$ . This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Area problems with related dimensions often lead to quadratic equations: if length is width plus a constant, set up area = width × (width + constant), resulting in w² + cw - area = 0, solved via factoring or quadratic formula, rejecting negative roots since dimensions are positive. For this patio, set up w(w + 4) = 96, expand to w² + 4w - 96 = 0, use quadratic formula w = [-4 ± √(16 + 384)]/2 = [-4 ± √400]/2 = [-4 ± 20]/2, so w = 8 or w = -12 (reject negative), meaning width 8 m and length 12 m. Choice A correctly sets up the quadratic with length as w + 4 and solves to find the dimensions 8 m by 12 m. A distractor like choice C mistakenly uses a linear equation by adding sides instead of multiplying for area, resulting in impossible dimensions. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A school club is ordering T-shirts. The printer charges a $45 setup fee plus $9 per shirt. The club has at most $250 to spend. Create an inequality for the cost and find the maximum number of shirts the club can order. Let $n$ = number of shirts.
Answer: Set up $45+9n\le 250$ . Then $n\le 22.78$ , so the maximum is 22 shirts. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Budget constraints with fixed and variable costs lead to linear inequalities: total cost = fixed + rate × quantity ≤ budget, solved by isolating the variable and considering real-world constraints like integer values. For this T-shirt order, set up 45 + 9n ≤ 250, subtract 45 to get 9n ≤ 205, divide by 9 for n ≤ 22.78, so maximum integer n = 22 shirts to stay within budget. Choice B correctly sets up the inequality with ≤ for 'at most' and interprets the maximum as 22 shirts. A common error, as in choice A, is using ≥ instead of ≤, which would incorrectly model exceeding the budget. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A medication dose starts at 180 mg and 20% is eliminated each hour. Model the amount remaining by an exponential equation and solve for how many hours until 60 mg remain. Let $t$ = time in hours.
Answer: Set up $180(0.80)^t=60$ . Then $t=\log_{0.8}(1/3)\approx 4.92$ hours. About $4.92$ hours are needed. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'eliminates 20% hourly' means 80% remains, so multiply by 0.80 each hour, giving formula amount = initial × $(0.80)^t$ . To find when it reaches a specific value, set up equation like 180 $(0.80)^t$ = 60 and solve using logarithms: $(0.80)^t$ = 1/3, so t = ln(1/3)/ln(0.80) ≈ 4.92 hours. The logarithm unlocks the exponent! Starting with 180 mg and losing 20% each hour means 80% remains, so the equation is 180 $(0.80)^t$ = 60. Dividing by 180: $(0.80)^t$ = 60/180 = 1/3. Taking logarithms: t = log₀.₈₀(1/3) = ln(1/3)/ln(0.80) ≈ -1.099/-0.223 ≈ 4.92 hours. Choice A correctly models exponential decay with 180 $(0.80)^t$ = 60 and solves to get approximately 4.92 hours, which makes sense as the medication needs to drop to one-third of its initial dose. Choice C incorrectly uses 0.20 as the base (as if 20% remains instead of 80%), which would show much faster decay. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A rectangle has an area of 96 cm $^2$ . Its length is 4 cm more than its width. Write and solve a quadratic equation to find the rectangle's width. Let $w$ = the width in cm.
Answer: Set up $w(w+4)=96$ . Then $w^2+4w-96=0$ and $w=8$ (reject $w=-12$ ). The width is 8 cm. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Area problems with related dimensions often lead to quadratic equations: if length is width plus a constant, then width × (width + constant) = area, expanding to w² + c w - area = 0, solved by quadratic formula or factoring, rejecting negative roots for positive dimensions. For this rectangle, let w be width, then length = w + 4, so w(w + 4) = 96, expanding to w² + 4w - 96 = 0, quadratic formula w = [-4 ± √(16 + 384)]/2 = [-4 ± √400]/2 = [-4 ± 20]/2, giving w = 8 or w = -12 (rejected), so width 8 cm, length 12 cm, area 96 cm² checks out. Choice A correctly sets up the quadratic from the area relationship and solves, rejecting the negative root to find width 8 cm. A common error, as in Choice B, treats it as perimeter instead of area, adding sides linearly, but area requires multiplication, leading to incorrect large dimensions. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
An investment of $1200 earns 6% interest compounded annually. Model the balance with an exponential equation and solve for how many years it takes for the investment to reach $2000. Let $t$ = the number of years.
Answer: Set up $1200(1.06)^t=2000$ . Then $t=\log_{1.06}\!\left(\frac{5}{3}\right)\approx 8.78$ . It takes about 8.78 years. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'grows 6% yearly' means multiply by 1.06 each year, giving formula amount = initial × $(1.06)^t$ . To find when it reaches a specific value, set up equation like 1200 $(1.06)^t$ = 2000 and solve using logarithms: $(1.06)^t$ = 5/3, so t = ln(5/3)/ln(1.06) ≈ 8.78 years. The logarithm unlocks the exponent! Starting with $1200 growing at 6% annually, after t years we have 1200 $(1.06)^t$ . Setting this equal to $2000: 1200 $(1.06)^t$ = 2000, so $(1.06)^t$ = 2000/1200 = 5/3. Taking logarithms: t = log₁.₀₆(5/3) = ln(5/3)/ln(1.06) ≈ 8.78 years. Choice A correctly sets up the exponential growth equation and solves using logarithms to get approximately 8.78 years. Choice D incorrectly uses linear growth (1200 + 0.06t = 2000), which would mean adding only $0.06 per year instead of 6% of the current balance - this gives an absurd 13,333 years! Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A ball is thrown from the top of a building. Its height $h$ (in feet) above the ground after $t$ seconds is given by $h = -16t^2 + 32t + 128$ . The ball will hit the ground when its height equals zero. Which equation should be solved to find when this occurs?
Answer: $-16t^2 + 32t + 128 = 0$ The correct answer is A. The ball hits the ground when $h = 0$ , so we need $-16t^2 + 32t + 128 = 0$ . Choice B multiplies the entire equation by $-1$ , which is mathematically equivalent but not the most direct form. Choice C divides the original equation by $-16$ , getting $t^2 - 2t - 8 = 0$ , which is also equivalent but represents an additional step. Choice D incorrectly changes signs when dividing, resulting in $t^2 + 2t + 8 = 0$ , which has no real solutions.
A streaming service has 5000 subscribers and grows by 12% each month. Set up an exponential equation and solve for the number of months $t$ until it reaches 10,000 subscribers. Let $t$ = time in months.
Answer: Set up $5000(1.12)^t=10000$ . Then $t=\log_{1.12}(2)\approx 6.1$ months, so it takes about 6 months. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'grows 8% yearly' means multiply by 1.08 each year, giving formula amount = initial × $(1.08)^t$ . To find when it reaches a specific value, set up equation like 500 $(1.08)^t$ = 1000 and solve using logarithms: $(1.08)^t$ = 2, so t = ln(2)/ln(1.08) ≈ 9 years. The logarithm unlocks the exponent! For this subscriber growth, set up 5000 $(1.12)^t$ = 10000, simplify to $(1.12)^t$ = 2, then t = ln(2)/ln(1.12) ≈ 6.1 months, meaning it takes about 6 months to double subscribers. Choice A correctly sets up the exponential growth equation and solves using logarithms to find approximately 6.1 months. A common error, as in choice C, models it linearly, leading to an absurdly long time frame. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A streaming service has 1,200 subscribers and the number of subscribers increases by 15% each month. Model the situation with an exponential equation and solve for how many months it will take to reach 2,400 subscribers. Let $t$ = the number of months.
Answer: Set up $1200(1.15)^t=2400$ . Then $t=\log_{1.15}(2)\approx 4.96$ months. It takes about $5.0$ months. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'increases 15% monthly' means multiply by 1.15 each month, giving formula subscribers = initial × $(1.15)^t$ . To find when it reaches a specific value, set up equation like 1200 $(1.15)^t$ = 2400 and solve using logarithms: $(1.15)^t$ = 2, so t = ln(2)/ln(1.15) ≈ 4.96 months. The logarithm unlocks the exponent! Starting with 1,200 subscribers and growing 15% monthly gives the equation 1200 $(1.15)^t$ = 2400. Dividing both sides by 1200: $(1.15)^t$ = 2. Taking logarithms: t = log₁.₁₅(2) = ln(2)/ln(1.15) ≈ 0.693/0.140 ≈ 4.96 months. Choice A correctly sets up the exponential growth equation 1200 $(1.15)^t$ = 2400 and solves using logarithms to get approximately 4.96 months (about 5 months), which makes sense as the subscribers need to double. Choice D incorrectly computes t = ln(1.15)/ln(2) (inverting the logarithm ratio), which gives about 0.20 months—far too quick for doubling with 15% monthly growth. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A medication dose starts at 200 mg. Each hour, 15% of the medication is eliminated (so 85% remains). Model the amount remaining after $t$ hours with an exponential equation and solve for when 50 mg remains. Let $t$ = time in hours.
Answer: Model: $200(0.85)^t=50$ . Then $t=\log_{0.85}(0.25)\approx 8.54$ hours. It takes about 8.54 hours for 50 mg to remain. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'eliminates 15% hourly' means 85% remains, so multiply by 0.85 each hour, giving formula amount = initial × $(0.85)^t$ . To find when it reaches a specific value, set up equation like 200 $(0.85)^t$ = 50 and solve using logarithms: $(0.85)^t$ = 50/200 = 0.25, so t = ln(0.25)/ln(0.85) ≈ 8.54 hours. The logarithm unlocks the exponent! Since 15% is eliminated each hour, 85% remains, so the medication amount after t hours is 200 $(0.85)^t$ mg, and we solve 200 $(0.85)^t$ = 50 by dividing both sides by 200 to get $(0.85)^t$ = 0.25, then taking logarithms: t = log₀.₈₅(0.25) ≈ 8.54 hours. Choice C correctly models exponential decay as 200 $(0.85)^t$ (since 85% remains each hour) and solves using logarithms to find t ≈ 8.54 hours. Choice A incorrectly uses 0.15 as the base (which would mean only 15% remains each hour), B uses 1.15 which would represent 15% growth not decay, and D treats it as linear decay subtracting 0.15t instead of multiplying by $(0.85)^t$ . Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A rectangular patio has an area of 96 square feet. Its length is 4 feet longer than its width. Write and solve an equation to find the patio's width. Let $w$ = the width (in feet).
Answer: Set up $w(w+4)=96$ . Then $w=8$ (reject $w=-12$ ). The width is 8 ft. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Area problems often lead to quadratic equations: when a rectangle has area A and one dimension is expressed in terms of another (like length = width + 4), multiplying dimensions gives a quadratic. Here, width × length = area becomes w(w+4) = 96, which expands to w² + 4w = 96. Setting up the equation: if width = w and length = w + 4, then area = w(w+4) = 96. Expanding: w² + 4w = 96. Rearranging: w² + 4w - 96 = 0. Using the quadratic formula or factoring: (w+12)(w-8) = 0, giving w = -12 or w = 8. Since width must be positive, w = 8 feet. Choice A correctly sets up w(w+4) = 96 and solves to get w = 8 feet (rejecting the negative solution w = -12), giving a patio that's 8 feet by 12 feet with area 96 square feet. Choice C incorrectly claims w = 12, but this would give a 12×16 patio with area 192 square feet, not 96. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A medication dose begins at 300 mg in the bloodstream. Each hour, 20% of the medication is eliminated (so 80% remains). Set up an exponential equation and solve for the time $t$ (in hours) when 90 mg remains. Let $t$ = time in hours.
Answer: Set up $300(0.8)^t=90$ . Then $t=\log_{0.8}(0.3)\approx 5.40$ . About 5.4 hours. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'grows 8% yearly' means multiply by 1.08 each year, giving formula amount = initial × $(1.08)^t$ . To find when it reaches a specific value, set up equation like 500 $(1.08)^t$ = 1000 and solve using logarithms: $(1.08)^t$ = 2, so t = ln(2)/ln(1.08) ≈ 9 years. The logarithm unlocks the exponent! For this medication decay, the equation is 300 $(0.8)^t$ = 90; dividing by 300 gives $(0.8)^t$ = 0.3, then t = ln(0.3)/ln(0.8) ≈ 5.40 hours, meaning it takes about 5.4 hours for 90 mg to remain. Choice A correctly sets up the exponential decay model and solves with logarithms for the positive time value. A common mistake like in choice D uses a linear subtraction instead of exponential, resulting in an absurdly long time that doesn't match decay behavior. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
An investment account starts with $\$2{,}500$ and grows by $6\%$ each year. Model the situation with an equation and solve for how many years it will take for the account to reach $\$4{,}000$ . Let $t$ = the number of years.
Answer: Set up $2500(1.06)^t=4000$ . Then $t=\log_{1.06}\!\left(\frac{4000}{2500}\right)\approx 8.1$ years. It takes about $8.1$ years. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'grows 6% yearly' means multiply by 1.06 each year, giving formula amount = initial × $(1.06)^t$ . To find when it reaches a specific value, set up equation like 2500 $(1.06)^t$ = 4000 and solve using logarithms: $(1.06)^t$ = 1.6, so t = ln(1.6)/ln(1.06) ≈ 8.1 years. The logarithm unlocks the exponent! Starting with $2,500 and growing 6% annually gives the equation 2500 $(1.06)^t$ = 4000. Dividing both sides by 2500: $(1.06)^t$ = 1.6. Taking logarithms: t = log₁.₀₆(1.6) = ln(1.6)/ln(1.06) ≈ 0.470/0.058 ≈ 8.1 years. Choice A correctly sets up the exponential growth equation 2500 $(1.06)^t$ = 4000 and solves using logarithms to get approximately 8.1 years, which makes sense as the investment needs to grow by 60%. Choice B incorrectly uses linear growth (1.06t instead of $(1.06)^t$ ), which would mean adding 6% of the original amount each year rather than compounding. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
A plumber can drain a tank in 5 hours. A leak refills the tank and would fill it in 20 hours if the drain were closed. If the drain is open and the leak continues, set up a rational equation and solve for the net time to empty the tank. Let $t$ = time in hours to empty the tank with both processes happening.
Answer: Set up $\frac{1}{5}-\frac{1}{20}=\frac{1}{t}$ . Then $t=\frac{20}{3}$ hours $\approx 6.67$ hours. The tank empties in about 6.67 hours. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Work-rate problems lead to rational equations: if one worker completes a job in time t₁ and another in t₂, their combined rate is 1/t₁ + 1/t₂ (adding rates), which equals 1/t_combined. The reciprocals represent 'fraction of job per hour,' and adding these fractions gives the combined rate. Solving these rational equations requires finding LCD and often produces fractional time answers that make sense: 3.4 hours = 3 hours 24 minutes. Here, drain rate 1/5 (emptying) minus leak rate 1/20 (filling) gives net 1/5 - 1/20 = (4-1)/20 = 3/20 = 1/t, so t = 20/3 ≈ 6.67 hours, meaning the leak slows down the draining process realistically. Choice B correctly sets up the equation by subtracting the opposing rates and solves for t ≈ 6.67 hours, giving the net time to empty. A mistake like in Choice C reverses the subtraction, leading to negative rate and time, but we gently note the drain is faster, so it should be dominant for positive net emptying time. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
An investment account starts with $800 and earns 6% interest compounded annually. Model the situation with an exponential equation and solve for the number of years $t$ it takes for the account to reach $1200. Let $t$ = number of years.
Answer: Set up $800(1.06)^t=1200$ . Then $t=\log_{1.06}(1.5)\approx 6.96$ . It takes about 7 years. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'grows 8% yearly' means multiply by 1.08 each year, giving formula amount = initial × $(1.08)^t$ . To find when it reaches a specific value, set up equation like 500 $(1.08)^t$ = 1000 and solve using logarithms: $(1.08)^t$ = 2, so t = ln(2)/ln(1.08) ≈ 9 years. The logarithm unlocks the exponent! In this investment scenario, the equation is 800 $(1.06)^t$ = 1200; dividing by 800 gives $(1.06)^t$ = 1.5, then t = ln(1.5)/ln(1.06) ≈ 6.96 years, indicating it takes nearly 7 years to reach $1200. Choice A correctly sets up the exponential growth equation and solves using logarithms to find approximately 7 years. A distractor like choice D mistakenly uses a linear model instead of exponential, leading to an unrealistically long time that ignores compounding. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!
Pipe A can fill a tank in 5 hours. Pipe B can fill the same tank in 8 hours. Write and solve an equation to find how long it takes to fill the tank if both pipes run together. Let $t$ = the time in hours for both pipes together.
Answer: Set up $\frac{1}{5}+\frac{1}{8}=\frac{1}{t}$ . Then $t=\frac{40}{13}\approx 3.08$ hours. Together they fill the tank in about $3.08$ hours. This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Work-rate problems lead to rational equations: if one worker completes a job in time t₁ and another in t₂, their combined rate is 1/t₁ + 1/t₂ (adding rates), which equals 1/t_combined. The reciprocals represent 'fraction of job per hour,' and adding these fractions gives the combined rate. Solving these rational equations requires finding LCD and often produces fractional time answers that make sense: 3.08 hours = 3 hours 5 minutes. Pipe A's rate is 1/5 (of the tank per hour) and Pipe B's rate is 1/8, so together: 1/5 + 1/8 = 1/t. Finding LCD of 40: 8/40 + 5/40 = 13/40 = 1/t, so t = 40/13 ≈ 3.08 hours. Choice B correctly sets up the equation as rates adding to combined rate (1/5 + 1/8 = 1/t) and solves to get t = 40/13 ≈ 3.08 hours, which means they fill the tank in about 3 hours and 5 minutes working together. Choice D incorrectly sets up the equation as 1/5 + 1/8 = t (missing the reciprocal relationship), getting 0.325 hours which would mean they work faster together than either works alone—impossible! Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

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