Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

Algebra

Algebra Help: Creating Solving One Variable Equations Inequalities

Review real example questions for Creating Solving One Variable Equations Inequalities in Algebra.

Question 1

Tickets to a school play cost 9each,andthereisaone−timeonlinefeeof9 each, and there is a one-time online fee of 9each,andthereisaone−timeonlinefeeof4 per order. Jordan has at most $40 to spend. Write an inequality representing the number of tickets Jordan can buy.

Let ttt = the number of tickets.

  1. 9t+4≥409t + 4 \ge 409t+4≥40
  2. 9t−4≤409t - 4 \le 409t−4≤40
  3. 4t+9≤404t + 9 \le 404t+9≤40
  4. 9t+4≤409t + 4 \le 409t+4≤40
Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. For inequalities, words like 'at most,' 'maximum,' 'no more than' signal ≤ (less than or equal), while 'at least,' 'minimum,' 'no less than' signal ≥ (greater than or equal). 'More than' means > (strict), and 'less than' means <. These key phrases tell you which inequality symbol to use! The context 'tickets cost 9eachwitha9 each with a 9eachwitha4 fee, at most 40′usesthephrase′atmost,′whichsignals≤.Settingup:costpertickettimesnumberplusfee≤totalavailable,so9t+4≤40.Solving:subtract4→9t≤36,divideby9→t≤4.ThismeansJordancanbuyamaximumof4wholetickets.ChoiceAiscorrectbecauseitproperlysetsuptheinequalityfromthecontextwiththefeeaddedanduses≤for′atmost,′givingt≤4ticketswithinbudget.ChoiceDsetsuptheinequalityincorrectly:itswitchesthevariablesto4t+9≤40,whichwouldbelike40' uses the phrase 'at most,' which signals ≤. Setting up: cost per ticket times number plus fee ≤ total available, so 9t + 4 ≤ 40. Solving: subtract 4 → 9t ≤ 36, divide by 9 → t ≤ 4. This means Jordan can buy a maximum of 4 whole tickets. Choice A is correct because it properly sets up the inequality from the context with the fee added and uses ≤ for 'at most,' giving t ≤ 4 tickets within budget. Choice D sets up the inequality incorrectly: it switches the variables to 4t + 9 ≤ 40, which would be like 40′usesthephrase′atmost,′whichsignals≤.Settingup:costpertickettimesnumberplusfee≤totalavailable,so9t+4≤40.Solving:subtract4→9t≤36,divideby9→t≤4.ThismeansJordancanbuyamaximumof4wholetickets.ChoiceAiscorrectbecauseitproperlysetsuptheinequalityfromthecontextwiththefeeaddedanduses≤for′atmost,′givingt≤4ticketswithinbudget.ChoiceDsetsuptheinequalityincorrectly:itswitchesthevariablesto4t+9≤40,whichwouldbelike4 per ticket and 9fee,buttheproblemsays9 fee, but the problem says 9fee,buttheproblemsays9 tickets and $4 fee—reading carefully for relationships is key! For inequalities, make a quick reference card: 'at most/maximum/no more than' → ≤ (can equal or be less), 'at least/minimum/no less than' → ≥ (can equal or be more), 'more than/over' → > (strictly greater), 'less than/under' → < (strictly less). Having these memorized means you'll never use the wrong symbol!

Question 2

Two pumps fill a tank together in 6 hours. Pump A alone can fill the tank in 10 hours. How long would it take Pump B alone to fill the tank? Let xxx = Pump B's time in hours.

  1. x=12x=12x=12
  2. x=15x=15x=15
  3. x=8x=8x=8
  4. x=16x=16x=16
Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. For rate problems like this, we set up: when working together, rates add. Pump A's rate is 1/10 tank per hour, Pump B's rate is 1/x tank per hour, and together they fill at 1/6 tank per hour. This gives us 1/10 + 1/x = 1/6. Solving: 1/x = 1/6 - 1/10 = 5/30 - 3/30 = 2/30 = 1/15. Therefore x = 15. Interpreting: Pump B alone takes 15 hours to fill the tank. Choice B is correct because it properly sets up the rate equation from context and solves correctly, giving x = 15 hours for Pump B alone. Choice A would mean Pump B works faster than the combined rate, which is impossible—two pumps together must work faster than either alone! When solving rate problems, remember that rates add when working together. Different contexts = different equation types: constant rates and simple relationships → linear, area and projectile motion → quadratic, working together and rate problems → rational, growth/decay over time → exponential. Recognizing these patterns helps you set up the right type of equation immediately!

Question 3

A rectangular garden has a length that is 5 feet more than its width. The area is 84 square feet. Write and solve an equation to find the width of the garden.

Let www = the width (in feet).

  1. w=14w = 14w=14
  2. w=12w = 12w=12
  3. w=7w = 7w=7
  4. w=9w = 9w=9
Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. This is an area problem, which means quadratic! Let w = width. The relationship 'length is 5 more than width, area 84' translates to w(w + 5) = 84, or w² + 5w - 84 = 0. Using quadratic formula, discriminant 25 + 336 = 361 = 19², so w = [-5 ± 19]/2, giving w = 7 or w = -12 (discard negative). So valid solution w = 7 with interpretation: the garden is 7 feet wide and 12 feet long. Choice A is correct because it properly sets up the quadratic from the area context, solves correctly, and interprets appropriately, giving w = 7 feet that makes sense. Choice C solves the equation correctly but doesn't check the context: w = 12 would imply length = 17, but that's swapping variables— the question asks for width, which is the smaller one. Always ask: does my answer make sense in the real world? This catches a lot of mistakes! The 'reality check' is your best friend: after solving, substitute your answer back into the original equation (math check), then ask 'does this make sense?' (reality check). Can dimensions be negative? Can you buy 7.3 shirts? Can there be -4 hours? The context tells you what's possible and what's not!

Question 4

A ball is thrown upward from the ground. Its height (in feet) after ttt seconds is given by h=−16t2+64th = -16t^2 + 64th=−16t2+64t. Solve for the time(s) when the ball is on the ground.

Let ttt = time in seconds.

  1. t=2t = 2t=2 only
  2. t=0t = 0t=0 and t=4t = 4t=4
  3. t=0t = 0t=0 and t=2t = 2t=2
  4. t=4t = 4t=4 only
Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. This is a motion problem, which means quadratic! Let t = time in seconds. The relationship 'height after t seconds is given by h = -16t² + 64t' and 'when the ball is on the ground' translates to set h=0: -16t² + 64t = 0. Using factoring, -16t(t - 4) = 0, so t=0 or t=4. Checking context: both make sense—at t=0 (start) and t=4 (lands). Interpreting: the ball is on the ground at 0 seconds and after 4 seconds. Choice B is correct because it properly sets up the quadratic equation from context, solves correctly, and interprets appropriately, giving t=0 and t=4 seconds. Choice D finds one solution to the quadratic but misses the other: solving gives two values, but perhaps they ignored t=0 since it's the start. For quadratics, always check both solutions against the real-world situation! The 'reality check' is your best friend: after solving, substitute your answer back into the original equation (math check), then ask 'does this make sense?' (reality check). Can dimensions be negative? Can you buy 7.3 shirts? Can there be -4 hours? The context tells you what's possible and what's not!

Question 5

A rectangular garden has a length that is 5 feet more than its width. The area of the garden is 84 square feet. Write and solve an equation to find the garden’s width.

Let www = the width (in feet).

  1. w=6w = 6w=6
  2. w=9w = 9w=9
  3. w=7w = 7w=7
  4. w=12w = 12w=12
Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. This is an area problem, which means quadratic! Let w = width. The relationship 'length that is 5 feet more than its width' and 'area of the garden is 84 square feet' translates to w(w + 5) = 84, or w² + 5w - 84 = 0. Using factoring, we get (w + 12)(w - 7) = 0, so w = -12 or w = 7. But wait—checking context: negative width doesn't make sense, so w = 7 feet. Interpreting: the width is 7 feet, length 12 feet, area 84 sq ft. Choice A is correct because it properly sets up the quadratic equation from context, solves correctly, and interprets appropriately, giving w=7 feet. Choice B finds one solution to the quadratic but misses the other (or keeps an extraneous one): solving gives w=7 and w=-12, but perhaps they took the positive non-valid or switched length/width. For quadratics, always check both solutions against the real-world situation! The 'reality check' is your best friend: after solving, substitute your answer back into the original equation (math check), then ask 'does this make sense?' (reality check). Can dimensions be negative? Can you buy 7.3 shirts? Can there be -4 hours? The context tells you what's possible and what's not!

Question 6

You have at most \50tospendonnotebooksthatcostto spend on notebooks that costtospendonnotebooksthatcost$4each.Writeaninequalityrepresentingthisconstraintandfindthemaximumnumberofnotebooksyoucanbuy.Leteach. Write an inequality representing this constraint and find the maximum number of notebooks you can buy. Leteach.Writeaninequalityrepresentingthisconstraintandfindthemaximumnumberofnotebooksyoucanbuy.Letn$ = number of notebooks.

  1. Maximum n=11n=11n=11
  2. Maximum n=12n=12n=12
  3. Maximum n=13n=13n=13
  4. Maximum n=14n=14n=14
Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. For inequalities, words like 'at most,' 'maximum,' 'no more than' signal ≤ (less than or equal), while 'at least,' 'minimum,' 'no less than' signal ≥ (greater than or equal). 'More than' means > (strict), and 'less than' means <. These key phrases tell you which inequality symbol to use! The context 'at most 50tospend′usesthephrase′atmost,′whichsignals≤.Settingup:costofnotebooks≤moneyavailable,so4n≤50.Solving:n≤12.5.Thismeansamaximumof12wholenotebooks(can′tbuyhalfanotebook!).ChoiceBiscorrectbecauseitproperlysetsuptheinequalityfromcontext,solvescorrectly,andinterpretsappropriatelyintherealworld,givingmaximumn=12notebooks.ChoiceCwouldcost4(13)=50 to spend' uses the phrase 'at most,' which signals ≤. Setting up: cost of notebooks ≤ money available, so 4n ≤ 50. Solving: n ≤ 12.5. This means a maximum of 12 whole notebooks (can't buy half a notebook!). Choice B is correct because it properly sets up the inequality from context, solves correctly, and interprets appropriately in the real world, giving maximum n = 12 notebooks. Choice C would cost 4(13) = 50tospend′usesthephrase′atmost,′whichsignals≤.Settingup:costofnotebooks≤moneyavailable,so4n≤50.Solving:n≤12.5.Thismeansamaximumof12wholenotebooks(can′tbuyhalfanotebook!).ChoiceBiscorrectbecauseitproperlysetsuptheinequalityfromcontext,solvescorrectly,andinterpretsappropriatelyintherealworld,givingmaximumn=12notebooks.ChoiceCwouldcost4(13)=52, which exceeds the $50 budget. When dealing with discrete items like notebooks, always round down to stay within the constraint! For inequalities, make a quick reference card: 'at most/maximum/no more than' → ≤ (can equal or be less), 'at least/minimum/no less than' → ≥ (can equal or be more), 'more than/over' → > (strictly greater), 'less than/under' → < (strictly less). Having these memorized means you'll never use the wrong symbol!

Question 7

A rectangular garden has a length that is 5 feet more than its width. The area of the garden is 84 square feet. Write and solve an equation to find the width. Let www = the width in feet.

  1. w=9w=9w=9
  2. w=6w=6w=6
  3. w=12w=12w=12
  4. w=7w=7w=7
Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. This is an area problem, which means quadratic! Let w = width in feet. The relationship 'length is 5 feet more than width' translates to length = w + 5. Since area = length × width = 84, we get w(w + 5) = 84. Expanding: w² + 5w = 84, so w² + 5w - 84 = 0. Using factoring: (w + 12)(w - 7) = 0, we get w = -12 or w = 7. But wait—checking context: width can't be negative! So w = 7 feet with interpretation: the garden is 7 feet wide and 12 feet long. Choice A is correct because it properly sets up the quadratic equation from the area context, solves correctly, and eliminates the extraneous negative solution, giving w = 7 feet. Choice B would give a length of 14 feet and area of 9 × 14 = 126 square feet, not 84. When solving quadratics from real-world contexts, always check both solutions—one often doesn't make physical sense! The 'reality check' is your best friend: after solving, substitute your answer back into the original equation (math check), then ask 'does this make sense?' (reality check). Can dimensions be negative? Can you buy 7.3 shirts? Can there be -4 hours? The context tells you what's possible and what's not!

Question 8

Pump A can fill a pool in 12 hours. Pump B can fill the same pool in xxx hours. Working together, they fill the pool in 8 hours. Set up and solve an equation to find how long Pump B takes to fill the pool alone.

Let xxx = Pump B’s time (in hours).

  1. x=20x = 20x=20
  2. x=24x = 24x=24
  3. x=18x = 18x=18
  4. x=16x = 16x=16
Explanation: This question tests your ability to translate a real-world situation into a mathematical equation or inequality, solve it, and interpret the result in the original context. Different contexts lead to different equation types: constant rates give linear equations (like cost = rate × quantity + fee), area problems often give quadratics (like length × width = area), and growth over time gives exponentials (like population = initial × (growth rate)^time). The context clues tell you which form to use. For rate problems like this, we set up: rates add when working together, so Pump A rate 1/12 + Pump B rate 1/x = combined rate 1/8. This gives us 1/12 + 1/x = 1/8. Solving: subtract 1/12 → 1/x = 1/8 - 1/12 = (3-2)/24 = 1/24, so x=24. Interpreting: Pump B takes 24 hours alone. Choice B is correct because it properly sets up the rational equation from context, solves correctly, and interprets appropriately, giving x=24 hours. Choice A makes an arithmetic error: perhaps in subtracting fractions, doing 1/8 - 1/12 = (3-2)/24=1/24 correctly but then misinterpreting. With all the steps in solving word problems—setting up, solving, interpreting—it's easy for calculation errors to slip in. Double-checking arithmetic is always worth it! The foolproof word problem strategy: (1) Read carefully and identify what's unknown—that's your variable, (2) Find what you know—those are your numbers, (3) Look for relationships—how are quantities connected? This gives you the equation, (4) Solve the equation using appropriate methods, (5) Check: does your answer satisfy the equation AND make sense in context? Following these steps systematically prevents most mistakes!

Question 9

An investment account grows according to the formula A=2500(1.06)tA = 2500(1.06)^tA=2500(1.06)t, where AAA is the account value and ttt is time in years. At the same time, the investor makes annual withdrawals of $200. Which equation represents the net account balance when accounting for both growth and withdrawals over ttt years?

  1. 2500(1.06)t−200=A2500(1.06)^t - 200 = A2500(1.06)t−200=A
  2. 2500(1.06)t−200t=A2500(1.06)^t - 200t = A2500(1.06)t−200t=A
  3. 2500(1.06−0.08)t=A2500(1.06 - 0.08)^t = A2500(1.06−0.08)t=A
  4. 2500(1.06)t⋅0.92t=A2500(1.06)^t \cdot 0.92^t = A2500(1.06)t⋅0.92t=A
Explanation: When you encounter problems involving exponential growth with regular withdrawals, you need to think about how these two separate processes affect the account differently. The growth compounds over time, while withdrawals happen at regular intervals and accumulate linearly. The original formula A=2500(1.06)tA = 2500(1.06)^tA=2500(1.06)t shows exponential growth at 6% annually. However, the investor also withdraws $200 each year for ttt years, which means total withdrawals equal 200t200t200t. Since withdrawals reduce the account balance, you subtract this linear term from the exponential growth: 2500(1.06)t−200t=A2500(1.06)^t - 200t = A2500(1.06)t−200t=A. Choice A (2500(1.06)t−200=A2500(1.06)^t - 200 = A2500(1.06)t−200=A) incorrectly assumes only one withdrawal of 200total,ratherthan200 total, rather than 200total,ratherthan200 per year for ttt years. Choice C (2500(1.06−0.08)t=A2500(1.06 - 0.08)^t = A2500(1.06−0.08)t=A) mistakenly tries to incorporate the 200withdrawalasapercentageratedecrease,but200 withdrawal as a percentage rate decrease, but 200withdrawalasapercentageratedecrease,but200 isn't 8% of the initial investment, and withdrawals don't work as percentage reductions anyway. Choice D (2500(1.06)t⋅0.92t=A2500(1.06)^t \cdot 0.92^t = A2500(1.06)t⋅0.92t=A) makes a similar error, treating withdrawals as if they reduce the growth rate by 8% annually, which incorrectly converts the fixed $200 into a percentage. Remember: when combining exponential processes with linear processes, keep them separate in your equation. Don't try to convert fixed amounts into percentages or combine them into a single exponential term. Exponential parts stay exponential, linear parts stay linear.

Question 10

A store offers a membership where customers pay $30 annually and then receive a 15% discount on all purchases. Without membership, customers pay full price. For what annual spending amount xxx (in dollars) would the membership cost exactly break even with non-membership shopping?

  1. 0.85x=x−300.85x = x - 300.85x=x−30
  2. 30+0.15x=x30 + 0.15x = x30+0.15x=x
  3. 30=0.15x30 = 0.15x30=0.15x
  4. 30+0.85x=x30 + 0.85x = x30+0.85x=x
Explanation: When you encounter break-even problems, you need to set up an equation where the total costs of both options are equal. Here, you're comparing the total annual cost of membership shopping versus non-membership shopping. With membership, you pay 303030 upfront plus 85% of your purchases (since you get a 15% discount, you pay 100% - 15% = 85%). So the total cost is 30+0.85x30 + 0.85x30+0.85x. Without membership, you simply pay the full price xxx for your purchases. At the break-even point, these costs are equal: 30+0.85x=x30 + 0.85x = x30+0.85x=x. This is answer choice D. Let's examine why the other options are incorrect: Choice A (0.85x=x−300.85x = x - 300.85x=x−30) incorrectly suggests that your discounted purchases equal your full purchases minus 303030. This doesn't account for the membership fee you actually paid. Choice B (30+0.15x=x30 + 0.15x = x30+0.15x=x) mistakenly uses 0.15 as what you pay, but 0.15 represents your savings, not your payment. You pay 85% of the original price, not 15%. Choice C (30=0.15x30 = 0.15x30=0.15x) only considers the membership fee equaling your savings, ignoring that you still have to pay for the discounted items. Study tip: In discount problems, always identify what percentage you actually pay (100% minus the discount percentage) rather than the discount itself. Set up break-even equations by making total costs equal, ensuring you include all fees and payments for each option.