How Do You Know if a Math Problem Has No Solution
Special Cases and Applications
Learning Objective(s)
· Solve equations that have one solution, no solution, or an infinite number of solutions.
· Solve application issues past using an equation in 1 variable.
Introduction
When you follow the steps to solve an equation, you endeavour to isolate the variable. You take a solution when you get the equation x = some value. In that location are equations, however, that accept no solution, and other equations that have an infinite number of solutions. How does this piece of work?
Algebraic Equations with No Solution
Allow's utilize the steps for solving an algebraic equation to the equation below.
Example | ||
| Problem | Solve for x. 12 + 2ten – eight = sevenx + 5 – 5x | |
| | Combine similar terms on both sides of the equation. Isolate the x term by subtracting 2x from both sides. | |
This is not a solution! Yous did not find a value for ten. Solving for ten the way y'all know how, you lot arrive at the false statement 4 = v. Surely 4 cannot be equal to 5!
This may make sense when you consider the second line in the solution where similar terms were combined. If you multiply a number by 2 and add together four you would never get the aforementioned answer every bit when y'all multiply that same number past 2 and add together v. Since at that place is no value of 10 that will always make this a truthful statement, the solution to the equation above is "no solution".
Exist careful that you exercise non confuse the solution 10 = 0 with "no solution". The solution x = 0 ways that the value 0 satisfies the equation, so at that place is a solution. "No solution" means that there is no value, not fifty-fifty 0, which would satisfy the equation.
Also, be careful non to make the fault of thinking that the equation four = 5 means that four and v are values for 10 that are solutions. If you substitute these values into the original equation, you'll see that they do not satisfy the equation. This is because there is truly no solution—in that location are no values for x that will brand the equation 12 + 2x – 8 = viix + 5 – five10 true.
| Example | |||
| Problem | Solve for x. 3x + 8 = 3(10 + 2) | ||
| | Use the distributive property to simplify. Isolate the variable term. Since you lot know that eight = 6 is false, there is no solution. | ||
| Answer | There is no solution. | ||
| Avant-garde Example | ||
| Problem | Solve for y. 8y = 2[3(y + 4) + y] | |
| | Apply the distributive property to simplify. When two sets of group symbols are used, evaluate the inner fix and and so evaluate the outer set. | |
| | t he variable term by subtracting 8y from both sides of the equation. Since yous know that 0 = 24 is false, there is no solution. | |
| Answer | There is no solution. | |
Algebraic Equations with an Space Number of Solutions
You accept seen that if an equation has no solution, you end upwardly with a false statement instead of a value for x. You can probably guess that at that place might be a manner y'all could end upward with a true statement instead of a value for 10.
| Case | ||
| Problem | Solve for x. 5ten + 3 – four10 = 3 + ten | |
| | Combine like terms on both sides of the equation. Isolate the 10 term by subtracting ten from both sides. | |
You arrive at the truthful argument "iii = 3". When you finish up with a true statement similar this, information technology means that the solution to the equation is "all real numbers". Try substituting ten = 0 into the original equation—you lot volition go a true statement! Try 
This equation happens to have an infinite number of solutions. Any value for ten that you tin remember of will brand this equation truthful. When you think about the context of the trouble, this makes sense—the equation ten + 3 = 3 + ten means "some number plus three is equal to 3 plus that aforementioned number." We know that this is always truthful—information technology's the commutative property of add-on!
| Case | ||
| Trouble | Solve for ten. v(x – 7) + 42 = iiix + 7 + 2ten | |
| | Apply the distributive belongings and combine like terms to simplify. Isolate the ten term by subtracting 5ten from both sides. You get the true statement 7 = 7, so you know that x tin can exist all real numbers. | |
| Respond | x = all real numbers | |
When solving an equation, multiplying both sides of the equation by zero is not a good choice. Multiplying both side of an equation past 0 volition always result in an equation of 0 = 0, but an equation of 0 = 0 does not assistance you lot know what the solution to the original equation is.
| Example | |||
| Problem | Solve for x. x = 10 + 2 | ||
| | Multiply both sides by zero. While it is truthful that 0 = 0, and you may be tempted to conclude that ten is true of all real numbers, that is not the example. | ||
| Bank check: Ameliorate Method: | For example, check and see if ten = 3 will solve the equation. Clearly three never equals 5, so x = 3 is not a solution. The equation has no solutions. It was not helpful to have multiplied both sides of the equation by zero. It would have been better to have started by subtracting ten from both sides, resulting in 0 = two, resulting in a false statement telling usa that at that place are no solutions. | ||
| Answer | In that location is no solution. | ||
In solving the algebraic equation ii(x – v) = 2x + 10, you cease up with −10 = ten. What does this mean?
A) ten = − 10 and 10
B) There is no solution to the equation.
C) You must have made a mistake in solving the equation.
D) x = all existent numbers
Evidence/Hide Reply
A) ten = − 10 and ten
Incorrect. Whatever solution to an equation must satisfy the equation. If yous substitute −10 into the original equation, you get −30 = −10. If you substitute 10 for x in the original equation, you go 10 = 30. The right reply is: There is no solution to the equation.
B) There is no solution to the equation .
Correct. Whenever you end up with a false statement like −x = 10 it means there is no solution to the equation.
C) Y'all must have made a error in solving the equation.
Wrong. A fake argument like this looks like a fault and information technology'southward always good to check the answer. In this case, though, there is not a mistake in the algebra. The correct reply is: There is no solution to the equation.
D) 10 = all existent numbers
Wrong. If you lot substitute some real numbers into the equation, you volition see that they practise non satisfy the equation. The correct respond is: There is no solution to the equation.
Advanced Question
How many solutions are there for the equation:
A) In that location is 1 solution.
B) At that place are two solutions.
C) At that place are an infinite number of solutions.
D) In that location are no solutions.
Show/Hide Answer
A) There is one solution.
Wrong. Endeavour substituting any value in for y in this equation and think about what you lot find. The right answer is: At that place are an infinite number of solutions to the equation.
B) At that place are two solutions.
Incorrect. Attempt substituting whatever two values in for y in this equation and think about what y'all find. When dealing with sets of parentheses, brand certain to evaluate the inner parentheses kickoff, and so move to the outer set. The correct reply is: There are an infinite number of solutions to the equation.
C) There are an space number of solutions.
Correct. When you lot evaluate the expressions on either side of the equals sign, you lot go 
D) There are no solutions.
Incorrect. Call back that statements such equally 3 = 5 are indicative of an equation having no solutions . The right answer is: There are an infinite number of solutions to the equation.
The power of algebra is how it tin can help y'all model existent situations in order to answer questions virtually them. This requires you to be able to translate real-world bug into the language of algebra, and and then be able to interpret the results correctly. Let's start by exploring a simple discussion trouble that uses algebra for its solution.
Amanda'due south dad is twice as sometime every bit she is today. The sum of their ages is 66. Use an algebraic equation to find the ages of Amanda and her dad.
One way to solve this problem is to use trial and error—you tin pick some numbers for Amanda's age, calculate her father'south age (which is twice Amanda's age), and and then combine them to see if they work in the equation. For example, if Amanda is 20, then her begetter would be 40 considering he is twice as old as she is, but then their combined historic period is lx, non 66. What if she is 12? 15? 20? Equally you tin run across, picking random numbers is a very inefficient strategy!
Yous can represent this situation algebraically, which provides some other fashion to detect the reply.
| Example | ||||||
| Problem | Amanda's dad is twice as old as she is today. The sum of their ages is 66. Find the ages of Amanda and her dad. | |||||
| We demand to detect Amanda's age and her male parent's age. | What is the problem asking? | |||||
| | Assign a variable to the unknown. The father's age is 2 times Amanda'due south age. | |||||
| | Amanda's age added to her male parent's age is equal to 66. | |||||
| | Solve the equation for the variable. | |||||
| | Apply Amanda's age to find her male parent's age. | |||||
| | Do the answers make sense? | |||||
| Answer | Amanda is 22 years onetime, and her begetter is 44 years one-time. | |||||
Let's try a new problem. Consider that the rental fee for a landscaping automobile includes a erstwhile fee plus an hourly fee. You lot could utilize algebra to create an expression that helps you determine the total toll for a variety of rental situations. An equation containing this expression would be useful for trying to stay within a fixed expense budget.
| Example | ||
| Trouble | A landscaper wants to rent a tree stump grinder to prepare an area for a garden. The rental company charges a $26 one-fourth dimension rental fee plus $48 for each hour the automobile is rented. Write an expression for the rental cost for any number of hours. | |
| The problem asks for an algebraic expression for the rental cost of the stump grinder for any number of hours. An expression will have terms, one of which will contain a variable, just it volition not contain an equal sign. | What is the trouble request? | |
| Look at the values in the problem: $26 = i-time fee $48 = per-hour fee Call back almost what this means, and effort to identify a pattern. i hr rental: $26 + $48 2 hr rental: $26 + $48 + $48 3 60 minutes rental: $26 + $48 + $48 + $48 Discover that the number of "+ $48" in the problem is the same every bit the number of hours the machine is being rented. Since multiplication is repeated addition, you lot could as well represent it similar this: one hr rental: $26 + $48(1) ii 60 minutes rental: $26 + $48(ii) 3 hr rental: $26 + $48(3) | What information is of import to finding an answer? | |
| Now let'due south employ a variable, h, to represent the number of hours the machine is rented. Rental for h hours: 26 + 48h | What is the variable? What expression models this situation? The total rental fee is determined by multiplying the number of hours by $48 and adding $26. | |
| Respond | The rental price for h hours is 26 + 48h. | |
Using the information provided in the trouble, you lot were able to create a general expression for this relationship. This means that you can find the rental cost of the machine for any number of hours!
Let's apply this new expression to solve another problem.
| Case | ||
| Trouble | A landscaper wants to hire a tree stump grinder to prepare an surface area for a garden. The rental company charges a $26 one-fourth dimension rental fee plus $48 for each hour the machine is rented. What is the maximum number of hours the landscaper tin rent the tree stump grinder, if he can spend no more than than $290? (The machine cannot be rented for role of an 60 minutes.) | |
| 26 + 48h, where h = the number of hours. | What expression models this state of affairs? | |
| | Write an equation to help you notice out when the expense equals $290. Solve the equation. | |
| | Check the solution. | |
| | Translate the answer. | |
| Respond | The landscaper can rent the machine for 5 hours. | |
Information technology is oftentimes helpful to follow a listing of steps to organize and solve application bug.
Solving Awarding Problems
Follow these steps to translate problem situations into algebraic equations you tin solve.
1. Read and empathise the problem.
2. Determine the constants and variables in the trouble.
3. Write an equation to represent the problem.
4. Solve the equation.
5. Cheque your answer.
6. Write a sentence that answers the question in the awarding problem.
Let's try applying the problem-solving steps with some new examples.
| Example | |||
| Problem | Gina has found a great price on paper towels. She wants to stock upward on these for her cleaning business. Paper towels cost $1.25 per bundle. If she has $60 to spend, how many packages of newspaper towels tin can she buy? Write an equation that Gina could use to solve this problem and show the solution. | ||
| The problem asks for how many packages of paper towels Gina can purchase. | What is the problem asking yous? | ||
| The paper towels cost $one.25 per package. Gina has $lx to spend on paper towels. | What are the constants? | ||
| Let p = the number of packages of paper towels. | What is the variable? | ||
| | What equation represents this situation? Solve for p . Carve up both sides of the equation by i.25 60 ÷ 1.25 = half-dozen,000 ÷ 125 five 00 1,000 1,000 0 | ||
| | Check your solution. Substitute 48 in for p in your equation. | ||
| Answer | Gina can purchase 48 packages of paper towels. | ||
| Example | ||
| Trouble | Levon and Maria were shopping for candles to decorate tables at a eating house. Levon bought 5 packages of candles plus three single candles. Maria bought 11 single candles plus 4 packages of candles. Each package of candles contains the same number of candles. After finishing shopping, Maria and Levon realized that they had each purchased the same exact number of candles. How many candles are in a package? | |
| The problem asks how many candles are independent in 1 packet. | What is the problem asking yous? | |
| Levon bought five packages and 3 single candles. Maria bought 4 packages and eleven single candles. | What are the constants? | |
| Let c = the number of candles in ane packet. | What is the variable? | |
| | What expression represents the number of candles Levon purchased? What expression represents the number of candles Maria purchased? What equation represents the state of affairs? | |
| Maria and Levon bought the same number of candles. | ||
| | Solve for c . Subtract 4c from both sides. Subtract 3 from both sides. | |
| | Cheque your solution . Substitute viii for c in the original equation. | |
| Answer | There are 8 candles in one package of candles. | |
| Advanced Example | ||
| Problem | The coin from two vending machines is being nerveless. One machine contains 30 dollar bills and a agglomeration of dimes. The other machine contains 38 dollar bills and a bunch of nickels. The number of coins in both machines is equal, and the amount of money that the machines collected is also equal. How many coins are in each machine? | |
| The problem asks how many coins are in each machine. | What is the problem request you? | |
| I automobile has thirty dollar bills and a bunch of dimes. Some other car has 38 dollar bills and a bunch of nickels—the same number of coins as the get-go machine. | What are the constants and what are the unknowns? | |
| Let c = the number of coins in each machine. | What is the variable? | |
| | What expression represents the corporeality of coin in the first machine? | |
| | What expression represents the amount of money in the second automobile? | |
| | What equation represents the situation? The corporeality of money in both machines is the same. | |
| | Solve for c . | |
| | Check your solution . Substitute 160 for c in the original equation. | |
| Answer | In that location are 160 coins in each car. | |
Avant-garde Question
Albert and Bryn are buying candy at the corner store. Albert buys v numberless and 3 individual pieces; Bryn buys 3 bags then eats 2 pieces of candy from one of the numberless. Each bag has the same number of pieces of processed.
Later on Bryn eats the ii pieces, she has exactly half the number of pieces of candy as Albert. How many pieces of candy are in each handbag?
Selection the equation that could exist used to solve the problem higher up. Employ the variable b to represent the number of pieces of processed in one bag.
A) 
B) 
C) 
D) 
Show/Hibernate Respond
A)
Incorrect. For the equation
B)
Incorrect. For the equation
C)
Correct. The corporeality of processed that Albert has tin can exist represented by 5b + 3, and the amount of processed Bryn has tin can exist represented by iiib – 2. Since Bryn has one-half as much every bit Albert, the concluding equation is
D)
Wrong. For the equation
Summary
Some equations are considered special cases. These are equations that have no solution and equations whose solution is the set of all existent numbers. When you apply the steps for solving an equation, and y'all become a faux statement rather than a value for the variable, there is no solution. When you utilise the steps for solving an equation, have avoided multiplying both sides of the equation past zero, and you go a true statement rather than a value for the variable, the solution is all real numbers. Algebra is a powerful tool for modeling and solving real-world problems.
Source: http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U10_L1_T3_text_final.html
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