Solve Equations with Inverse Operations
This page will provide you with all of the information to get a 100% solid foundation in solving equations in one-variable. This is a fundamentally important concept, and one where you also need procedural proficiency. Teacher-Speak aside, that means you have to know what you’re doing, and you have to be good at doing it!
Take your time working through each section. Read the article for each, try the problems, give yourself time to think before moving on to the next section. If you have questions, please let me know. Send an email to: [email protected]
If you haven’t worked through the information explaining what Algebra is, or how to solve equations logically, please consider doing so. You can find the links here: What is Algebra, Solving Equations Logically.
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On this page we will cover the basics of how to apply inverse operations to solve equations. Work through each section carefully, and patiently. Learning math takes time. Try the sets of practice problems offered with each section.
Algebra Unit
In the first section we saw how mathematics is a language, complete with its own set of rules. The rules allow us to communicate ideas, not at all unlike how punctuation and phonetics allow us to communicate with written English. In both languages we use symbols to represent ideas and the spatial arrangement of those symbols is how we organize those ideas.
Q: What is an equation?
An equation is a literal statement where both sides of the equal sign have the same value. You are as tall as I am, or, we both have twelve dollars, are examples of equations written in English. Typically, we don’t use words like you and I, or height, or we, in Algebra. We would use symbols to represent those things.
Let’s see how You are as tall as I am, would work in a mathematical statement. We’d first select variables for “you,” and for “I.” Let x = I, and y = you.
x cm = y cm
This is what would likely be written. The units, centimeters (cm), indicate that we are talking about a one-dimensional measurement. A short written statement contextualizing the equation often accompanies the equation. It might read something like,
Given that x is my height, in centimeters, and y is your height, in centimeters …
Here’s the take-away: In math we don’t write fiction. The statements are literal statements of truth. We use the properties of those statements to find deeper meaning. Solving equations is often how we do that. (A solution is a value that makes a statement true.)
If you’re going to solve an equation there are two things you can try.
- Read the equation and with knowledge of arithmetic, figure it out.
- That’s what we practiced in the previous section.
- Apply inverse operations.
- That’s what we’ll focus on here.
In the previous section we talked about how Algebra is the application of the linguistic rules of mathematics to find solutions through regression (going backwards). The rules of mathematics are just the order in which operations are performed. We multiply before we subtract, we perform any grouped operations first, and so on.
There are four operations: × ÷ + –, multiply, divide, add and subtract. We perform those operations in that order. How, we do groups first, and if there is an exponent or square root, that goes after grouped operations, but before the other operations. That’s what PEMDAS means.
P: Groups E: Exponents M: Multiplication, D: Division, A: Addition, S: Subtraction
When we want to solve an equation using the rules of math, we perform those operations in reverse: SADMEP. Let’s see an example.
Solve the equation: 3x – 5 = 7.
Read it carefully. What’s the solution?
Let’s look at the operations here.
FACT: We have a solution when the desired unknown or variable is isolated. Here, if x is by itself, has an exponent of one, and no other terms containing a factor of x exist, we have solved for x.
In 3x – 5 = 7, the seven does not need any attention. It is on the other side of the equals sign. The three and five need attention. We just need to identify the mathematical operation that is being performed between those numbers and x, then do the opposite operation, in inverse order.
The 3 is multiplying and the five is subtracting. You can write them next to the appropriate letter in SADMEP. This will keep your order of inverse operations correct while you’re learning.
Now, if you perform those inverse operations in the correct order, we will end up with “one times x plus zero equals …”
This might seem obvious, but you likely hold a conceptual gap you’ve always taken for granted. Read this part carefully.
The reason we add five to both sides is because 5 – 5 is zero. The step makes us feel like we moved five to the other side. What we actually did is add five to both sides. For that matter, you can add or subtract any number from both sides without issue.
We don’t use the division symbol (÷), often in math because it is difficult to manipulate. The spatial arrangement of mathematics allows us to perform some operations easily. That’s why we use fractions for division in Algebra. Here we just divide 3 by 3, and 12 by 3 to arrive at the following.
This is a very important lesson that carefully walks you through so you come away knowing WHAT it is you’re doing, and WHY. Click the picture above to download the PowerPoint.
With a $1.00 purchase of the packet you help support this website. What you get in this packet is:
- Complete, easy to follow notes
- PowerPoint
- Assignment
- Answer Key with worked out problems and explanations
The second installment of how to use inverse operations to solve equations is slightly more complicated. The ideas at play are the same, but you need to be sure you’re 100% solid with part 1 before continuing. Use the tabs to read the text, and practice the problems.
We will begin with another example where we write out SADMEP and use that to help us identify the operations and their order. Using this structure will help you to stay on track and not make as many mistakes. Over time, you will not need to write out SADMEP, and you may not need to already.
As you can see, you cannot use properties of arithmetic to logically figure out the value of x. We are forced to use inverse operations here. In the space below, write the numbers or values that are keeping the x from being isolated, and the operations they are performing with the x.
Did you see that the 5b is one number, that is subtracting from x? The three is an exponent, the a is multiplying and the c is dividing. Did you also catch that ax3 – 5b is in a group? That means we will be multiplying both sides by c first!
On the left of the equals sign the cs reduce to one. On the right, we wrote cy because it is in alphabetical order. This is not super important, the product of y and c does not change. It is a good habit to write the variables alphabetically when easy.
On the left, 5b – 5b = 0, and we don’t need to write + 0. On the right we placed the terms alphabetically (which descending order).
When dividing both sides of the equal sign by a, the as reduce to one on the left. On the right we cannot reduce anything because there are two terms in the numerator and they do not share a common factor with the term in the denominator.
A quick work about exponents. When we say we have solved for x when x is isolated, that means we really have:
We have an x3. The reason cube root inverse of cube is because cube root is looking for the number cubed that is the radicand. The radicand is a cubed number. That’s almost as tricky as answering, what color was George Washington’s white horse? The answer is self-evident, but it seems like a trap.
Another reason that you can see why cube and cube-root are inverses is when we consider cube root as a rational exponent. Remember that the exponent of one-third is the same as cube root.
If we start with x3 and raise that to the power of 1/3, we have the follow.
When “a power is raised to a power, we multiply the exponents,” is the rule. The product of three and one-third is just one.
Let’s apply that to our equation.
Q: How do you know when you’re done solving for x?
If x, or another variable you’re solving for, is isolated, and the other side of the equal sign does not have a factor of x, you’re done. Basically, you need x = something without an x.
Q: How do you know if you’re right?
If you can substitute your solution back into the equation, for every x that exists, and through simplification you find the equation to be true, then you’re done.
The cube root being cubed leaves an exponent of one for the entire group.
The coefficient of a reduces with the denominator. The removes the need for parenthesis.
Combining like terms, which is an application of the commutative property of addition.
The cs reduce to one.
Since solving uses inverse operations, which is the application of the math backwards, checking can be worth your efforts. It will improve your understanding of the application of the rules of mathematics (the order of operations) which will help you apply inverse operations.
There are three things to know here, so far.
- What to do, in what order.
- How to do it.
- Why?
The what to do, and in what order, is the inverse of each operation being performed between the variable you wish to isolate and other terms. The order in which to do those operations is the inverse of the order of operations, SADMEP.
How to do it involves basic arithmetic. If you understand the arithmetic, you should be okay. If you get stuck, you will likely need to review whatever it is that caused your trouble.
The reason why we do this is we are using the rules of mathematics in reverse to find hidden meaning. Mathematics uses equations and inequalities to demonstrate relationships. We can manipulate those relationships to find more information than is obvious.
Last Issue: What do you do if the variable you need is in a fraction, in the denominator? For example:
Eventually we will need to get the x out of the denominator. The term “3 over x,” really means what, mathematically? If you can answer that question, then you can apply inverse operations to the situation and get x in the numerator.
In “3 over x,” the three is being divided by x. Be warned, this is not x divided by 3. You cannot switch the order at all with division. The opposite of division by x is multiplication by x. If you do that initially, you’ll need to make sure to multiply all terms, on both sides of the equal sign by x, not just the “three over x.”
If we look at the right process where we end up with 3 + x = 9x, we have two xs, and they’re separated by the equal sign. We need to use inverse operations to get them on the same side. It is a best practice to move the term with x that has the smaller coefficient.
The PowerPoint that contains the lesson for Part 1 begins by reviewing a problem from the homework assignment in the previous section, Solve by Reading. The lesson walks you carefully through application of inverse operations and leaves you with plenty of problems to practice on your own.
To download the PowerPoint, please click the icon below.
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The third part of solving equations with inverse operations is where things can begin to get tricky. Here, the variable we need will be stuck in a group. Your steps will depend on the situation. We will cover several options here, but experience will be key to your success! Practice lots of these.
Groups
There are a couple of specific situations that can further muddy the waters for you that we can discuss now. As you move forward in math you’ll develop knowledge and ability with more complicated mathematics, but the principles here are always at the heart of solving equations, no matter the application.
The trickiest equations to solve involve situations where x, or whatever you’re solving for, is inside of a grouped operation. Before you can even begin to use inverse operations to isolate the unknown value you have to perform operations to eliminate that grouping. Sometimes the operations you’ll use with be inverse, sometimes just the normal order of operations. Experience is key. Let’s run through some examples.
The first situation that is tricky is when the unknown you need to isolate is in the denominator of an expression. The second is when more than one of those unknowns exist. Let’s get into it, shall we.
There are short-cuts that you can apply here, namely the one called, “Cross Multiply.” However, that is a trick, it is not mathematical at all. It is a pattern describing what happens with exactly ONE type of situation. Let’s learn it right so that we don’t have to try to track a million different tricks, so we only have to know one thing!
Since x is in the denominator, we need to multiply both sides by the group containing x.
On the left, the 2 – x will reduce with 2 – x, leaving just 3. On the right, we have x now in the numerator, but it is in a group. To remove the group, we must take care of the 5.
3 = 5(2 – x)
There are two ways to take care of the 5. We can distribute and get 10 – 5x, or we can divide by sides by 5, giving us 3/5 = 2 – x. It doesn’t matter, the choice is yours. If the 3/5 reduced, I’d say try that method, but since it doesn’t I’d go with the first option.
3 = 10 – 5x
If you’re this far in this section, you should be 100% rock solid with solving an equation like this. The answer is – 7/5.
You try this one.
If you multiplied both sides by x first, you’re off to a good start. However, if you distributed the x, then you probably got stuck. It is not a wrong step, but it is a step that doesn’t align with your objective of getting x by itself. Try it again if you got stuck, but don’t distribute the x in your second step. Recognize that (8 + b) is multiplying with x. It is grouped together, so you can treat it as a single number. Your answer will be (a + b) ÷ (8 + b). Do you know why that cannot be reduced? All terms must have a common factor.
Our second example where x is stuck in a group looks like the equation below. We have multiple terms (parts) with x as a factor, but there are also a pair of xs inside parenthesis. For example:
In these situations, we often have to manipulate the equations mathematically before we can start using Algebra. When manipulating the equations, which will involve distributing and combining like terms, we must follow the order of operations (PEMDAS). Then, when we start to solve, we use inverse operations (SADMEP). Let’s get into this example.
Here we distributed the 9 and the 7x. This allowed us to get the xs out of their respective groups. We could not really divide by 9 and by 7 here unless we divided each and every term by 63 (9 × 7). That would be ugly!
There aren’t any like terms to be combined yet, but we do have a negative 7x2 on each side. So let’s add a 7x2 to each side, giving us the following.
We can do that because of the commutative property of addition. That’s why combining like terms works, too.
Now, we need to collect those terms with xs together. We have two options. We can subtract a 9x from each side, or subtract a 7x from each side. It is almost always easier to “move,” the smaller term, the term with the smallest coefficient. It leaves us with a positive coefficient, and positives are easier to deal with. Let’s look at both examples so you can see.
The example on the right needs to be simplified, it has one more step. In addition, when dealing with negative numbers we are inviting sign errors due to carelessness. It happens too frequently. So it is best to move the x with the smaller coefficient.
To be clear, if you’re unknown value is inside the parenthesis, with a coefficient to be distributed, you have two options. You can distribute, or you can divide ALL terms by that coefficient. The division is GREAT if all terms will reduce. We just saw an example where reducing would not happen. Let’s see an example where it would happen.
Do you see how we can divide ALL terms by 3?
The third type of problem we will discuss that has our variable trapped inside of a group involves radical expressions.
A typical mistake is to multiply both sides by seven. That does not follow the rules of the language of mathematics because that 7 is in a group, and the entire group is inside the cube-root. The inverse of cube-root is to cube the expression. So both sides must be cubed.
A variation of this problem that can get sticky is below.
While it is possible to correctly cube both sides of the equation during the first step, it is most likely that operation will be done incorrectly. To do the cubing of the left side correctly would be quite difficult. Let’s see why.
Remember, exponents do not distribute. The entire base will be multiplied by itself three times. So the left side of the equal sign would be:
So in a case like this we will need to add one to both sides. Then, we have the same equation we had to begin with. Do you see how experience is key? There are so many possible variations that it is impossible to have rehearsed them all. Instead you need a solid understanding of mathematics and a sense of how to manipulate things algebraically.
Q: How many solutions could exist?
Infinite Solutions: Let’s go back to the basics, reading and interpreting the mathematics for what it means. Consider the equation x + 1 = x + 1. What could x be? Anything, right? Any number you choose for x will hold true!
What if you used inverse operations and subtracted x from both sides? You get a true statement, regardless of the existence of x. If you don’t catch that there are infinitely many solutions while performing inverse operations and you end up with a true statement like 3 = 3, then your equation has infinitely many solutions.
No Solutions, Indeterminate, Undefined: Consider the equation x + 2 = x + 1. Can you think of a value for x that would resolve this equation, make it true? What if you used inverse operations and subtracted x from both sides? You end up with 2 = 1. This is nonsense of course, and exactly the nonsense that inspired the store Alice in Wonderland, seriously!
If you end up with a situation that is false, like 2 = 1, double check your working. If you were right, then you have an impossible equation, one that does not have a solution.
Step Back
If you are an Algebra 1 student, we have covered 95% (a made up statistic, like 78.2% of all statistics), of the types of equations you’ll be required to solve. However, it is often the case you’ll be expected to manipulate equations and expressions algebraically to uncover a hidden truth. The skill of being able to solve an equation algebraically is not a stand-alone ability. It is widely applied.
As your knowledge of mathematics grows you will be applying the ideas and skills learned here in new ways.
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Here we will solve basic rational equations. The problems covered will not involve quadratic equations or factoring of polynomials. That will come later.
Note: The problems covered in this article will not involve quadratics or factoring of polynomials.
The most difficult topic to date has most likely been Algebraic Fractions. When you had to manipulate something like the expression below, it was tricky.
Common denominators can be daunting with variables because they involve finding the LCM of abstractions like x. Let’s review a few fundamental facts to help make this easier.
First: The variable x is prime because we can only factor it as 1 × x. However, x2 is not prime. Its factors are: 1 × x × x.
Second: The LCM is the smallest number that two other numbers (or more) divide into evenly. For x and x2, the LCM is x2. This true because x divides into x2, x times. (x × x = x2). And of course, x2 divides into itself one time. The LCM will be used A LOT with this unit. It is used to get common denominators, and for a new “trick,” that you’ll see shortly.
Third: The GCF is the largest factor that a set of numbers share. For x and x2, the GCF is x. Remember that to reduce an Algebraic Fraction, all terms must have a common factor.
The New Idea: One thing that almost always will help make these types of problems simpler is getting a common denominator, at least on one side of the equal sign. If not that, then perhaps all terms on both sides of the equal sign can be multiplied by the LCM of the denominators. This will give you an opportunity to reduce and eliminate all of the denominators.
WARNING: Just because the process on the right column was easier for this problem does NOT mean that is always the easiest path to a solution. Each problem will present a different opportunity, so you need to know as many approaches as possible.
Let’s work through a few problems. Some we will see multiple approaches to, others we will just explore the most efficient approach.
Example 2: Solve for x.
For this example, let’s get a common denominator on the left side of the equal sign.
Notice the parenthesis around the x + 4? Why are they there? To help you answer that question, which of the following two are the correct next step?
The original problem involves subtraction of algebraic fractions. These negatives are the most problematic of all negatives! We are subtracting x + 4. Mathematically that is written – (x + 4), which is –x – 4. The first one was right.
This is still a super tricky problem because we have an x stuck in the denominator. But, the term 14x is dividing into the numerator. We can just multiply all terms, on both sides of the equal sign, by 14x. On the left, the 14x will reduce.
Once you’re at this point, collect the xs on one side and you’re almost home.
Let’s not get so focused on the details of individual problems that we lose sight of the new, big idea here. When solving these rational equations, you will need to find the LCM. Whether that LCM is used to find common denominators or multiply by all terms to reduce the denominators to one depends on the decision you make. Sometimes one method is easier than the other and you’ll not always know until you try.
At this point it unlikely that more examples will help, until you try some on your own. Try these two problems. Solve for x.
If you decided to try a different approach, your approach might be fine. Just because you didn’t perform the steps below does not mean you are wrong. If you arrived at a different answer, it is likely you made a procedural error.
Here we will do this problem by multiplying all terms by the LCM, and then reducing.
At this point you should be good to go! x = – 1/5
Let’s look at problem B)
For this problem we will start with a common denominator.
Because we will need to multiply both sides by the denominator, we will distribute the 4. Then, combining like terms we get the following.
Now to multiply both sides by 4x – 8, to get the term with the x out of the denominator.
Be sure to distribute and always watch for sign errors when distributing.
At this point you should be good to go.
Summary
Equations with algebraic fractions are called rational equations. If the unknown you’re solving for is in the denominator, you’ll have to take one of two steps. You can either get a common denominator and combine like terms, which will allow you to easily multiply both sides by the denominator, or you can multiply all terms by the LCM of the denominators.
For the equation on the left, multiplying all terms by the LCM of the denominators would be an easy approach. For the equation on the right, multiplying by the LCM might be more difficult than getting a common denominator. You need to be good at both approaches.
One last word on negatives. It is super easy to look past a negative sign and make a sign error, especially in cases where the fractions are subtracting and the second numerator has more than one term. Here’s an example.
The entire numerator, 3x + 1, is being subtracted, making it -3x – 1. To help make this clear, consider
The one is subtracted from the three, nothing happens with the denominator. You can read more about why in the section about order of operations.
This is the introduction to solving rational equations. We start carefully so that we can build a solid foundation for this important topic.
In the lesson you’ll learn two ways that you can use the LCM of the denominators to help you solve the equations. The idea is that we want to manipulate the terms in order to eliminate the denominators, at which point we can more easily navigate inverse operations.
To download the PowerPoint, click the picture above, or click this link: PowerPoint.
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