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Algebra Topics  - Simplifying Expressions

Algebra topics  -, simplifying expressions, algebra topics simplifying expressions.

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Algebra Topics: Simplifying Expressions

Lesson 7: simplifying expressions.

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Simplifying expressions

Simplifying an expression is just another way to say solving a math problem . When you simplify an expression, you're basically trying to write it in the simplest way possible. At the end, there shouldn't be any more adding, subtracting, multiplying, or dividing left to do. For example, take this expression:

If you simplified it by combining the terms until there was nothing left to do, the expression would look like this:

In other words, 15 is the simplest way to write 4 + 6 + 5 . Both versions of the expression equal the exact same amount; one is just much shorter.

Simplifying algebraic expressions is the same idea, except you have variables (or letters) in your expression. Basically, you're turning a long expression into something you can easily make sense of. So an expression like this...

(13x + -3x) / 2

...could be simplified like this:

If this seems like a big leap, don't worry! All you need to simplify most expressions is basic arithmetic -- addition, subtraction, multiplication, and division -- and the order of operations.

The order of operations

Like with any problem, you'll need to follow the order of operations when simplifying an algebraic expression. The order of operations is a rule that tells you the correct order for performing calculations. According to the order of operations, you should solve the problem in this order:

  • Parentheses
  • Multiplication and division
  • Addition and subtraction

Let's look at a problem to see how this works.

In this equation, you'd start by simplifying the part of the expression in parentheses : 24 - 20 .

2 ⋅ (24 - 20) 2 + 18 / 6 - 30

24 minus 20 is 4 . According to the order of operations, next we'll simplify any exponents . There's one exponent in this equation: 4 2 , or four to the second power .

2 ⋅ 4 2 + 18 / 6 - 30

4 2 is 16 . Next, we need to take care of the multiplication and division . We'll do those from left to right: 2 ⋅ 16 and 18 / 6 .

2 ⋅ 16 + 18 / 6 - 30

2 ⋅ 16 is 32 , and 18 / 6 is 3 . All that's left is the last step in the order of operations: addition and subtraction .

32 + 3 - 30

32 + 3 is 35 , and 35 - 30 is 5 . Our expression has been simplified—there's nothing left to do.

That's all it takes! Remember, you must follow the order of operations when you're performing calculations—otherwise, you may not get the correct answer.

Still a little confused or need more practice? We wrote an entire lesson on the order of operations. You can check it out here .

Adding like variables

To add variables that are the same, you can simply add the coefficients . So 3 x + 6 x is equal to 9 x . Subtraction works the same way, so 5 y - 4 y = 1 y , or just y .

5 y - 4 y = 1 y

You can also multiply and divide variables with coefficients. To multiply variables with coefficients, first multiply the coefficients, then write the variables next to each other. So 3 x ⋅ 4 y is 12 xy .

3x ⋅ 4y = 12xy

The Distributive Property

Sometimes when simplifying expressions, you might see something like this:

Normally with the Order of Operations, we would simplify what is inside the parentheses first. In this case, however, x+7 can't be simplified since we can't add a variable and a number. So what's our first step?

As you might remember, the 3 on the outside of the parentheses means that we need to multiply everything inside the parentheses by 3. There are two things inside the parentheses: x and 7 . We'll need to multiply them both by 3.

3(x) + 3(7) - 5

3 · x is 3x and 3 · 7 is 21 . We can rewrite the expression as:

3x + 21 - 5

Next, we can simplify the subtraction 21 - 5. 21 - 5 is 16 .

Since it's impossible to add variables and numbers, we can't simplify this expression any further. Our answer is 3x + 16 . In other words, 3(x+7) - 5 = 3x+16.

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Simplifying Algebraic Expressions - Practice Problems

Now that you've studied the three detailed examples for Simplfying Algebraic Expressions, you are ready to try some on your own! If you haven't studied this lesson yet, click here.

Be very careful as you simplify your terms and make sure that you always take the sign in front of the term as you move things around!

Practice Problems

Simplify each expression to lowest terms.

1. 2a - 4b +3ab -5a +2b

2. 4(2x+1) - 3x

3. 4(p - 5) +3(p +1)

4. 6(p +3q) - (7 +4q)

5. 4rs -2s - 3(rs +1) - 2s

1. -3a - 2b + 3ab

4. 6p + 14q - 7

5. rs - 4s - 3

You're doing a great job! Remember... Algebra skills build on one another, so make sure that you really understand each concept!

problem solving simplifying algebraic expressions

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The algebra section of QuickMath allows you to manipulate mathematical expressions in all sorts of useful ways. At the moment, QuickMath can expand, factor or simplify virtually any expression, cancel common factors within fractions, split fractions up into smaller ('partial') fractions and join two or more fractions together into a single fraction. More specialized commands are on the way.

What is algebra?

Algebra is the branch of elementary mathematics which uses symbols to stand for unknown quantities. In a more basic sense, it consists of solving equations or manipulating expressions which contain symbols (usually letters, like x, y or z) as well as numbers and functions. Although solving equations is really a part of algebra, it is such a big area that it has its own section in QuickMath.

This part of QuickMath deals only with algebraic expressions. These are mathematical statements which contain letters, numbers and functions, but no equals signs. Here are a few examples of simple algebraic expressions :

The expand command is used mainly to rewrite polynomials with all brackets and whole number powers multiplied out and all like terms collected together. In the advanced section, you also have the option of expanding trigonometric functions, expanding modulo any integer and leaving certain parts of the expression untouched whilst expanding the rest.

Go to the Expand page

The factor command will try to rewrite an expression as a product of smaller expressions. It takes care of such things as taking out common factors, factoring by pairs, quadratic trinomials, differences of two squares, sums and differences of two cubes, and a whole lot more. The advanced section includes options for factoring trigonometric functions, factoring modulo any integer, factoring over the field of Gaussian integers (just the thing for those tricky sums of squares), and even extending the field over which factoring occurs with your own custom extensions.

Go to the Factor page

Simplifying is perhaps the most difficult of all the commands to describe. The way simplification is performed in QuickMath involves looking at many different combinations of transformations of an expression and choosing the one which has the smallest number of parts. Amongst other things, the Simplify command will take care of canceling common factors from the top and bottom of a fraction and collecting like terms. The advanced options allow you to simplify trigonometric functions or to instruct QuickMath to try harder to find a simplified expression.

Go to the Simplify page

The cancel command allows you to cancel out common factors in the denominator and numerator of any fraction appearing in an expression. This command works by canceling the greatest common divisor of the denominator and numerator.

Go to the Cancel page

Partial Fractions

The partial fractions command allows you to split a rational function into a sum or difference of fractions. A rational function is simply a quotient of two polynomials. Any rational function can be written as a sum of fractions, where the denominators of the fractions are powers of the factors of the denominator of the original expression. This command is especially useful if you need to integrate a rational function. By splitting it into partial fractions first, the integration can often be made much simpler.

Go to the Partial Fractions page

Join Fractions

The join fractions command essentially does the reverse of the partial fractions command. It will rewrite a number of fractions which are added or subtracted as a single fraction. The denominator of this single fraction will usually be the lowest common multiple of the denominators of all the fractions being added or subtracted. Any common factors in the numerator and denominator of the answer will automatically be cancelled out.

Go to the Join Fractions page

Introduction to Algebraic Functions

The notion of correspondence is encountered frequently in everyday life. For example, to each book in a library there corresponds the number of pages in the book. As another example, to each human being there corresponds a birth date. To cite a third example, if the temperature of the air is recorded throughout a day, then at each instant of time there is a corresponding temperature.

The examples of correspondences we have given involve two sets X and Y. In our first example, X denotes the set of books in a library and Y the set of positive integers. For each book x in X there corresponds a positive integer y, namely the number of pages in the book. In the second example, if we let X denote the set of all human beings and Y the set of all possible dates, then to each person x in X there corresponds a birth date y.

We sometimes represent correspondences by diagrams of the type shown in Figure 1.17, where the sets X and Y are represented by points within regions in a plane. The curved arrow indicates that the element y of Y corresponds to the element x of X. We have pictured X and Y as different sets. However, X and Y may have elements in common. As a matter of fact, we often have X = Y.

problem solving simplifying algebraic expressions

A function f from a set X to a set Y is a correspondence that assigns to each element x of X a unique element y of Y. The element y is called the image of x under f and is denoted by f(x). The set X is called the domain of the function. The range of the function consists of all images of elements of X.

Earlier, we introduced the notation f(x) for the element of Y which corresponds to x. This is usually read "f of x." We also call f(x) the value of f at x. In terms of the pictorial representation given earlier, we may now sketch a diagram as in Figure 1.18. The curved arrows indicate that the elements f(x), f(w), f(z), and f(a) of Y correspond to the elements x, y, z and a of X. Let us repeat the important fact that to each x in X there is assigned precisely one image f(x) in Y; however, different elements of X such as w and z in Figure 1.18 may have the same image in Y.

problem solving simplifying algebraic expressions

Solution As in Example 1, finding images under f is simply a matter of substituting the appropriate number for x in the expression for f(x). Thus:

problem solving simplifying algebraic expressions

Many formulas which occur in mathematics and the sciences determine functions. As an illustration, the formula A = pi*r 2 for the area A of a circle of radius r assigns to each positive real number r a unique value of A. This determines a function f, where f(r) = pi*r 2 , and we may write A= f(r). The letter r, which represents an arbitrary number from the domain off, is often called an independent variable. The letter A, which represents a number from the range off, is called a dependent variable, since its value depends on the number assigned tor. When two variables r and A are related in this manner, it is customary to use the phrase A is a function of r. To cite another example, if an automobile travels at a uniform rate of 50 miles per hour, then the distance d (miles) traveled in time t (hours) is given by d = 50t and hence the distance d is a function of time t.

We have seen that different elements in the domain of a function may have the same image. If images are always different, then, as in the next definition, the function is called one-to-one.  

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How to Simplify Algebraic Expressions

Last Updated: April 7, 2023 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 11 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 612,471 times.

Learning how to simplify algebraic expressions is a key part of mastering basic algebra and an extremely valuable tool for all mathematicians to have under their belt. Simplification allows a mathematician to change a complex, long, and/or awkward expression into a simpler or more convenient one that's equivalent. Basic simplification skills are fairly easy to learn - even for the math-averse. By following a few simple steps it is possible to simplify many of the most common types of algebraic expressions without any sort of special mathematical knowledge at all. See Step 1 below to begin!

Understanding Important Concepts

Step 1 Define

  • For example, 3x 2 and 4x 2 are like terms because each contains the variable x raised to the second power. However, x and x 2 are not like terms because each term has x raised to a different power. Similarly, -3yx and 5xz are not like terms because each term has a different set of variables.

Step 2 Factor...

  • For example, if we wanted to factor 20, we might write it as 4 × 5 .
  • Note that variable terms can also be factored - 20x, for instance, can be written as 4(5x) .
  • Prime numbers can't be factored because they are only evenly divisible by themselves and 1.

Step 3 Use the acronym PEMDAS to remember the order of operations.

  • P arentheses
  • M ultiplication
  • S ubtraction

Combining Like Terms

Step 1 Write your equation.

  • As an example problem, for the next few steps, let's consider the expression 1 + 2x - 3 + 4x .

Step 2 Identify like terms.

  • For example, let's identify like terms in our equation 1 + 2x - 3 + 4x. 2x and 4x both have the same variable raised to the same exponent (in this case, the x's aren't raised to any exponent at all). In addition, 1 and -3 are like terms, as neither has any variables. So, in our equation, 2x and 4x and 1 and -3 are like terms.

Step 3 Combine like terms.

  • 2x + 4x = 6x
  • 1 + -3 = -2

Step 4 Create a simplified expression from your simplified terms.

  • In our example, our simplified terms are 6x and -2, so our new expression is 6x - 2 . This simplified expression is equal to the original (1 + 2x - 3 + 4x), but is shorter and easier to manage. It's also easier to factor, which, as we'll see below, is another important simplifying skill.

Step 5 Obey the order of operation when combining like terms.

  • 5(3x-1) + x((2x)/(2)) + 8 - 3x
  • 15x - 5 + x(x) + 8 - 3x
  • 15x - 5 + x 2 + 8 - 3x. Now , since the only operations left are addition and subtraction, we can combine like terms.
  • x 2 + (15x - 3x) + (8 - 5)
  • x 2 + 12x + 3

Step 1 Identify the greatest...

  • Let's use the equation 9x 2 + 27x - 3. Notice that every term in this equation is divisible by 3. Since the terms aren't all evenly divisible by any larger number, we can say that 3 is our expression's greatest common factor.

Step 2 Divide the terms in the expression by the greatest common factor.

  • 9x 2 /3 = 3x 2
  • Thus, our new expression is 3x 2 + 9x - 1 .

Step 3 Represent your expression as the product of the greatest common factor and the remaining terms.

  • For our example expression, 3x 2 + 9x - 1, we would enclose the expression in parentheses and multiply by the greatest common factor of the original equation to get 3(3x 2 + 9x - 1) . This equation is equal to the original, 9x 2 + 27x - 3.

Step 4 Use factoring to simplify fractions.

  • Let's substitute the factored form of our original expression for the expression in the numerator: (3(3x 2 + 9x - 1))/3
  • Notice that now, both the numerator and the denominator share the coefficient 3. Dividing the numerator and denominator by 3, we get: (3x 2 + 9x - 1)/1.
  • Since any fraction with "1" in the denominator is equal to the terms in the numerator, we can say that our original fraction can be simplified to 3x 2 + 9x - 1 .

Applying Additional Simplification Skills

Step 1 Simplify fractions by dividing through by common factors.

  • Doing so gets us ((5x 2 )/10) + x + 2. If we like, we may want to rewrite the first term as (1/2)x 2 to get (1/2)x 2 + x + 2.

Step 2 Use square factors to simplify radicals.

  • (√(9) × √(10))

Step 3 Add exponents when multiplying two exponential terms; subtract when dividing.

  • 6x 3 × 8x 4 + (x 17 /x 15 )
  • (6 × 8)x 3 + 4 + (x 17 - 15 )
  • 48x 7 + x 2
  • Multiplying exponential terms is essentially like multiplying long strings of non-exponential terms. For example, since x 3 = x × x × x and x 5 = x × x × x × x × x, x 3 × x 5 = (x × x × x) × (x × x × x × x × x), or x 8 .
  • Similarly, dividing exponential terms is like dividing long strings of non-exponential terms. x 5 /x 3 = (x × x × x × x × x)/(x × x × x). Since each term in the numerator can be canceled out by a matching term in the denominator, we're left with two x's in the numerator and none in the bottom, giving us an answer of x 2

Community Q&A

Community Answer

Video . By using this service, some information may be shared with YouTube.

  • Always remember that you have to think of these numbers as having positive and negative signs. Many people get stuck thinking "What sign should I put here?" Thanks Helpful 47 Not Helpful 29
  • Ask for help when needed! Thanks Helpful 41 Not Helpful 34
  • Simplifying Algebraic Expressions is nothing easy, but once you get the hang of it, you will use it all of your life. Thanks Helpful 36 Not Helpful 44

problem solving simplifying algebraic expressions

  • Always look for like terms and don't get tricked by exponents. Thanks Helpful 19 Not Helpful 21
  • Make sure you haven't accidentally added in some extra number, exponent, or operation that doesn't belong. Thanks Helpful 19 Not Helpful 23

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Solve Two Step Algebraic Equations

  • ↑ https://www.mathsisfun.com/definitions/like-terms.html
  • ↑ https://www.mathsisfun.com/algebra/factoring.html
  • ↑ https://www.purplemath.com/modules/orderops.htm
  • ↑ https://www.softschools.com/math/topics/combining_like_terms/
  • ↑ https://www.mathsisfun.com/algebra/like-terms.html
  • ↑ https://www.mathwarehouse.com/algebra/like-terms/how-to-combine-like-terms-in-math.php
  • ↑ https://www.freemathhelp.com/combining-like-terms.html
  • ↑ https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:rational-functions/x9e81a4f98389efdf:reducing-rational-expressions-to-lowest-terms/a/reducing-rational-expressions-to-lowest-terms
  • ↑ https://www.bbc.com/bitesize/guides/zwv9y4j/revision/1
  • ↑ https://mathbitsnotebook.com/Algebra1/Radicals/RADSimplifyingRadicals.html
  • ↑ https://www.montereyinstitute.org/courses/DevelopmentalMath/TEXTGROUP-9-14_RESOURCE/U11_L1_T2_text_final.html

About This Article

David Jia

To simplify algebraic expressions, start by identifying the like terms, which are terms that have the same variables and exponents. Then, combine the like terms by adding them together to get the simplified expression. You can also simplify the expression further by finding the greatest common factor and then dividing all of the terms in the expression by that number. After you've done that, put the expression in parentheses with the greatest common factor on the outside. To learn more ways you can simplify algebraic expressions, scroll down! Did this summary help you? Yes No

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Simplify: to make simpler!

One of the big jobs we do in Algebra is simplification .

You will often be asked to put something "in simplest form"

What is the Simplest Form?

In general, it is simpler when it is easier to use .

It is now a little easier to use.

"Half" is definitely simpler than "three sixths", unless it is important to know that something was cut into sixths.

That last example can be argued with! Some people say to remove parentheses to make it "simpler", but (x−3)(x+1) is usually a lot easier to use.

The moral of the story:

"Simplified" is sometimes obvious, but can also depend on what you want to do.

How to Simplify

There are many ways to simplify!

When we simplify we use similar skills to solving equations , and that page has some good advice.

Some of these things might help:

  • Combine Like Terms
  • Expand (the opposite of factoring)
  • Clear out fractions by multiplying
  • Find some pattern you have seen before, like the difference of squares .

And Which Is Simpler Here?

Here is one more interesting case:

Which is simpler? You decide!

problem solving simplifying algebraic expressions

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Table of Contents

Last modified on December 18th, 2023

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Simplifying algebraic expressions.

Simplifying algebraic expressions means rewriting the algebraic expressions in the simplest form with no like terms and without any operators like addition, subtraction, multiplication, and division.

By Combining Like Terms

The algebraic expressions involving the variables with whole number coefficients (no fractions, radicals, etc.) can be simplified by combining and solving the like terms (terms with the same variables and exponents). Then, the expression will only be left with the unlike terms that can not be simplified further.

For example,

Let us simplify the expression 15x – 1 – 3x + 8

Now, identifying the like terms, we get 15x and -3x, -1 and 8 as two pairs of the like terms.

Then, by combining them, we get 

= (15x – 3x) + (-1 + 8)

Solving them, we get, 

Thus, 15x – 1 – 3x + 8 can be simplified as 12x + 7.

By Order of Operations

To simplify the above algebraic expression, we follow the general order of operations PEMDAS – which stands for 

P – Parentheses, E – Exponents, M – Multiplication, D – Division, A – Addition, and S – Subtraction.

Parentheses

Now let us simplify the algebraic expression ${\left( x\times x^{2}\right) -\left( 4x\div 2\right) +3\times 9x-4\div 2}$

First, we solve the parentheses by dividing the terms inside the bracket; we get

= ${\left( x\times x^{2}\right) -2x+3\times 9x-4\div 2}$

Now, simplifying the terms containing exponents by the exponent rule (${a^{m}\div a^{n}=a^{m-n}}$, when ${a\neq 0}$), we get,

= ${x^{3}-2x+3\times 9x-4\div 2}$

Multiplication and Division

Next, we multiply and then divide according to the order of operations, and we get

= ${x^{3}-2x+27x-4\div 2}$

= ${x^{3}-2x+27x-2}$

Addition and Subtraction

Now, by combining the like terms, adding or subtracting them, and writing the unlike terms as they are, we get

= ${x^{3}+25x-2}$

While simplifying an expression, the result must be in the standard form (from the highest to the lowest power).

Thus, the given expression ${\left( x\times x^{2}\right) -\left( 4x\div 2\right) +3\times 9x-4\div 2}$ is simplified to ${x^{3}+25x-2}$.

Sometimes, if we have a variable and a number inside the bracket, then, using parentheses, we can not solve the terms inside the bracket. Then, we need to expand the expressions.

Expanding Using Distributive Property

The distributive property states that an expression given in the form of: 

x(y + z) can be simplified as xy + xz 

Or, x(y – z) can be simplified as xy – xz. 

Let us expand and simplify the expression 2(x + 4 + 3x) + 3(x – 5 + 7) – 2y 

First, we combine the like terms,

2{(x + 3x) + 4} + 3{x + (- 5 + 7)} – 2y

= 2(4x + 4) + 3(x + 2) – 2y

Now, using the distributive property, we get,

2(4x) + (2 x 4) + 3(x) + (3 x 2) – 2y

= 8x + 8 + 3x + 6 – 2y

Finally, by combining the like terms, we get

(8x + 3x) – 2y + (8 + 6)

= 11x – 2y + 14

Thus, 2(x + 4 + 3x) + 3(x – 5 + 7) – 2y can be simplified as 11x – 2y + 14.

Simplifying with Fractions

If fractions are given in any expression, then we use the exponent rules and the distributive property to simplify such expressions. 

 Let us simplify ${\dfrac{5}{2}\left( 2x+8\right) +\dfrac{y}{4}\left( 4+x\right)}$

Using distributive property, we get,

${\left( \dfrac{5}{2}\times 2x\right) +\left( \dfrac{5}{2}\times 8\right) +\left( \dfrac{y}{4}\times 4\right) +\left( \dfrac{y}{4}\times x\right)}$

= ${5x+\dfrac{xy}{4}+y+20}$

Thus, ${\dfrac{5}{2}\left( 2x+8\right) +\dfrac{y}{4}\left( 4+x\right)}$ can be simplified as ${5x+\dfrac{xy}{4}+y+20}$.

While simplifying an algebraic expression with a fraction, the fraction must be in the simplest form, and only the unlike terms are kept using any change.

Simplifying with Factors

Some expressions require factoring for simplification. In such an expression, we remove the common factors among all the terms and keep the remaining ones unchanged. 

Now, let us simplify the expression ${\dfrac{2}{5}\left( 10x^{2}+5x-35\right)}$

= ${\dfrac{2\times 5\left( 2x^{2}+x-7\right) }{5}}$

= ${2\left( 2x^{2}+x-7\right)}$

= ${4x^{2}+2x-14}$

Thus, the simplified algebraic expression is ${4x^{2}+2x-14}$.

However, we must remember some other rules to simplify an algebraic expression.

Other Rules

  • To add/subtract the like terms, we add/subtract the coefficients of those terms and write the common variable with it.
  • If there is a negative sign just before the parentheses, we reverse the signs of the terms inside the brackets.
  • If there is a positive sign just before the parentheses, we remove the brackets and keep the signs of the terms unchanged.

Solved Examples

Simplify the following expressions: a) 5x – ${\left( -2x^{2}+3x-1\right)}$ b) ${4ab-2b+3\left( ab+1\right) -2b}$ c) ${\dfrac{5x^{2}}{10x^{2}+5x^{3}}}$ d) ${\dfrac{1}{2}\left( 10x^{2}-34\right)}$

a) The given algebraic expression is 5x – ${\left( -2x^{2}+3x-1\right)}$ Solving the parentheses, we get, 5x + ${2x^{2}}$ – 3x + 1 Now, by combining all the like terms, we get, ${2x^{2}}$ + (5x – 3x) + 1 = ${2x^{2}}$ + 2x + 1 Thus, the simplified algebraic expression is ${2x^{2}}$ + 2x + 1. b) The given algebraic expression is ${4ab-2b+3\left( ab+1\right) -2b}$ Solving the parentheses, we get, 4ab – 2b + 3ab + 3 – 2b Now, by combining all the like terms, we get, (4ab + 3ab) + (-2b – 2b) + 3 = 7ab – 4b + 3 Thus, the simplified algebraic expression is 7ab – 4b + 3. c) The given algebraic expression is${\dfrac{5x^{2}}{10x^{2}+5x^{3}}}$ By finding the common factors, we get, ${\dfrac{5x^{2}}{5x^{2}\left( 2+x\right) }}$ = ${\dfrac{1}{\left( 2+x\right) }}$ Thus, the simplified algebraic expression is ${\dfrac{1}{\left( 2+x\right) }}$. d) The given algebraic expression is ${\dfrac{1}{2}\left( 10x^{2}-34\right)}$ By finding the common factors, we get, ${\dfrac{2\left( 5x^{2}-17\right) }{2}}$ = ${5x^{2}-17}$ Thus, the simplified algebraic expression is ${5x^{2}-17}$.

Simplify the algebraic expression 15xy-13+4x+3y+xy+21

The given algebraic expression is 15xy-13+4x+3y+xy+21 By combining all the like terms, we get, 4x + (15xy + xy) + 3y + (21 – 13) = 4x + 16xy + 3y + 8 Thus, the simplified algebraic expression is 4x + 16xy + 3y + 8.

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Simplifying Expressions

Simplifying expressions mean rewriting the same algebraic expression with no like terms and in a compact manner. To simplify expressions, we combine all the like terms and solve all the given brackets, if any, and then in the simplified expression, we will be only left with unlike terms that cannot be reduced further. Let us learn more about simplifying expressions in this article.

How to Simplify Expressions?

Before learning about simplifying expressions, let us quickly go through the meaning of expressions in math. Expressions refer to mathematical statements having a minimum of two terms containing either numbers , variables, or both connected through an addition/subtraction operator in between. The general rule to simplify expressions is PEMDAS - stands for Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. In this article, we will be focussing more on how to simplify algebraic expressions. Let's begin!

We need to learn how to simplify expressions as it allows us to work more efficiently with algebraic expressions and ease out our calculations. To simplify algebraic expressions, follow the steps given below:

  • Step 1: Solve parentheses by adding/subtracting like terms inside and by multiplying the terms inside the brackets with the factor written outside. For example, 2x (x + y) can be simplified as 2x 2 + 2xy.
  • Step 2: Use the exponent rules to simplify terms containing exponents .
  • Step 3: Add or subtract the like terms.
  • Step 4: At last, write the expression obtained in the standard form (from highest power to the lowest power).

Let us take an example for a better understanding. Simplify the expression: x (6 – x) – x (3 – x). Here, there are two parentheses both having two unlike terms. So, we will be solving the brackets first by multiplying x to the terms written inside. x(6 - x) can be simplified as 6x - x 2 , and -x(3 - x) can be simplified as -3x + x 2 . Now, combining all the terms will result in 6x - x 2 - 3x + x 2 . In this expression, 6x and -3x are like terms, and -x 2 and x 2 are like terms. So, adding these two pairs of like terms will result in (6x - 3x) + (-x 2 + x 2 ). By simplifying it further, we will get 3x, which will be the final answer. Therefore, x (6 – x) – x (3 – x) = 3x.

Look at the image given below showing another simplifying expression example.

simplifying expressions

Rules for Simplifying Algebraic Expressions

The basic rule for simplifying expressions is to combine like terms together and write unlike terms as it is. Some of the rules for simplifying expressions are listed below:

  • To add two or more like terms, add their coefficients and write the common variable with it.
  • Use the distributive property to open up brackets in the expression which says that a (b + c) = ab + ac.
  • If there is a negative sign just outside parentheses, change the sign of all the terms written inside that bracket to simplify it.
  • If there is a 'plus' or a positive sign outside the bracket, just remove the bracket and write the terms as it is, retaining their original signs.

Simplifying Expressions with Exponents

To simplify expressions with exponents is done by applying the rules of exponents on the terms. For example, (3x 2 )(2x) can be simplified as 6x 3 . The exponent rules chart that can be used for simplifying algebraic expressions is given below:

Example: Simplify: 2ab + 4b (b 2 - 2a).

To simplify this expression, let us first open the bracket by multiplying 4b to both the terms written inside. This implies, 2ab + 4b (b 2 ) - 4b (2a). By using the product rule of exponents, it can be written as 2ab + 4b 3 - 8ab, which is equal to 4b 3 - 6ab.

This is how we can simplify expressions with exponents using the rules of exponents.

Simplifying Expressions with Distributive Property

Distributive property states that an expression given in the form of x (y + z) can be simplified as xy + xz. It can be very useful while simplifying expressions. Look at the above examples, and see whether and how we have used this property for the simplification of expressions. Let us take another example of simplifying 4(2a + 3a + 4) + 6b using the distributive property.

Simplifying expressions with distributive property

Therefore, 4(2a + 3a + 4) + 6b is simplified as 20a + 6b + 16. Now, let us learn how to use the distributive property to simplify expressions with fractions.

Simplifying Expressions with Fractions

When fractions are given in an expression, then we can use the distributive property and the exponent rules to simplify such expression. For example, 1/2 (x + 4) can be simplified as x/2 + 2. Let us take one more example to understand it.

Example: Simplify the expression: 3/4x + y/2 (4x + 7).

By using the distributive property, the given expression can be written as 3/4x + y/2 (4x) + y/2 (7). Now, to multiply fractions , we multiply the numerators and the denominators separately. So, y/2 × 4x/1 = (y × 4x)/2 = 4xy/2 = 2xy. And, y/2 × 7/1 = 7y/2. Therefore, 3/4x + y/2 (4x + 7) = 3/4x + 2xy + 7y/2. All three are unlike terms, so it is the simplified form of the given expression.

While simplifying expressions with fractions , we have to make sure that the fractions should be in the simplest form and only unlike terms should be present in the simplified expression. For an instance, (2/4)x + 3/6y is not the simplified expression, as fractions are not reduced to their lowest form. On the other hand, x/2 + 1/2y is in a simplified form as fractions are in the reduced form and both are unlike terms.

► Related Topics:

Check these interesting articles related to the concept of simplifying expressions in math.

  • Simplifying Expressions Calculator
  • Simplifying Rational Expressions
  • Simplifying Radical Expressions

Simplifying Expressions Examples

Example 1: Find the simplified form of the expression formed by the following statement: "Addition of k and 8 multiplied by the subtraction of k from 16".

Solution: From the given statement, the expression formed is (k + 8)(16 - k). To simplify this expression, we need to use the concept of multiplication of algebraic expressions . By using the distributive property of simplifying expression, it can be simplified as,

⇒ k (16 - k) + 8 (16 - k)

⇒ 16k - k 2 + 128 - 8k

⇒ - k 2 + 16k - 8k + 128

⇒ - k 2 + 8k + 128

Therefore, - k 2 + 8k + 128 is the simplified form of the given expression.

Example 2: Simplify the expression: 4ps - 2s - 3(ps +1) - 2s .

Solution: By using the rules of simplifying expressions, 4ps - 2s - 3(ps +1) - 2s can be simplified as,

⇒ 4ps - 2s - 3(ps +1) - 2s

⇒ 4ps - 2s - 3ps - 3 - 2s

⇒ 4ps - 3ps - 2s - 2s - 3

⇒ ps - 4s - 3

Therefore, 4ps - 2s - 3(ps +1) - 2s = ps - 4s - 3.

Example 3: Daniel bought 5 pencils each costing $x, and Victoria bought 6 pencils each costing $x. Find the total cost of buying pencils by both of them.

Solution: Given, Daniel bought 5 pencils each for $x. The cost of all 5 pencils = $5x. And, Victoria bought 6 pencils each for $x, so the cost of 6 pencils = $6x. Therefore, the total cost of pencils bought by them = $5x + $6x = $11x.

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Practice Questions on Simplifying Algebraic Expressions

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FAQs on Simplifying Expressions

What is simplifying expressions in math.

In math, simplifying expressions is a way to write an expression in its lowest form by combining all like terms together. It requires one to be familiar with the concepts of arithmetic operations on algebraic expressions, fractions, and exponents. We follow the same PEMDAS rule to simplify algebraic expressions as we do for simple arithmetic expressions. Along with PEMDAS, exponent rules, and the knowledge about operations on expressions also need to be used while simplifying algebraic expressions.

What Mathematical Concepts are Important in Simplifying Expressions?

The mathematical concepts that are important in simplifying algebraic expressions are given below:

  • Familiarity with like and unlike algebraic terms .
  • Basic knowledge of algebraic expressions is required.
  • Addition and subtraction of algebraic expressions .
  • Multiplication and division of expressions .
  • Understanding of terms with exponents and exponent rules.
  • Algebraic identities and properties.

What are the Rules for Simplifying Expressions?

The rules for simplifying expressions are given below:

  • Follow the PEMDAS rule to determine the order of terms to be simplified in an expression.
  • Distributive property can be used to simplify the multiplication of two terms in an algebraic expression.
  • Exponent rules can be used to simplify terms with exponents.
  • First, we open the brackets, if any. Then we simplify the terms containing exponents.
  • After that, combine all the like terms.
  • The simplified expression will only have unlike terms connected by addition/subtraction operators that cannot be simplified further.

How to do Simplifying Expressions?

Follow the steps given below to learn how to simplify expressions:

  • Open up brackets, if any. If there is a positive sign outside the bracket, then remove the bracket and write all the terms retaining their original signs. If there is a negative sign outside the bracket, then remove the bracket and change the signs of all the terms written inside from + to -, and - to +. And if there is a number or variable written just outside the bracket, then multiply it with all the terms inside using the distributive property.
  • Use exponent rules to simplify terms with exponents, if any.
  • Add/subtract all like terms.
  • Write the simplified expression in the standard form (from the highest power term to the lowest power term).

How do Simplifying Expressions and Solving Equations Differ?

Equations refer to those statements that have an equal to "=" sign between the term(s) written on the left side and the term(s) written on the right side. Solving equations mean finding the value of the unknown variable given. On the other hand, simplifying expressions mean only reducing the expression to its lowest form. It does not intend to find the value of an unknown quantity.

What is an Example of Simplifying Expressions?

Simplifying algebraic expressions refer to the process of reducing the expression to its lowest form. An example of simplifying algebraic expressions is given below:

2x + 6x (y - 7) - 8

= 2x + 6xy - 42x - 8

= 6xy - 40x - 8

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Course: praxis core math   >   unit 1.

  • Algebraic properties | Lesson
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What are algebraic word problems?

What skills are needed.

  • Translating sentences to equations
  • Solving linear equations with one variable
  • Evaluating algebraic expressions
  • Solving problems using Venn diagrams

How do we solve algebraic word problems?

  • Define a variable.
  • Write an equation using the variable.
  • Solve the equation.
  • If the variable is not the answer to the word problem, use the variable to calculate the answer.

What's a Venn diagram?

  • Your answer should be
  • an integer, like 6 ‍  
  • a simplified proper fraction, like 3 / 5 ‍  
  • a simplified improper fraction, like 7 / 4 ‍  
  • a mixed number, like 1   3 / 4 ‍  
  • an exact decimal, like 0.75 ‍  
  • a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  
  • (Choice A)   $ 4 ‍   A $ 4 ‍  
  • (Choice B)   $ 5 ‍   B $ 5 ‍  
  • (Choice C)   $ 9 ‍   C $ 9 ‍  
  • (Choice D)   $ 14 ‍   D $ 14 ‍  
  • (Choice E)   $ 20 ‍   E $ 20 ‍  
  • (Choice A)   10 ‍   A 10 ‍  
  • (Choice B)   12 ‍   B 12 ‍  
  • (Choice C)   24 ‍   C 24 ‍  
  • (Choice D)   30 ‍   D 30 ‍  
  • (Choice E)   32 ‍   E 32 ‍  
  • (Choice A)   4 ‍   A 4 ‍  
  • (Choice B)   10 ‍   B 10 ‍  
  • (Choice C)   14 ‍   C 14 ‍  
  • (Choice D)   18 ‍   D 18 ‍  
  • (Choice E)   22 ‍   E 22 ‍  

Things to remember

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GMAT Math : Simplifying Algebraic Expressions

Study concepts, example questions & explanations for gmat math, all gmat math resources, example questions, example question #1 : simplifying algebraic expressions.

\frac{x^{2}+6x+5}{x^{2}+10x+25}.

Let's first look at the numerator and denominator separately.

x^{2}+6x+5

no solution

problem solving simplifying algebraic expressions

all real numbers

Let's combine like terms.

problem solving simplifying algebraic expressions

Example Question #3 : Simplifying Algebraic Expressions

problem solving simplifying algebraic expressions

Example Question #4 : Simplifying Algebraic Expressions

problem solving simplifying algebraic expressions

However, note the second group in parentheses is being subtracted. So we must invert all the signs in the group to simplify properly. So the previous expression simplifies to 

problem solving simplifying algebraic expressions

Finally we reorder and combine like terms to get

problem solving simplifying algebraic expressions

A number is divided by 4; its decimal point is then moved to the right 3 places. This is the same as doing what to the number?

Multiplying it by 2,500.

Dividing it by 400.

Multiplying it by 250.

Dividing it by 250.

Dividing it by 4,000.

The best way to illustrate the answer to this question is to do these operations to the number 1.

First, divide by 4:

problem solving simplifying algebraic expressions

Now move the decimal point right three spaces:

problem solving simplifying algebraic expressions

This has the effect of multiplying the number by 250.

Example Question #6 : Simplifying Algebraic Expressions

problem solving simplifying algebraic expressions

Example Question #7 : Simplifying Algebraic Expressions

The sum of three consecutive integers is 12.  What is the value of the middle integer?

problem solving simplifying algebraic expressions

Example Question #8 : Simplifying Algebraic Expressions

problem solving simplifying algebraic expressions

Example Question #9 : Simplifying Algebraic Expressions

problem solving simplifying algebraic expressions

Tired of practice problems?

Try live online GMAT prep today.

problem solving simplifying algebraic expressions

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

[FREE] Math Equations Check for Understanding Quiz (Grade 6 to 8)

Use this quiz to check your grade 6 to 8 students’ understanding of math equations. 10+ questions with answers covering a range of 6th, 7th and 8th grade math equations topics to identify areas of strength and support!

How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

  • Math equations
  • Rearranging equations
  • How to find the equation of a line
  • Solve equations with fractions
  • Linear equations
  • Writing linear equations
  • Substitution
  • Identity math
  • One step equation

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

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Simple Algebra Problems – Easy Exercises with Solutions for Beginners

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Understanding Algebraic Expressions

Breaking down algebra problems, solving algebraic equations, tackling algebra word problems, types of algebraic equations, algebra for different grades.

Simple Algebra Problems Easy Exercises with Solutions for Beginners

For instance, solving the equation (3x = 7) for (x) helps us understand how to isolate the variable to find its value.

Illustration of Simple Algebra Problems

I always find it fascinating how algebra serves as the foundation for more advanced topics in mathematics and science. Starting with basic problems such as ( $(x-1)^2 = [4\sqrt{(x-4)}]^2$ ) allows us to grasp key concepts and build the skills necessary for tackling more complex challenges.

So whether you’re refreshing your algebra skills or just beginning to explore this mathematical language, let’s dive into some examples and solutions to demystify the subject. Trust me, with a bit of practice, you’ll see algebra not just as a series of problems, but as a powerful tool that helps us solve everyday puzzles.

Simple Algebra Problems and Strategies

When I approach simple algebra problems, one of the first things I do is identify the variable.

The variable is like a placeholder for a number that I’m trying to find—a mystery I’m keen to solve. Typically represented by letters like ( x ) or ( y ), variables allow me to translate real-world situations into algebraic expressions and equations.

An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like ( x ) or ( y )), and operators (like add, subtract, multiply, and divide). For example, ( 4x + 7 ) is an algebraic expression where ( x ) is the variable and the numbers ( 4 ) and ( 7 ) are terms. It’s important to manipulate these properly to maintain the equation’s balance.

Solving algebra problems often starts with simplifying expressions. Here’s a simple method to follow:

  • Combine like terms : Terms that have the same variable can be combined. For instance, ( 3x + 4x = 7x ).
  • Isolate the variable : Move the variable to one side of the equation. If the equation is ( 2x + 5 = 13 ), my job is to get ( x ) by itself by subtracting ( 5 ) from both sides, giving me ( 2x = 8 ).

With algebraic equations, the goal is to solve for the variable by performing the same operation on both sides. Here’s a table with an example:

Algebra word problems require translating sentences into equations. If a word problem says “I have six less than twice the number of apples than Bob,” and Bob has ( b ) apples, then I’d write the expression as ( 2b – 6 ).

Understanding these strategies helps me tackle basic algebra problems efficiently. Remember, practice makes perfect, and each problem is an opportunity to improve.

In algebra, we encounter a variety of equation types and each serves a unique role in problem-solving. Here, I’ll brief you about some typical forms.

Linear Equations : These are the simplest form, where the highest power of the variable is one. They take the general form ( ax + b = 0 ), where ( a ) and ( b ) are constants, and ( x ) is the variable. For example, ( 2x + 3 = 0 ) is a linear equation.

Polynomial Equations : Unlike for linear equations, polynomial equations can have variables raised to higher powers. The general form of a polynomial equation is ( $a_nx^n + a_{n-1}x^{n-1} + … + a_2x^2 + a_1x + a_0 = 0$ ). In this equation, ( n ) is the highest power, and ( $a_n$ ), ( $a_{n-1} $), …, ( $a_0$ ) represent the coefficients which can be any real number.

  • Binomial Equations : They are a specific type of polynomial where there are exactly two terms. Like ($ x^2 – 4 $), which is also the difference of squares, a common format encountered in factoring.

To understand how equations can be solved by factoring, consider the quadratic equation ( $x^2$ – 5x + 6 = 0 ). I can factor this into ( (x-2)(x-3) = 0 ), which allows me to find the roots of the equation.

Here’s how some equations look when classified by degree:

Remember, identification and proper handling of these equations are essential in algebra as they form the basis for complex problem-solving.

In my experience with algebra, I’ve found that the journey begins as early as the 6th grade, where students get their first taste of this fascinating subject with the introduction of variables representing an unknown quantity.

I’ve created worksheets and activities aimed specifically at making this early transition engaging and educational.

6th Grade :

Moving forward, the complexity of algebraic problems increases:

7th and 8th Grades :

  • Mastery of negative numbers: students practice operations like ( -3 – 4 ) or ( -5 $\times$ 2 ).
  • Exploring the rules of basic arithmetic operations with negative numbers.
  • Worksheets often contain numeric and literal expressions that help solidify their concepts.

Advanced topics like linear algebra are typically reserved for higher education. However, the solid foundation set in these early grades is crucial. I’ve developed materials to encourage students to understand and enjoy algebra’s logic and structure.

Remember, algebra is a tool that helps us quantify and solve problems, both numerical and abstract. My goal is to make learning these concepts, from numbers to numeric operations, as accessible as possible, while always maintaining a friendly approach to education.

I’ve walked through various simple algebra problems to help establish a foundational understanding of algebraic concepts. Through practice, you’ll find that these problems become more intuitive, allowing you to tackle more complex equations with confidence.

Remember, the key steps in solving any algebra problem include:

  • Identifying variables and what they represent.
  • Setting up the equation that reflects the problem statement.
  • Applying algebraic rules such as the distributive property ($a(b + c) = ab + ac$), combining like terms, and inverse operations.
  • Checking your solutions by substituting them back into the original equations to ensure they work.

As you continue to engage with algebra, consistently revisiting these steps will deepen your understanding and increase your proficiency. Don’t get discouraged by mistakes; they’re an important part of the learning process.

I hope that the straightforward problems I’ve presented have made algebra feel more manageable and a little less daunting. Happy solving!

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Mathematics LibreTexts

7.1: Simplifying Rational Expressions

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Learning Objectives

  • Determine the restrictions to the domain of a rational expression.
  • Simplify rational expressions.
  • Simplify expressions with opposite binomial factors.
  • Simplify and evaluate rational functions.

Rational Expressions, Evaluation, and Restrictions

A rational number, or fraction \(\frac{a}{b}\), is a real number defined as a quotient of two integers a and b , where \(b≠0\). Similarly, we define a rational expression, or algebraic fraction \(\frac{P}{Q}\), as the quotient of two polynomials P and Q , where \(Q≠0\). Some examples of rational expressions follow:

The example \(\frac{x+3}{x-5}\) consists of linear expressions in both the numerator and denominator. Because the denominator contains a variable, this expression is not defined for all values of x .

Example \(\PageIndex{1}\)

Evaluate \(\frac{x+3}{x-5}\) for the set of x -values \(\{-3,4,5\}\).

Substitute the values in for x .

\(\begin{array}{c|c}{x=-3} & {x=4} & {x=5} \\ \hline \frac{x+3}{x-5}=\frac{(\color{OliveGreen}{-3}\color{black}{)}+3}{(\color{OliveGreen}{-3}\color{black}{)}-5} & {\frac{x+3}{x-5}=\frac{(\color{OliveGreen}{4}\color{black}{)}+3}{(\color{OliveGreen}{4}\color{black}{)}-5}} & {\frac{x+3}{x-5}=\frac{(\color{OliveGreen}{5}\color{black}{)}+3}{(\color{OliveGreen}{5}\color{black}{)}-5}} \\ {=\frac{0}{-8}}&{=\frac{7}{-1}}&{=\frac{8}{0}\:\:\color{Cerulean}{Undefined}}\\{=0}&{=-7}&{}\end{array}\)

When \(x=−3\), the value of the rational expression is \(0\); when \(x=4\), the value of the rational expression is \(−7\); and when \(x=5\), the value of the rational expression is undefined.

This example illustrates that variables are restricted to values that do not make the denominator equal to 0. The domain of a rational expression is the set of real numbers for which it is defined, and restrictions are the real numbers for which the expression is not defined. We often express the domain of a rational expression in terms of its restrictions.

Example \(\PageIndex{2}\)

Find the domain of the following:

\(\frac{x+7}{2 x^{2}+x-6}\)

In this example, the numerator \(x+7\) is a linear expression and the denominator \(2x^{2}+x−6\) is a quadratic expression. If we factor the denominator, then we will obtain an equivalent expression.

\(\frac{x+7}{2 x^{2}+x-6}=\frac{x+7}{(2 x-3)(x+2)}\)

Because rational expressions are undefined when the denominator is 0, we wish to find the values for x that make it 0. To do this, apply the zero product property. Set each factor in the denominator equal to 0 and solve.

\((2 x-3)(x+2)=0\)

\(\begin{array}{ll}{2 x-3=0} & {\text { or } \quad x+2=0} \\ {2 x=3} &\qquad\quad {x=-2} \\ {x=\frac{3}{2}}\end{array}\)

We conclude that the original expression is defined for any real number except \(\frac{3}{2}\) and \(−2\). These two values are the restrictions to the domain. It is important to note that \(−7\) is not a restriction to the domain because the expression is defined as 0 when the numerator is 0.

\(\begin{aligned} \frac{x+7}{2 x^{2}+x-6} &=\frac{(\color{OliveGreen}{-7}\color{black}{)}+7}{2(\color{OliveGreen}{-7}\color{black}{)}^{2}+(\color{OliveGreen}{-7}\color{black}{)}-6} \\ &=\frac{0}{98-7-6} \\ &=\frac{0}{85} \\ &=0 \end{aligned}\)

The domain consists of any real number x , where \(x≠\frac{3}{2}\) and \(x≠−2\).

We can express the domain of the previous example using notation as follows:

\(\begin{array}{cc}{\color{Cerulean} { Set-builder\: notation}} & {\color{Cerulean} {Interval\: notation}} \\ {\left\{x | x \neq-2, \frac{3}{2}\right\}} & {(-\infty,-2) \cup\left(-2, \frac{3}{2}\right) \cup\left(\frac{3}{2}, \infty\right)}\end{array}\)

The restrictions to the domain of a rational expression are determined by the denominator. Ignore the numerator when finding those restrictions.

Example \(\PageIndex{3}\)

Determine the domain:

\(\frac{x^{4}+x^{3}-2x^{2}-x}{x^{2}-1}\)

To find the restrictions to the domain, set the denominator equal to 0 and solve:

\(\begin{array}{r}{x^{2}-1=0} \\ {(x+1)(x-1)=0}\end{array}\)

\(\begin{array}{rlrl}{x+1} & {=0} & {\text { or }} & {x-1=0} \\ {x} & {=-1} & &{x=1}\end{array}\)

These two values cause the denominator to be 0. Hence they are restricted from the domain.

The domain consists of any real number x , where \(x≠±1\).

Example \(\PageIndex{4}\)

\(\frac{x^{2}-25}{4}\)

There is no variable in the denominator and thus no restriction to the domain.

The domain consists of all real numbers, R .

Simplifying Rational Expressions

When simplifying fractions, look for common factors that cancel. For example,

\(\frac{12}{60}=\frac{1 \cdot \color{Cerulean}{\cancel{\color{black}{12}}}}{5 \cdot \color{Cerulean}{\cancel{\color{black}{12}}}}\color{black}{=}\frac{1}{5}\)

We say that the fraction \(\frac{12}{60}\) is equivalent to \(\frac{1}{5}\). Fractions are in simplest form if the numerator and denominator share no common factor other than \(1\). Similarly, when working with rational expressions, look for factors to cancel. For example,

\(\frac{x+4}{(x-3)(x+4)}=\frac{1 \cdot\color{Cerulean}{\cancel{\color{black}{(x+4)}}}}{(x-3)\color{Cerulean}{\cancel{\color{black}{(x+4)}}}}=\frac{1}{x-3}\)

The resulting rational expression is equivalent if it shares the same domain. Therefore, we must make note of the restrictions and write

\(\frac{x+4}{(x-3)(x+4)}=\frac{1}{x-3}, \text { where } x \neq 3 \text { and } x \neq-4\)

In words, \(\frac{x+4}{(x-3)(x+4)}\) is equivalent to\(\frac{1}{x-3}\), if \(x≠3\) and \(x≠−4\). We can verify this by choosing a few values with which to evaluate both expressions to see if the results are the same. Here we choose \(x=7\) and evaluate as follows:

\(\begin{aligned} \frac{x+4}{(x-3)(x+4)} &=\frac{1}{x-3} \\ \frac{(\color{OliveGreen}{7}\color{black}{)}+4}{(\color{OliveGreen}{7}\color{black}{-}3)(\color{OliveGreen}{7}\color{black}{+}4)} &=\frac{1}{(\color{OliveGreen}{7}\color{black}{)}-3} \\ \frac{11}{(4)(11)} &=\frac{1}{4} \\ \frac{1}{4} &=\frac{1}{4}\:\:\color{Cerulean}{\checkmark} \end{aligned}\)

It is important to state the restrictions before simplifying rational expressions because the simplified expression may be defined for restrictions of the original. In this case, the expressions are not equivalent. Here −4 is defined for the simplified equivalent but not for the original, as illustrated below:

\(\begin{aligned} \frac{x+4}{(x-3)(x+4)} &=\frac{1}{x-3} \\ \frac{(\color{OliveGreen}{-4}\color{black}{)}+4}{(\color{OliveGreen}{-4}\color{black}{-}3)(\color{OliveGreen}{-4}\color{black}{+}4)} &=\frac{1}{(\color{OliveGreen}{-4}\color{black}{)}-3} \\ \frac{0}{(-7)(0)} &=\frac{1}{-7} \\ \frac{0}{0} &=-\frac{1}{7} \quad \color{red}{x} \end{aligned}\)

Example \(\PageIndex{5}\)

Simplify and state the restriction:

\(\frac{25 x^{2}}{15 x^{3}}\)

In this example, the expression is undefined when x is 0.

\(\frac{25 x^{2}}{15 x^{3}}=\frac{25(\color{OliveGreen}{0}\color{black}{)}^{2}}{15(\color{OliveGreen}{0}\color{black}{)}^{3}}=\frac{0}{0}\:\:\color{Cerulean}{Undefined}\)

Therefore, the domain consists of all real numbers x , where \(x≠0\). With this understanding, we can cancel common factors.

\(\begin{aligned} \frac{25 x^{2}}{15 x^{3}} &=\frac{5 \cdot \color{Cerulean}{\cancel{\color{black}{5 x^{2}}}}}{3 x \cdot \color{Cerulean}{\cancel{\color{black}{5 x^{2}}}}} \\ &=\frac{5}{3 x} \end{aligned}\)

\(\frac{5}{3 x}\), where \(x\neq 0\)

Example \(\PageIndex{6}\)

State the restrictions and simplify:

\(\frac{3x(x-5)}{(2x-1)(x-5)}\)

To determine the restrictions, set the denominator equal to 0 and solve.

\((2 x+1)(x-5)=0\)

\(\begin{array}{rlrl}{2 x+1} & {=0} & {\text { or }} & {x-5=0} \\ {2 x} & {=-1} & &{x=5} \\ {x} & {=-\frac{1}{2}}\end{array}\)

The domain consists of all real numbers except for \(−\frac{1}{2}\) and \(5\). Next, we find an equivalent expression by canceling common factors.

\(\begin{aligned} \frac{3 x(x-5)}{(2 x+1)(x-5)} &=\frac{3 x \color{Cerulean}{\cancel{\color{black}{(x-5)}}}}{(2 x+1)\color{Cerulean}{\cancel{\color{black}{(x-5)}}}} \\ &=\frac{3 x}{2 x+1} \end{aligned}\)

\(\frac{3 x}{2 x+1}\), where \(x\neq -\frac{1}{2}\) and \(x\neq 5\)

Typically, rational expressions are not given in factored form. If this is the case, factor first and then cancel. The steps are outlined in the following example.

Example \(\PageIndex{7}\)

\(\frac{3 x+6}{x^{2}+x-2}\)

Step 1 : Completely factor the numerator and denominator.

\(\frac{3 x+6}{x^{2}+x-2}=\frac{3(x+2)}{(x-1)(x+2)}\)

Step 2 : Determine the restrictions to the domain. To do this, set the denominator equal to 0 and solve.

\((x-1)(x+2)=0\)

\(\begin{array}{rlrl}{x-1} & {=0} & {\text { or }} & {x+2=0} \\ {x} & {=1} && {x=-2}\end{array}\)

The domain consists of all real numbers except \(−2\) and \(1\).

Step 3 : Cancel common factors, if any.

\(\begin{aligned} \frac{3 x+6}{x^{2}+x-2} &=\frac{3\color{Cerulean}{\cancel{\color{black}{(x+2)}}}}{(x-1)\color{Cerulean}{\cancel{\color{black}{(x+2)}}}} \\ &=\frac{3}{x-1} \end{aligned}\)

\(\frac{3}{x-1}\), where \(x\neq 1\) and \(x\neq -2\)

Example \(\PageIndex{8}\)

First, factor the numerator and denominator.

Any value of x that results in a value of \(0\) in the denominator is a restriction. By inspection, we determine that the domain consists of all real numbers except \(4\) and \(3\). Next, cancel common factors.

\(\begin{array}{l}{=\color{black}{\frac{\color{Cerulean}{\cancel{\color{black}{(x-3)}}}\color{black}{(x+10)}}{(x-4)\color{Cerulean}{\cancel{\color{black}{(x-3)}}}}}} \\ {=\frac{x+10}{x-4}}\end{array}\)

\(\frac{x+10}{x-4}\), where \(x\neq 3\) and \(x\neq 4\)

It is important to remember that we can only cancel factors of a product. A common mistake is to cancel terms. For example,

Exercise \(\PageIndex{1}\)

\(\frac{x^{2}-16}{5x^{2}-20x}\)

\(\frac{x+4}{5 x}\), where \(x\neq 0\) and \(x\neq 4\)

In some examples, we will make a broad assumption that the denominator is nonzero. When we make that assumption, we do not need to determine the restrictions.

Example \(\PageIndex{9}\)

(Assume all denominators are nonzero.)

Factor the numerator by grouping. Factor the denominator using the formula for a difference of squares.

Next, cancel common factors.

\(\begin{array}{l}{=\frac{\color{Cerulean}{\cancel{\color{black}{(x+y)}}}\color{black}{(y-3)}}{\color{Cerulean}{\cancel{\color{black}{(x+y)}}}\color{black}{(x-y)}}} \\ {=\frac{y-3}{x-y}}\end{array}\)

\(\frac{y-3}{x-y}\)

Opposite Binomial Factors

Recall that the opposite of the real number a is −a . Similarly, we can define the opposite of a polynomial P to be −P . We first consider the opposite of the binomial \(a−b\):

\[-(a-b)=-a+b=b-a\]

This leads us to the opposite binomial property:

\[-(a-b)=(b-a)\]

This is equivalent to factoring out a \(–1\).

\[(b-a)=-1(a-b)\]

If \(a≠b\), then we can divide both sides by \((a−b)\) and obtain the following:

\(\frac{b-a}{a-b}=-1\)

Example \(\PageIndex{10}\)

\(\frac{3-x}{x-3}\)

By inspection, we can see that the denominator is \(0\) if \(x=3\). Therefore, \(3\) is the restriction to the domain. Apply the opposite binomial property to the numerator and then cancel.

\(\frac{3-x}{x-3}\), where \(x\neq 3\)

Since addition is commutative, we have

\[(a+b)=(b+a)\]

\(\frac{b+a}{a+b}=1\)

Take care not to confuse this with the opposite binomial property. Also, it is important to recall that

\(\frac{-a}{b}=-\frac{a}{b}=\frac{a}{-b}\)

In other words, show a negative fraction by placing the negative sign in the numerator, in front of the fraction bar, or in the denominator. Generally, negative denominators are avoided.

Example \(\PageIndex{11}\)

Simplify and state the restrictions:

\(\frac{4-x^{2}}{x^{2}+3 x-10}\)

Begin by factoring the numerator and denominator.

\(\begin{aligned} \frac{4-x^{2}}{x^{2}+3 x-10} &=\frac{(2+x)\color{Cerulean}{(2-x)}}{\color{Cerulean}{(x-2)}\color{black}{(}x+5)}\qquad\qquad\quad\color{Cerulean}{The\:restrictions\:are\:x\neq2\:and\:x\neq-5.} \\ &=\frac{(2+x) \cdot(-1)\color{Cerulean}{\cancel{\color{black}{(x-2)}}}}{\color{Cerulean}{\cancel{\color{black}{(x-2)}}}\color{black}{(x+5)}}\quad\:\color{Cerulean}{Apply\:the\:opposite\:binomial\:property,\:then\:cancel.} \\ &=\frac{(2+x) \cdot(-1)}{(x+5)} \\ &=-\frac{2+x}{x+5} \quad \text { or } \quad=-\frac{x+2}{x+5} \end{aligned}\)

\(-\frac{x+2}{x+5}\), where \(x\neq 2\) and \(x\neq -5\)

Exercise \(\PageIndex{2}\)

\(\frac{-2x+3}{x+5}\), where \(x\neq\pm 5\)

Rational Functions

Rational functions have the form

\[r(x)=\frac{p(x)}{q(x)}\],

where p(x) and q(x) are polynomials and q(x)≠0 . The domain of a rational function consists of all real numbers x such that the denominator q(x)≠0 .

Example \(\PageIndex{12}\)

  • Simplify: \(r(x)=\frac{2 x^{2}+5 x-3}{6 x^{2}+18 x}\).
  • State the domain.
  • Calculate \(r(-2)\).

a. To simplify the rational function, first factor and then cancel.

\(\begin{aligned} r(x) &=\frac{2 x^{2}+5 x-3}{6 x^{2}+18 x} \\ &=\frac{(2 x-1)\color{Cerulean}{\cancel{\color{black}{(x+3)}}}}{6 x\color{Cerulean}{\cancel{\color{black}{(x+3)}}}} \\ &=\frac{2 x-1}{6 x} \end{aligned}\)

b. To determine the restrictions, set the denominator of the original function equal to 0 and solve.

\(\begin{array}{l}{6 x^{2}+18 x=0} \\ {6 x(x+3)=0}\end{array}\)

\(\begin{array}{cc}{6 x=0} & {\text { or } \quad x+3=0} \\ {x=0} & {x=-3}\end{array}\)

The domain consists of all real numbers x , where \(x≠0\) and \(x≠−3\).

c. Since \(−2\) is not a restriction, substitute it for the variable x using the simplified form.

\(\begin{aligned} r(x) &=\frac{2 x-1}{6 x} \\ r(-2) &=\frac{2(\color{OliveGreen}{-2}\color{black}{)}-1}{6(\color{OliveGreen}{-2}\color{black}{)}} \\ &=\frac{-4-1}{-12} \\ &=\frac{-5}{-12} \\ &=\frac{5}{12} \end{aligned}\)

a. \(\frac{2 x-1}{6 x}\) b. The domain is all real numbers except \(0\) and \(−3\). c. \(r(-2) = \frac{5}{12}\)

If a cost function \(C(x)\) represents the cost of producing x units, then the average cost \(c(x)\) is the cost divided by the number of units produced.

\[\color{Cerulean}{Average\:cost}\color{black}{:} c(x)=\frac{C(x)}{x}\]

Example \(\PageIndex{13}\)

The cost in dollars of producing t-shirts with a company logo is given by \(C(x)=7x+200\), where x represents the number of shirts produced. Determine the average cost of producing

  • 40 t-shirts
  • 250 t-shirts
  • 1,000 t-shirts

Set up a function representing the average cost.

\(c(x)=\frac{C(x)}{x}=\frac{7 x+200}{x}\)

Next, calculate c(40), c(250), and c(1000).

\(\begin{aligned} c(40) &=\frac{7(\color{OliveGreen}{40}\color{black}{)}+200}{\color{OliveGreen}{40}}=\frac{280+200}{40}=\frac{480}{40}=12.00 \\ c(250) &=\frac{7(\color{OliveGreen}{250}\color{black}{)}+200}{\color{OliveGreen}{250}}=\frac{1750+200}{250}=\frac{1950}{250}=7.80 \\ c(1000) &=\frac{7(\color{OliveGreen}{1000}\color{black}{)}+200}{\color{OliveGreen}{1000}}=\frac{7000+200}{1000}=\frac{7200}{1000}=7.20 \end{aligned}\)

  • If 40 t-shirts are produced, then the average cost per t-shirt is $12.00.
  • If 250 t-shirts are produced, then the average cost per t-shirt is $7.80.
  • If 1,000 t-shirts are produced, then the average cost per t-shirt is $7.20.

Key Takeaways

  • Rational expressions usually are not defined for all real numbers. The real numbers that give a value of 0 in the denominator are not part of the domain. These values are called restrictions.
  • Simplifying rational expressions is similar to simplifying fractions. First, factor the numerator and denominator and then cancel the common factors. Rational expressions are simplified if there are no common factors other than 1 in the numerator and the denominator.
  • Simplified rational expressions are equivalent for values in the domain of the original expression. Be sure to state the restrictions if the denominators are not assumed to be nonzero.
  • Use the opposite binomial property to cancel binomial factors that involve subtraction. Use \(−(a−b)=b−a\) to replace factors that will then cancel. Do not confuse this with factors that involve addition, such as \((a+b)=(b+a)\).

Exercise \(\PageIndex{3}\) Rational Expressions

Evaluate for the given set of x -values.

  • \(5x; {−1, 0, 1}\)
  • \(\frac{4x^{3}}{x^{2}}; {−1, 0, 1} \)
  • \(\frac{1}{x+9}; {−10, −9, 0} \)
  • \(\frac{x+6}{x−5}; {−6, 0, 5}\)
  • \(\frac{3x(x−2)}{2x−1}; {0, \frac{1}{2}, 2}\)
  • \(\frac{9x^{2}−1}{x−7}; {0, \frac{1}{3}, 7} \)
  • \(5x^{2}−9; {−3, 0, 3}\)
  • \(\frac{x^{2}−2}{5x^{2}−3x−10}; {−5, −4, 5}\)

Screenshot (308).png

1. \(−5\), undefined, \(5\)

3. \(−1\), undefined, \(\frac{1}{9}\)

5. \(0\), undefined, \(0\)

7. Undefined, \(−\frac{5}{9}\), undefined

Screenshot (312).png

Exercise \(\PageIndex{4}\) Rational Expressions

An object’s weight depends on its height above the surface of earth. If an object weighs 120 pounds on the surface of earth, then its weight in pounds, W , x miles above the surface is approximated by the formula

\(W=\frac{120\cdot 4000^{2}}{(4000+x)^{2}}\)

For each problem below, approximate the weight of a 120-pound object at the given height above the surface of earth. (1 mile = 5,280 feet)

  • 1,000 miles
  • 44,350 feet
  • 90,000 feet

1. 114 pounds

3. 119.5 pounds

Exercise \(\PageIndex{5}\) Rational Expressions

The price to earnings ratio (P/E) is a metric used to compare the valuations of similar publicly traded companies. The P/E ratio is calculated using the stock price and the earnings per share (EPS) over the previous 12-month period as follows:

P/E=price per share earnings per share

If each share of a company stock is priced at $22.40, then calculate the P/E ratio given the following values for the earnings per share.

  • What happens to the P/E ratio when earnings decrease?
  • What happens to the P/E ratio when earnings increase?

3. The P/E ratio increases.

Exercise \(\PageIndex{6}\) Rational Expressions

State the restrictions to the domain.

  • \(\frac{1}{3x}\)
  • \(\frac{3x^{2}}{7x^{5}}\)
  • \(\frac{3x(x+1)}{x+4}\)
  • \(\frac{2x^{2}(x−3)}{x−1}\)
  • \(\frac{1}{5x−1}\)
  • \(\frac{x−2}{3x−2}\)
  • \(\frac{x−9}{5x(x−2)}\)
  • \(\frac{1}{(x−3)(x+6)}\)
  • \(\frac{x}{1−x^{2}}\)
  • \(\frac{x^{2}−9}{x^{2}−36}\)
  • \(\frac{1}{2x(x+3)(2x−1)}\)
  • \(\frac{x−3}{(3x−1)(2x+3)}\)
  • \(\frac{4x(2x+1)}{12x^{2}+x−1}\)
  • \(\frac{x−5}{3x^{2}−15x}\)

1. \(x≠0\)

3. \(x≠−4\)

5. \(x≠\frac{1}{5}\)

7. \(x≠0\) and \(x≠2\)

9. \(x≠±1\)

11. \(x≠0\), \(x≠−3\), and \(x≠\frac{1}{2}\)

13. \(x≠−\frac{1}{3}\) and \(x≠\frac{1}{4}\)

Exercise \(\PageIndex{7}\) Simplifying Rational Expressions

State the restrictions and then simplify.

  • \(\frac{5x^{2}}{20x^{3}}\)
  • \(\frac{12x^{6}}{60x}\)
  • \(\frac{3x^{2}(x−2)}{9x(x−2)}\)
  • \(\frac{20(x−3)(x−5)}{6(x−3)(x+1)}\)
  • \(\frac{6x^{2}(x−8)}{36x(x+9)(x−8)}\)
  • \(\frac{16x^{2}−1}{(4x+1)^{2}}\)
  • \(\frac{9x^{2}−6x+1}{(3x−1)^{2}}\)
  • \(\frac{x−7}{x^{2}−49}\)
  • \(\frac{x^{2}−64}{x^{2}+8x}\)
  • \(\frac{x+10}{x^{2}−100}\)
  • \(\frac{2x^{3}−12x^{2}}{5x^{2}−30x}\)
  • \(\frac{30x^{5}+60x^{4}}{2x^{3}−8x}\)
  • \(\frac{2x−12}{x^{2}+x−6}\)
  • \(\frac{x^{2}−x−6}{3x^{2}−8x−3}\)
  • \(\frac{6x^{2}−25x+25}{3x^{2}+16x−35}\)
  • \(\frac{3x^{2}+4x−1}{5x^{2}−9}\)
  • \(\frac{x^{2}−10x+21}{x^{2}−4x−21}\)
  • \(\frac{x^{3}−1}{x^{2}−1}\)
  • \(\frac{x^{3}+8}{x^{2}−4}\)
  • \(\frac{x^{4}−1}{6x^{2}−4}\)

1. \(\frac{1}{4x}; x≠0\)

3. \(\frac{x}{3}; x≠0, 2\)

5. \(\frac{x}{6(x+9)}; x≠0,−9, 8\)

7. \(1; x≠\frac{1}{3}\)

9. \(\frac{x−8}{x}; x≠0,−8\)

11. \(\frac{2x}{5}; x≠0, 6\)

13. \(\frac{2x−12}{x^{2}+x−6}\; x≠−2, \frac{3}{2}\)

15. \(\frac{2x−5}{x+7}; x≠−7, \frac{5}{3}\)

17. \(\frac{x−3}{x+3}; x≠−3, 7\)

19. \(\frac{x^{2}-2 x+4}{x-2} ; x≠±2\)

Exercise \(\PageIndex{8}\) Simplifying Rational Expressions with Opposite Binomial Factors

  • \(\frac{x−9}{9−x}\)
  • \(\frac{3x−2}{2−3x}\)
  • \(\frac{x+6}{6+x}\)
  • \(\frac{3x+1}{1+3x}\)
  • \(\frac{(2x−5)(x−7)}{(7−x)(2x−1)}\)
  • \(\frac{(3x+2)(x+5)}{(x−5)(2+3x)}\)
  • \(\frac{x^{2}−4}{(2−x)^{2}}\)
  • \(\frac{16−9x^{2}}{(3x+4)^{2}}\)
  • \(\frac{4x^{2}(10−x)}{3x^{3}−300x}\)
  • \(−\frac{2x+1}{4x^{3}−49x}\)
  • \(\frac{2x^{2}−7x−4}{1−4x^{2}}\)
  • \(\frac{9x^{2}−4}{4x−6x^{2}}\)
  • \(\frac{x^{2}−5x−14}{7−15x+2x^{2}}\)
  • \(\frac{2x^{3}+x^{2}−2x−1}{1+x−2x^{2}}\)
  • \(\frac{x^{3}+2 x-3 x^{2}-6}{2+x^{2}}\)
  • \(\frac{27+x^{3}}{x^{2}+6x+9}\)
  • \(\frac{64−x^{3}}{x^{2}−8x+16}\)
  • \(\frac{x^{2}+4}{4−x^{2}}\)

1. \(−1; x≠9 \)

3. \(1; x≠−6 \)

5. \(\frac{−2x−5}{2x−1}; x≠\frac{1}{2},7\)

7. \(\frac{x+2}{x−2}; x≠2\)

9. \(−\frac{4x}{3(x+10)}; x≠±10, 0\)

11. \(\frac{x−4}{1−2x}; x≠±\frac{1}{2}\)

13. \(\frac{x+2}{2x−1}; x≠\frac{1}{2},7\)

15. \(x−3\); none

17. \(\frac{−16+4x+x^{2}}{x−4}; x≠4\)

Exercise \(\PageIndex{9}\) Simplifying Rational Expressions with Opposite Binomial Factors

Simplify. (Assume all denominators are nonzero.)

  • \(−\frac{15x^{3}y^{2}}{5xy^{2}(x+y)}\)
  • \(\frac{14x^{7}y^{2}(x−2y)^{4}}{7x^{8}y(x−2y)^{2}}\)
  • \(\frac{y+x}{x^{2}−y^{2}}\)
  • \(\frac{y−x}{x^{2}−y^{2}}\)
  • \(\frac{x^{2}−y^{2}}{(x−y)^{2}}\)
  • \(\frac{a^{2}−ab−6b^{2}}{a^{2}−6ab+9b^{2}}\)
  • \(\frac{2a^{2}−11a+12}{−32+2a^{2}}\)
  • \(\frac{a^{2}b−3a^{2}}{3a^{2}−3ab}\)
  • \(\frac{x y^{2}-x+y^{3}-y}{x-x y^{2}}\)
  • \(\frac{x^{3}−xy^{2}−x^{2}y+y}{3x^{2}−2xy+y^{2}}\)
  • \(\frac{x^{3}−27}{x^{2}+3x+9}\)
  • \(\frac{x^{2}−x+1}{x^{3}+1}\)

1. \(−\frac{3x^{2}}{x+y}\)

3. \(\frac{1}{x−y}\)

5. \(\frac{x+y}{x−y}\)

7. \(\frac{2 a-3}{2(a+4)}\)

9. \(-\frac{x+y}{x}\)

11. \(x−3\)

Exercise \(\PageIndex{10}\) Rational Functions

Calculate the following.

  • \(f(x)=\frac{5x}{x−3}; f(0), f(2), f(4)\)
  • \(f(x)=\frac{x+7}{x^{2}+1}; f(−1), f(0), f(1) \)
  • \(g(x)=\frac{x^{3}}{(x−2)^{2}}; g(0), g(2), g(−2) \)
  • \(g(x)=\frac{x^{2}−9}{9−x^{2}}; g(−2), g(0), g(2) \)
  • \(g(x)=\frac{x^{3}}{x^{2}+1}; g(−1), g(0), g(1)\)
  • \(g(x)=\frac{5x+1}{x^{2}−25}; g(−\frac{1}{5}), g(−1), g(−5)\)

1. f(0)=0, f(2)=−10, f(4)=20

3. g(0)=0, g(2) undefined, g(−2)=−\(\frac{1}{2}\)

5. g(−1)=−\(\frac{1}{2}\), g(0)=0, g(1)=\(\frac{1}{2}\)

Exercise \(\PageIndex{11}\) Rational Functions

State the restrictions to the domain and then simplify.

  • \(f(x)=−\frac{3x^{2}−6x}{x^{2}+4x+4}\)
  • \(f(x)=\frac{x^{2}+6x+9}{2x^{2}+5x−3 }\)
  • \(g(x)=\frac{9−x}{x^{2}−81}\)
  • \(g(x)=\frac{x^{3}−27}{3−x}\)
  • \(g(x)=\frac{3x−15}{10−2x}\)
  • \(g(x)=\frac{25−5x}{4x−20}\)
  • The cost in dollars of producing coffee mugs with a company logo is given by \(C(x)=x+40\), where x represents the number of mugs produced. Calculate the average cost of producing 100 mugs and the average cost of producing 500 mugs.
  • The cost in dollars of renting a moving truck for the day is given by \(C(x)=0.45x+90\), where x represents the number of miles driven. Calculate the average cost per mile if the truck is driven 250 miles in one day.
  • The cost in dollars of producing sweat shirts with a custom design on the back is given by \(C(x)=1200+(12−0.05x)x\), where x represents the number of sweat shirts produced. Calculate the average cost of producing 150 custom sweat shirts.
  • The cost in dollars of producing a custom injected molded part is given by \(C(x)=500+(3−0.001x)x\), where x represents the number of parts produced. Calculate the average cost of producing 1,000 custom parts.

1. \(f(x)=−\frac{3x}{x+2}; x≠−2 \)

3. \(g(x)=−\frac{1}{x+9}; x≠±9 \)

5. \(g(x)=−\frac{3}{2}; x≠5 \)

7. The average cost of producing 100 mugs is $1.40 per mug. The average cost of producing 500 mugs is $1.08 per mug.

Exercise \(\PageIndex{12}\) Discussion Board

  • Explain why \(\frac{b−a}{a−b}=−1\) and illustrate this fact by substituting some numbers for the variables.
  • Explain why \(\frac{b+a}{a+b}=1\) and illustrate this fact by substituting some numbers for the variables.
  • Explain why we cannot cancel x in the expression \(\frac{x}{x+1}\).

1. Answers may vary

3. Answers may vary

IMAGES

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VIDEO

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COMMENTS

  1. 2.2: Simplifying Algebraic Expressions

    Bookshelves Algebra Beginning Algebra 2: Linear Equations and Inequalities

  2. Simplify Calculator

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  3. Algebra Topics: Simplifying Expressions

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  7. Algebraic expressions

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  8. 4 Ways to Simplify Algebraic Expressions

    1 Define "like terms" by their variables and powers. In algebra, "like terms" have the same configuration of variables, raised to the same powers. In other words, for two terms to be "like", they must have the same variable or variables, or none at all, and each variable must be raised to the same power, or no power at all.

  9. Combining like terms review (article)

    A common technique for simplifying algebraic expressions. When combining like terms, such as 2x and 3x, we add their coefficients. For example, 2x + 3x = (2+3)x = 5x. What is combining like terms? We call terms "like terms" if they have the same variable part. For example, 4 x and 3 x are like terms, but 4 x and 3 w are not like terms.

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    Welcome to How to Simplify an Expression with Mr. J! Need help with how to simplify algebraic expressions? You're in the right place!Whether you're just star...

  13. Simplify in Algebra

    "Simplified" is sometimes obvious, but can also depend on what you want to do. How to Simplify There are many ways to simplify! When we simplify we use similar skills to solving equations, and that page has some good advice. Some of these things might help: Combine Like Terms Factor Expand (the opposite of factoring)

  14. Simplifying Algebraic Expressions

    Solution: The given algebraic expression is 15xy-13+4x+3y+xy+21. By combining all the like terms, we get, 4x + (15xy + xy) + 3y + (21 - 13) = 4x + 16xy + 3y + 8. Thus, the simplified algebraic expression is 4x + 16xy + 3y + 8. Comments are closed. Learn how to simplify algebraic expressions with formula and examples.

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    Step 1: Solve parentheses by adding/subtracting like terms inside and by multiplying the terms inside the brackets with the factor written outside. For example, 2x (x + y) can be simplified as 2x 2 + 2xy. Step 2: Use the exponent rules to simplify terms containing exponents. Step 3: Add or subtract the like terms.

  17. Math Practice Problems

    1. - 1 - 1 + 2 + x - 7 Simplified: 2. 2 x + 7 - 3 x + 5 x + x Simplified: Complexity=5, Mode=two-vars Simplify the expression and order your answer based on alphabetical letter and magnitude. Examples: 3x + y - 3, 2x + y2 - 4y + 4. For x2, type x^2. Example: x2 - 2x + 3 (type as: x^2 - 2x + 3). 1. 5 y + 4 + 3 x + 5 y + 4 y Simplified:

  18. Simplifying Expressions

    5. Write and simplify algebraic expressions. We can write algebraic expressions to help simplify problems. We will often be able to make a linear equation or a quadratic equation and solve it. Example of writing and simplifying expressions . Write an expression for the perimeter of the shape. Read the question carefully and highlight the key ...

  19. Algebraic word problems

    Algebraic word problems are questions that require translating sentences to equations, then solving those equations. The equations we need to write will only involve. basic arithmetic operations. and a single variable. Usually, the variable represents an unknown quantity in a real-life scenario.

  20. Simplifying Algebraic Expressions

    Possible Answers: x + 1 x − 5 1 (x − 5)2 1 (x + 5)2 x + 1 x + 5 Correct answer: x + 1 x + 5 Explanation: Let's first look at the numerator and denominator separately. x2 + 6x + 5: We need two numbers that multiply to 5 and add to 6. The numbers 1 and 5 work. So, x2 + 6x + 5 = (x + 5)(x + 1)

  21. Practice Simplifying Algebraic Expressions

    Instructions: Choose an answer and hit 'next'. You will receive your score and answers at the end. question 1 of 3 Simplify the expression: 2m + 4m - 8. 6m - 8 2m - 8 14m 6m^2 - 8 14 + m Next...

  22. Solving Equations

    Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

  23. Simple Algebra Problems

    For example, ( 4x + 7 ) is an algebraic expression where ( x ) is the variable and the numbers ( 4 ) and ( 7 ) are terms. It's important to manipulate these properly to maintain the equation's balance. Breaking Down Algebra Problems. Solving algebra problems often starts with simplifying expressions. Here's a simple method to follow:

  24. 7.1: Simplifying Rational Expressions

    Rational expressions usually are not defined for all real numbers. The real numbers that give a value of 0 in the denominator are not part of the domain. These values are called restrictions. Simplifying rational expressions is similar to simplifying fractions. First, factor the numerator and denominator and then cancel the common factors.