Simplifying expressions and the skillset needed to accurately simplify expressions are essential building blocks in algebra. Throughout your algebra and more advanced math classes, you\u2019ll be using these rules and techniques, so mastering the key steps in simplifying expressions will give you an edge later on.<\/p>\n\n\n\n
In this article, we\u2019ll break down the fundamentals of PEMDAS and refresh the different algebraic properties that will come in handy when simplifying expressions. You\u2019ll have a chance to try different problems to also test your understanding. By the end of our discussion, you\u2019ll feel confident to work on more complex expressions!<\/p>\n\n\n\n
When simplifying expressions, group appropriate terms together and apply the rules of operation in the correct order. Writing the final and simplified expression in its standard form is a great last step to follow. This means that there are different approaches when simplifying expressions, but here are some steps to help guide you:<\/p>\n\n\n\n
Step 1<\/strong>: <\/strong>Eliminate the parentheses and brackets by evaluating, distributing, and combining terms inside them.<\/p>\n\n\n\n Step 2<\/strong>: <\/strong>Rewrite terms so that they share the same form, so evaluate terms with exponents and rewrite mixed numbers to fractions.<\/p>\n\n\n\n Step 3<\/strong>: <\/strong>Multiply and divide terms when needed and when these operations are present.<\/p>\n\n\n\n Step 4<\/strong>: <\/strong>Add and subtract terms from left to right to combine like terms.<\/p>\n\n\n\n Step 5<\/strong>: <\/strong>When you can no longer combine any terms, stop and rewrite the final expression in standard form.<\/p>\n\n\n\n Do these steps sound familiar? Because you\u2019ve already encountered some of these rules in the form of the PEMDAS rule. Recall that PEMDAS<\/strong> is an arithmetic rule that stands for: P<\/strong>arenthesis, E<\/strong>xponents, M<\/strong>ultiplication, D<\/strong>ivision, A<\/strong>ddition, and S<\/strong>ubtraction. It\u2019s the general rule to follow when performing a series of arithmetic operations.<\/p>\n\n\n\n Take a look at how the expression shown above are evaluated and simplified to appreciate the PEMDAS rule even better. Aside from following the PEMDAS closely when simplifying expressions, it\u2019s important to learn how to identify like terms. Recall that like terms share the same variable and power<\/strong>. Here are some examples to help you refresh your skill in identifying like terms and unlike terms:<\/p>\n\n\n\n Like terms must share both variable and power. It\u2019s essential that both of these conditions are met when combining like terms. This is crucial when simplifying algebraic expressions, so let us show you an example of how these rules are applied with algebraic terms.<\/p>\n\n\n\n There are also different rules to remember when simplifying expressions and you\u2019d learn more about them in the next section. For now, test your understanding by answering the problems shown below. <\/p>\n\n\n\n To simplify the expression, [katex]3(4x \u2013 5y) + 2x \u2013 6y + 4^2[\/katex], what should you do first?<\/p>\n\n\n\n A. Rewrite 42<\/sup> into its whole number counterpart, 16.<\/p>\n\n\n\n B. Group all like terms by inspecting the variable and power shared.<\/p>\n\n\n\n C. Group the last three terms using a bracket.<\/p>\n\n\n\n D. Distribute 3 to eliminate the parenthesis first.<\/p>\n\n\n\n Recall that when simplifying algebraic expressions, the PEMDAS rule still applies. In PEMDAS, the first step is to eliminate existing brackets and parenthesis by distributing and simplifying terms. For [katex]3(4x \u2013 5y) + 2x \u2013 6y + 4^2[\/katex], the grouped terms are enclosed by a parenthesis with a constant factor of 3. <\/p>\n\n\n\n This means that when simplifying the given algebraic expression, begin by eliminating the parenthesis first. Hence, the first step is to distribute 3 to each of the term inside the parenthesis making D the correct answer<\/strong>.<\/p>\n\n\n\n Identify whether each pair of terms are like terms or unlike terms. <\/p>\n\n\n\n a. [katex]12mn, 10mn[\/katex]<\/p>\n\n\n\n When determining whether two terms are like terms or not, inspect the variables and the power of these variables for each of these terms. <\/p>\n\n\n\n From these, we can see that both terms share the same variables and each variable share the same power. Hence, the two terms are like terms<\/strong>. <\/p>\n\n\n\n b. [katex]2x^2y, 3xy^2[\/katex]<\/p>\n\n\n\n Using a similar process, inspect the terms\u2019 variables and powers to see whether the two are like terms or unlike terms.<\/p>\n\n\n\n Taking a look at the table, it\u2019s easy to see that they both share the same variables. However, each variable do not share the same power. Remember that for two terms to be like terms, each of the variables must share the same power. Hence, the two terms are unlike terms<\/strong>.<\/p>\n\n\n\n After simplifying an expression and you end up with unlike terms, that\u2019s the sign that you\u2019ve reached near the end of the process. An additional step would be to rewrite the resulting expression in standard form, if any, to make it easier to inspect or review later on. <\/p>\n\n\n\n At this point, you\u2019ve probably encountered the terms distribute, fractions, and even exponents throughout this discussion. That\u2019s because the process of simplifying expressions utilizes a wide range of rules and properties that you\u2019ve learned in the past. Depending on the complexity of the expression that needs to be simplified, you\u2019ll need to ensure that you\u2019re comfortable utilizing these rules.<\/p>\n\n\n\n There are additional rules that will help you in simplifying expressions and some of them are:<\/p>\n\n\n\n [katex]\\begin{aligned}k(a + b) &= k\\cdot a + k\\cdot b\\\\k(a – b) &= k\\cdot a – k\\cdot b\\end{aligned}[\/katex]<\/p>\n\n\n\n Remember these pointers when simplifying expressions. This section will show you different examples of where these rules apply. For now, take a quick refresher on how to combine like terms by adding or subtracting their coefficients.<\/p>\n\n\n\n When simplifying expressions with like terms, add or subtract the like terms\u2019 coefficients. For example, when working with [katex]4mn + 6mn[\/katex] in one expression, add the two terms by adding their coefficients and retaining the variable part.<\/p>\n\n\n\n [katex]\\begin{aligned}{\\color{Purple}4}mn + {\\color{Purple}6}mn &= {\\color{Purple}(4+ 6)}mn\\\\&= 10mn\\end{aligned}[\/katex]<\/p>\n\n\n\n Similarly, when subtracting two terms such as [katex]5a^2b \u2013 2a^2b[\/katex], retain the variable part and subtract the terms\u2019 coefficients. <\/p>\n\n\n\n [katex]\\begin{aligned}{\\color{Purple}5}a^2b – {\\color{Purple}2}a^2b &= {\\color{Purple}(5 – 2)}a^2b\\\\&= 3a^2b\\end{aligned}[\/katex]<\/p>\n\n\n\n These are steps that you\u2019ll often encounter when simplifying expressions, so it\u2019s important that you\u2019re comfortable with combining like terms. <\/p>\n\n\n\n Simplify the following expressions by combining like terms.<\/p>\n\n\n\n a. [katex]5x^2y + 6x^2y[\/katex]<\/p>\n\n\n\n By inspecting the variables and their powers, we can see that the two terms are like terms. When combining like terms, add (or subtract) the terms by retaining the shared variables and adding (or subtracting) the coefficients. <\/p>\n\n\n\n [katex]\\begin{aligned}{\\color{Teal}5}x^2y + {\\color{Teal}6}x^2y &= {\\color{Teal}(5 + 6)}x^2y\\\\&=11x^2y\\end{aligned}[\/katex]<\/p>\n\n\n\n By adding the terms\u2019 coefficients, we have added and simplified the two like terms into one simplified term. The expression, [katex]5x^2y + 6x^2y[\/katex], can still be simplified to [katex]11x^2y[\/katex].<\/p>\n\n\n\n b. [katex]12ab + 3mn \u2013 6ab + 2mn[\/katex]<\/p>\n\n\n\n When asked to simplify terms with three or more terms, always group like terms altogether first. <\/p>\n\n\n\n [katex]\\begin{aligned}{\\color{DarkBlue}12ab} + {\\color{DarkRed}3mn} {\\color{DarkBlue}-6ab} + {\\color{DarkRed}2mn}&= {\\color{DarkBlue}(12ab – 6ab)} + {\\color{DarkRed}(3mn + 2mn)} \\end{aligned}[\/katex]<\/p>\n\n\n\n Now that the like terms have been grouped together, simplify each group of terms by applying the same process. Focus on the coefficients for each pair of like terms when simplifying them.<\/p>\n\n\n\n [katex]\\begin{aligned}{\\color{DarkBlue}(12ab – 6ab)} + {\\color{DarkRed}(3mn + 2mn)} &= {\\color{DarkBlue}(12 – 6)ab} + {\\color{DarkRed}(3 +2)mn}\\\\&= 6ab + 5mn\\end{aligned}[\/katex]<\/p>\n\n\n\n Since the resulting terms no longer share the same variables and powers, there is no need to simplify the expression further. Hence, the simplified form of [katex]12ab + 3mn \u2013 6ab + 2mn[\/katex] is equal to [katex]6ab + 5mn[\/katex].<\/p>\n\n\n\n Simplifying expressions with distributive property is usually the first step you do when the expression contains coefficients before brackets or parenthesis. When distributing the coefficient or factor into the parenthesis, multiply each terms inside the parenthesis by the coefficient. <\/p>\n\n\n\n [katex]\\begin{aligned}k(a + b) &= k\\cdot a + k\\cdot b\\\\k(a – b) &= k\\cdot a – k\\cdot b\\end{aligned}[\/katex]<\/p>\n\n\n\n If the coefficient is negative, account for its negative sign and make sure to change the signs of the terms inside the parenthesis. After distributing any existing factors, remove the parenthesis enclosing each term. <\/p>\n\n\n\n [katex]\\begin{aligned}4(3x + 2y) &={\\color{Orchid}4}(3x) +{\\color{Orchid}4}(2y)\\\\&=12x +8y\\\\\\\\-2(5a – b) &={\\color{Orchid}-2}(5a) +{\\color{Orchid}-2}(-b)\\\\&=-10a + 2b\\end{aligned}[\/katex]<\/p>\n\n\n\n Now that you\u2019ve had a quick refresher on how to apply the distributive property, why don\u2019t you try simplifying the expression shown below? <\/p>\n\n\n\n Simplify the following expression, [katex]2(x \u2013 m) + 4(x + m)[\/katex], using the different rules that you\u2019ve learned.<\/p>\n\n\n\n When working with expressions with parenthesis or two, eliminate them first. When you see coefficients before the parenthesis, such as 2 and 4, apply the distributive property.<\/p>\n\n\n\n [katex]\\begin{aligned}2(x \u2013 m) + 4(x + m)&={\\color{Orchid}2}(x) -{\\color{Orchid}2}(m) +{\\color{Teal}4}(x) +{\\color{Teal}4}(m)\\\\&= 2x – 2m + 4x + 4m\\end{aligned}[\/katex]<\/p>\n\n\n\n By inspection, we can still that there are like terms that can still be combined. Group the like terms and evaluate them to simplify the expression further. <\/p>\n\n\n\n [katex]\\begin{aligned}2x – 2m + 4x + 4m &= {\\color{Orchid}(4m – 2m)} + {\\color{Teal}(2x +4x)}\\\\&={\\color{Orchid}(4-2)}m + {\\color{Teal}(2 + 4)}x\\\\&= 2m + 6x\\end{aligned}[\/katex]<\/p>\n\n\n\n By applying the correct properties and rules, [katex]2(x \u2013 m) + 4(x + m)[\/katex] has been simplified to [katex]2m + 6x[\/katex].<\/p>\n\n\n\n When simplifying expressions with exponents, it\u2019s important to follow the rules of exponents accurately. After eliminating the parenthesis, the next step in simplifying expressions is to evaluate terms with exponents. This is why it\u2019s important that you\u2019re familiar with the fundamental rules of exponents.<\/p>\n\n\n\n Apply these rules when simplifying expressions with exponents. Of course, eliminating the parenthesis is still the first step to work on when you have to. <\/p>\n\n\n\n Simplify the following expressions by applying the appropriate rules.<\/p>\n\n\n\n a.<\/p>\n\n\n\n [katex]4 + 2(1\/4 + 1\/2) \u2013 (1\/2)^2 +2^3 [\/katex]<\/p>\n\n\n\n Of course, begin by eliminating the parenthesis by distributing 2 into the terms contained by the parenthesis. Then evaluate the terms with exponents by applying the appropriate rules of exponents.<\/p>\n\n\n\n [katex]\\begin{aligned}4 + 2(1\/4 + 1\/2) \u2013 (1\/2)^2 +2^3 &= 4 + {\\color{DarkOrange}2}(1\/4) +{\\color{DarkOrange}2}(1\/2)\u2013 (1\/2)^2 +2^3\\\\&=4 + 1\/2 + 1-(1\/2)^2 +2^3\\\\&= 4 + 1\/2 + 1{\\color{DarkOrange}-1\/4 + 8}\\end{aligned}[\/katex]<\/p>\n\n\n\n Now, simplify the expression further by adding the terms from left to right.<\/p>\n\n\n\n [katex]\\begin{aligned}4 + 1\/2 + 1-(1\/2)^2 +2^3&= 4 + 1\/2 + 1-1\/4 + 8\\\\&=9\/2+ 1-1\/4 + 8\\\\&=11\/2 -1\/4 +8\\\\&=21\/4 +8\\\\&=53\/4\\end{aligned}[\/katex]<\/p>\n\n\n\n This means that we can simplify the expression to [katex]53\/4[\/katex].<\/p>\n\n\n\n b. [katex](ab)^2 + 2(a^2b^2 + m^2n^2) – m^2n^2[\/katex]<\/p>\n\n\n\n Now, apply a similar process to simplify the algebraic expression. Distribute 2 into the parenthesis then evaluate the first term by applying the power of a product rule. <\/p>\n\n\n\n [katex]\\begin{aligned}(ab)^2 + 2(a^2b^2 + m^2n^2) – m^2n^2&= (ab)^2 + {\\color{Purple}2}(a^2b^2)+{\\color{Purple}2}(m^2n^2) – m^2n^2\\\\&=(ab)^2 + 2a^2b^2 +2m^2n^2-m^2n^2\\\\&={\\color{Purple}a^2b^2}+ 2a^2b^2 +2m^2n^2-m^2n^2\\end{aligned}[\/katex]<\/p>\n\n\n\n Inspect for like terms (check the variable and make sure they also share the same powers). Group these terms then combine the like terms.<\/p>\n\n\n\n [katex]\\begin{aligned}a^2b^2+ 2a^2b^2 +2m^2n^2-m^2n^2&={\\color{Purple}(a^2b^2+ 2a^2b^2)} +{\\color{Teal} (2m^2n^2-m^2n^2)}\\\\&= {\\color{Purple}(1 + 2)}a^2b^2 + {\\color{Teal}(2 -1)}m^2n^2\\\\&=3a^2b^2+m^2n^2\\end{aligned}[\/katex]<\/p>\n\n\n\n Now that there are no more like terms that can be found from the simplified expression, we can now stop. Hence, the simplified expression is now [katex]3a^2b^2+m^2n^2[\/katex].<\/p>\n\n\n\n These examples have shown that by applying the correct properties and algebraic rule in the right order, simplifying expressions won\u2019t be as intimidating. By making a systematic approach, you avoid making crucial arithmetic mistakes. Review this article whenever you need to and practice the problems again if you want to!<\/p>\n","protected":false},"excerpt":{"rendered":" Simplifying expressions and the skillset needed to accurately simplify expressions are essential building blocks in algebra. In this article, we\u2019ll break down the fundamentals of PEMDAS and refresh the different algebraic properties that will come in handy when simplifying expressions.<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"class_list":["post-576","post","type-post","status-publish","format-standard","hentry","category-algebra"],"_links":{"self":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/576","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/comments?post=576"}],"version-history":[{"count":5,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/576\/revisions"}],"predecessor-version":[{"id":608,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/576\/revisions\/608"}],"wp:attachment":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/media?parent=576"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/categories?post=576"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/tags?post=576"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Applying PEMDAS Rules in Arithmetic Operations<\/strong><\/td><\/tr> [katex]\\begin{aligned}5\\times (12 -8) + 3(5 -2) – 3^2 &= 5 \\times {\\color{DarkGreen}4} + 3\\times{\\color{DarkGreen}3} – 3^2\\,\\,,\\color{DarkGreen}\\text{Parenthesis}\\\\&= 5\\times 4 + 3\\times 3 – {\\color{DarkGreen}9}\\,\\,,\\color{DarkGreen}\\text{Exponent}\\\\&={\\color{DarkGreen}20} + {\\color{DarkGreen}9} – 9\\,\\,,\\color{DarkGreen}\\text{Multiplication}\\\\&= {\\color{DarkGreen}29} – 9\\,\\,,\\color{DarkGreen}\\text{Addition and Subtraction}\\\\&=20\\end{aligned}[\/katex]<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Like Terms<\/strong><\/td> Unlike Terms<\/strong><\/td><\/tr> [katex]4x[\/katex] [katex]6x[\/katex]<\/td> [katex]2x[\/katex] [katex]5x^2[\/katex]<\/td><\/tr> [katex]-5mn[\/katex] [katex]10mn[\/katex]<\/td> [katex]-5mn^2[\/katex] [katex]-5m^2n[\/katex]<\/td><\/tr> [katex]2a^2b[\/katex] [katex]3a^2b[\/katex]<\/td> [katex]4ab^2[\/katex] [katex]3a^3b[\/katex]<\/td><\/tr> [katex]\\dfrac{1}{2}x^3y^2[\/katex] [katex]\\dfrac{2}{3}x^3y^2[\/katex]<\/td> [katex]\\dfrac{1}{5}xy^2z[\/katex] [katex]\\dfrac{1}{5}x^2y^3z^2[\/katex]<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Applying PEMDAS Rules With Algebraic Terms<\/strong><\/td><\/tr> [katex]\\begin{aligned}3mn – 2(4mn + n) + 6n&= 3mn {\\color{DarkOrange} -8mn – 2n} + 6n\\,\\,,\\color{DarkOrange}\\text{Parenthesis}\\\\&= {\\color{DarkOrange}(3mn – 8mn)} + {\\color{DarkOrange}(-2n + 6n)}\\,\\,,\\color{DarkOrange}\\text{Group Terms}\\\\&={\\color{DarkOrange}-5mn} + {\\color{DarkOrange}4n}\\,\\,,\\color{DarkOrange}\\text{Add and Subtract}\\\\&= 4n – 5mn\\end{aligned}[\/katex]<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Problem 1<\/h4>\n\n\n\n
Problem 2<\/h4>\n\n\n\n
Term<\/strong><\/td> Variable<\/strong><\/td> Power<\/strong><\/td><\/tr> [katex]12mn[\/katex]<\/td> [katex]m[\/katex]<\/td> [katex]n[\/katex]<\/td><\/tr> [katex]1[\/katex]<\/td> [katex]1[\/katex]<\/td><\/tr> [katex]10mn[\/katex]<\/td> [katex]m[\/katex]<\/td> [katex]n[\/katex]<\/td><\/tr> [katex]1[\/katex]<\/td> [katex]1[\/katex]<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Term<\/strong><\/td> Variable<\/strong><\/td> Power<\/strong><\/td><\/tr> [katex]2x^2y[\/katex]<\/td> [katex]x[\/katex]<\/td> [katex]y[\/katex]<\/td><\/tr> [katex]2[\/katex]<\/td> [katex]1[\/katex]<\/td><\/tr> [katex] 3xy^2 [\/katex]<\/td> [katex]x[\/katex]<\/td> [katex]y[\/katex]<\/td><\/tr> [katex]1[\/katex]<\/td> [katex]2[\/katex]<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n What Are Additional Rules for Simplifying Expressions?<\/h2>\n\n\n\n
Simplifying Expressions With Like Terms<\/h3>\n\n\n\n
Problem 3<\/h4>\n\n\n\n
Simplifying Expressions With Distributive Property <\/h3>\n\n\n\n
Problem 4<\/h4>\n\n\n\n
Simplifying Expressions With Exponents <\/h3>\n\n\n\n
Rules of Exponents<\/strong><\/td><\/tr> Zero Rule<\/td> [katex]b^0 = 1[\/katex]<\/td><\/tr> Identity Rule<\/td> [katex]b^1 = b[\/katex]<\/td><\/tr> Negative Exponent Rule<\/td> [katex]b^{-m} = 1\/ b^m[\/katex]<\/td><\/tr> Product Rule<\/td> [katex]b^m \\cdot b^n = b^{m +n}[\/katex]<\/td><\/tr> Quotient Rule<\/td> [katex]b^m\/b^n = b^{m \u2013n}[\/katex]<\/td><\/tr> Power of a Power Rule<\/td> [katex](b^m)^n = b^{mn}[\/katex]<\/td><\/tr> Power of a Product Rule<\/td> [katex](ab)^m = a^mb^m[\/katex]<\/td><\/tr> Power of a Quotient Rule<\/td> [katex](a\/b)^m = a^m\/ b^m[\/katex]<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Problem 5<\/h4>\n\n\n\n