{"id":1,"date":"2018-02-21T08:12:39","date_gmt":"2018-02-21T08:12:39","guid":{"rendered":"https:\/\/www.homeworkminutes.com\/blog\/?p=1"},"modified":"2022-12-23T01:42:32","modified_gmt":"2022-12-23T09:42:32","slug":"are-you-struggling-with-algebra","status":"publish","type":"post","link":"https:\/\/www.homeworkminutes.com\/blog\/are-you-struggling-with-algebra\/","title":{"rendered":"Are you struggling with Algebra?"},"content":{"rendered":"

 <\/p>\n

We are here for you. We will be your 24hr guide to teach you Algebra from a to z.<\/h3>\n

An important skill in algebra is the ability to perform mathematical manipulations of expressions and equalities. When we deal with simple numbers, such as 2 + 9 = 11, there is seldom a need to make changes to simplify or expand the expressions (2 + 9 and 11 are equivalent expressions). If we deal with an algebraic expression, such as 2 +\u00a0x<\/em>\u00a0= 9, then we must extend our ability to perform arithmetic operations to deal with variables as well as numbers.<\/p>\n

 <\/p>\n

Order of Operations<\/em><\/strong><\/p>\n

It is helpful to briefly review the\u00a0order of operations<\/strong>\u00a0in the context of algebra. When performing any series of operations expressed in mathematical form, perform them in the following order:<\/p>\n

1. Any operations in parentheses<\/strong><\/p>\n

2. Exponents<\/strong><\/p>\n

3. Multiplication\/division<\/strong><\/p>\n

4. Addition\/subtraction<\/strong><\/p>\n

Within a set of parentheses, there may be a series of operations-in this case, simply start back at the first step for the expression in the parentheses.<\/p>\n

Practice Problem:\u00a0<\/u>Evaluate the following expression using\u00a0x<\/em>\u00a0= 4: 3x<\/em>2<\/sup>\u00a0\u2013 (4x<\/em>+ 1).<\/p>\n

Solution:\u00a0<\/u>Use the rules for order of operations. First, evaluate the expression in parentheses, using\u00a0x<\/em>\u00a0= 4:<\/p>\n

4x<\/em>\u00a0+ 1 = 4(4) + 1 = 16 + 1 = 17<\/p>\n

Next, evaluate the exponent in the first term:<\/p>\n

x<\/em>2<\/sup>\u00a0= (4)2<\/sup>\u00a0= 16<\/p>\n

Let\u2019s now take a look at the full expression:<\/p>\n

3(16)\u00a0\u2013 17<\/p>\n

Now, evaluate the product of 3 and 16, then subtract 17.<\/p>\n

3\u00a0\"\"16 \u2013 17 = 48 \u2013 17 = 31<\/p>\n

The solution is then 31.<\/p>\n

 <\/p>\n

Equalities<\/em><\/strong><\/p>\n

First, we need to recognize the importance of the \u201cequal sign\u201d (=) in math: if two expressions are equal, then they are indeed\u00a0equal<\/em>. Although this fact may seem obvious, it has several important implications for algebra. Let\u2019s take the following equality as an example for our discussion:<\/p>\n

\"\"<\/sub><\/p>\n

Remember, both sides of this equality are equal to one another. As a result, if we perform the same operation,\u00a0whatever it is<\/em>, on\u00a0both<\/em>\u00a0sides of the equation, the resulting expressions will\u00a0still<\/em>\u00a0be equal. For instance, if we have two bananas on the left and two bananas on the right, then the numbers of bananas on each side are equal. Thus, if you take one banana away from the left\u00a0and also<\/em>\u00a0take one away from the right, then you still have the same number on both sides. Indeed, if you add\u00a0x<\/em>\u00a0bananas to the left and you add\u00a0x<\/em>\u00a0bananas to the right, you still have the same number on each side, even if you don\u2019t know what\u00a0x<\/em>\u00a0is.<\/p>\n

So, let\u2019s consider our algebraic expression example above. Since both sides are equal, we can do whatever we want to the expressions and still have an equality\u00a0as long as we perform the same operations, in the same sequence, on both sides<\/em>. (Note that the sequence is important\u2013the best way to approach this process is to perform one and only one operation on both sides before performing a different operation.) What if we decide to add 2 to both sides?<\/p>\n

\"\"<\/sub><\/p>\n

The resulting expressions are still equal, because we performed the same operation on both sides. We can even simplify the result as follows (using associativity and commutativity).<\/p>\n

\"\"<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Commutativity<\/p>\n

\"\"<\/sub>\u00a0Associativity<\/p>\n

\"\"<\/sub><\/p>\n

These expressions are all still equal. Although the expressions have changed, the equality has not. Note carefully that in this case we have modified the equality so that the constant number value only appears on the right side. Let\u2019s do the same for multiples of\u00a0x<\/em>\u00a0by subtracting 3x<\/em>\u00a0from both sides of the equality. We will then apply our rules of real numbers.<\/p>\n

\"\"<\/sub><\/p>\n

\"\"<\/sub>\u00a0\u00a0 \u00a0 \u00a0 Commutativity<\/p>\n

\"\"<\/sub>\u00a0Associativity<\/p>\n

\"\"<\/sub><\/p>\n

Interested in learning more? Why not take an\u00a0online class in Algebra<\/a>?<\/div>\n

So, we can perform operations on both sides of the equal sign regardless of whether they involve just numbers or numbers and variables, and still maintain the equality. Note that the final result above has only three terms total: this is a simpler equality than the one we started with (which had five terms).<\/p>\n

We can also perform multiplication and division operations in the same manner. For instance, consider the following equality..<\/p>\n

\"\"<\/sub><\/p>\n

Since these expressions are in the form of fractions, they can be difficult to work with. We can simplify them, however, by applying what we know about equalities. Let\u2019s first multiply both sides by 5x<\/em>; the equality will still hold. Again, we\u2019ll use our rules for real numbers: associativity, identity, and distributivity.<\/p>\n

\"\"<\/sub><\/p>\n

\"\"<\/sub>\u00a0\u00a0 Associativity<\/p>\n

\"\"<\/sub>\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Identity<\/p>\n

\"\"<\/sub>\u00a0\u00a0\u00a0\u00a0 Distributivity<\/p>\n

\"\"<\/sub><\/p>\n

So, multiplication and division (note that 5x<\/em>\u00a0divided by 5x<\/em>\u00a0is just 1) work the same way for algebraic expressions containing numbers as they do for simple numbers. As you become more familiar with these and other algebraic operations, you will become more able to perform them without needing to recognize the specific rules (for instance, associativity) that justify them.<\/p>\n

Although it may or may not seem obvious, there is one additional point that we need to consider regarding equalities. Let\u2019s say a particular equality contains a variable\u00a0x<\/em>. The values of\u00a0x<\/em>\u00a0that satisfy this equality\u00a0also<\/em>\u00a0satisfy other equalities that are derived from this equality using (most) legitimate algebraic operations. (There are some limitations to this statement, but they can typically be addressed case by case.) Let\u2019s consider an example: 3 +\u00a0x<\/em>\u00a0= 5. Obviously, the value\u00a0x<\/em>\u00a0= 2 satisfies this equality, since 3 + 2 = 5. What if we perform some algebraic operation, however? Let\u2019s add 2 to both sides, then multiply both sides by 5.<\/p>\n

3 +\u00a0x<\/em>\u00a0+ 2 = 5 + 2<\/p>\n

x<\/em>\u00a0+ 5 = 7<\/p>\n

5(x<\/em>\u00a0+ 5) = 5(7)<\/p>\n

5x<\/em>\u00a0+ 25 = 35<\/p>\n

We know that 10 + 25 = 35, so we know that 5x<\/em>\u00a0must be equal to 10. Thus,\u00a0x<\/em>\u00a0= 2 once more. We can note from this example that the correct value (or values) for\u00a0x<\/em>\u00a0does not change if the equality is manipulated in a legitimate manner.<\/p>\n

Practice Problem:<\/u>\u00a0Justify the labeled steps in the following sequence of algebraic manipulations:<\/p>\n

\"\"<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 \u00a0 \u00a0\u00a0 Given<\/p>\n

\"\"<\/sub>\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 \u00a0\u00a0 A. ?<\/p>\n

\"\"<\/sub>\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 \u00a0\u00a0 Equality<\/p>\n

\"\"<\/sub>\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 B. ?<\/p>\n

\"\"<\/sub>\u00a0C. ?<\/p>\n

\"\"<\/sub>\u00a0\u00a0 \u00a0\u00a0 Equality<\/p>\n

\"\"<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. ?<\/p>\n

\"\"<\/sub>\u00a0E. ?<\/p>\n

Solution:<\/u>\u00a0To solve this problem, we must recall the basic rules discussed in the previous chapter and apply them to algebra. In the case of A, two terms are reversed, so the justification is\u00a0commutativity<\/em>. For B, the factor\u00a0t<\/em>\u00a0is distributed as a factor to each term in parentheses, so this step is justified by\u00a0distributivity<\/em>. In C, the product of\u00a0t<\/em>\u00a0and\u00a0\"\"<\/sub>\u00a0is 1, which is the property of (multiplicative)\u00a0inverses<\/em>. The step corresponding to D can be justified by\u00a0associativity<\/em>, and the step corresponding to E can be justified by (additive)\u00a0inverses<\/em>.<\/p>\n","protected":false},"excerpt":{"rendered":"

  We are here for you. We will be your 24hr guide to teach you Algebra from a to z. An important skill in algebra…<\/p>\n","protected":false},"author":3,"featured_media":7995,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":"","_wp_rev_ctl_limit":""},"categories":[21],"tags":[406,407,408],"yoast_head":"\nAre You Struggling With Algebra?<\/title>\n<meta name=\"description\" content=\"Are you struggling with algebra? Read this blog and find a solution to all the questions that you stuck at. 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