Algebra 1 Articles - WyzAnt Tutor Blogshttp://www.wyzant.com/resources/blogs/algebra_1This is an aggregate of all of the Algebra 1 articles in WyzAnt.com's Tutors' Blogs. WyzAnt.com is your source for tutors and students.Thu, 17 Apr 2014 13:57:44 -0500http://www.wyzant.com/images/WyzAnt_white200.gifAlgebra 1 Articles - WyzAnt Tutor Blogshttp://www.wyzant.com/resources/blogs/algebra_1http://www.wyzant.com/resources/blogs/algebra_1262225http://www.wyzant.com/resources/blogs/262225/mathematical_journeys_the_unperformed_operationEllen S.http://www.wyzant.com/resources/users/view/75479140Mathematical Journeys: The Unperformed Operation<div>Come with me on a journey of division.<br /><br />I have here a bag of M&Ms, which you and I and two of your friends want to share equally. I'm going to pour the bag out on the table and split it into four equal piles. For this example, “one bag” is our whole, and the best number to represent that whole would be the number of M&Ms in the bag. Let's say there were 32. If I split those 32 M&Ms into four equal piles and asked you how many were in one pile, you could certainly just count them. But a quicker way would be to take that 32 and divide it by the number of piles I'd made, which in this case is 4. You'd probably write that as:<br /><br /> 32 ÷ 4 = 8<br /><br />So there are 8 candies in each pile.<br /><br />Seems easy enough with a large number of M&Ms, right? But what if there were less candies – what if our “whole” was less than the entire bag? Well, for a while we'd be okay – if there were 16, for example, we'd do the same thing and come up with piles of 4 instead of piles of 8. But what if there were less candies than we needed piles – if there were less than 4 candies in the whole? What if, in fact, we had only one candy? There's still four of us, and we still need to share equally – I guess we're each getting less than one candy, right? <br /><br />If we set up our equation the same way we did above, we'll get:<br /><br /> 1 ÷ 4 = ?<br /><br />Now hold on a minute. See that dividing sign there? That looks an awful lot like a fraction, doesn't it? If you just replaced the dots with numbers? That's because it is. The dividing sign they teach you first actually came after the invention of fractions. It's a way of indicating that the first number goes on top of the fraction and the second number goes on the bottom. But once you get into higher level math classes, that dividing sign disappears. Instead, we write division as a fraction – because that's what it is!<br /><br />The reason we stop using the dividing sign is because writing division as a fraction allows you to deal with our one-M&M scenario from earlier. You simply write the division itself as a fraction, and that fraction becomes the result of the division.<br /><br />So instead of writing that equation as:<br /><br /> 1 ÷ 4 = ?<br /><br />And being confused by your lack of an answer, you'd write:<br /><br /> <span style="text-decoration: underline;">1</span><br /> 4<br /><br />And that fraction would <em>become</em> the answer. <br /><br />What's important to remember here is that a fraction is not just a number. While it IS a number – there IS exactly one point on the number line that that fraction represents – it's also an indicator of an unperformed operation. By writing that number as a fraction, you are saying “I'm supposed to divide this number by this other number, but I don't want to do that calculation just yet.” <br /><br />This might seem a bit strange, especially given that ¼ is an easy calculation to make, and that its decimal form, 0.25, is equally easy to work with. Okay, fine – I'm going to make one of your friends disappear!<br /><br />Now there are only three of us fighting over that one candy. Your new fraction would be:<br /><br /> <span style="text-decoration: underline;">1</span><br /> 3<br /><br />If you try to convert that into decimal form by dividing one by three, you'll get 0.33333333333... on into infinity. Now, I don't know about you, but I don't particularly like the idea of trying to work with a number that stretches on into infinity – my arms aren't that long! So I just won't let it out of the box – I'll keep it as a fraction as long as I possibly can, thus acknowledging the existence of another operation while refraining from performing it until I'm really ready.<br /><br />You see this concept of the unperformed operation a lot once you get into higher level math concepts, particularly in the use of named constants. Take pi, for instance. Pi is a constant, described as the result of a specific calculation involving circles. No matter what dimensions you give a circle, when you perform this calculation you end up with the same number. So clearly it's important, and it makes sense that we should be able to work with it. Only one problem – it's an incredibly unwieldy number, a non-repeating, non-terminating decimal that stretches out into infinity. Working with such a number would be downright impossible unless we are willing to approximate and chop off most of the digits. So what do we do? We give it a name, assigning it to a letter of the greek alphabet and using this letter to represent the constant in full.<br /><br />To make sure we are always working with the entire non-terminating number and not an approximation, we leave operations involving this number unperformed. We simply carry the symbol through the problem, attached to whatever other number it was supposed to be multiplied or divided by. Only at the very end do we ever actually perform the operation, and even then only if we need a numerical estimation. Much more frequently we simply express our answer “in terms of” this constant, leaving the symbol intact for the next mathematician to pick up and work with later.</div>Tue, 18 Feb 2014 18:36:42 -06002014-02-18T18:36:42-06:00260028http://www.wyzant.com/resources/blogs/260028/factoring_without_the_guess_and_check_methodShawil D.http://www.wyzant.com/resources/users/view/84848050Factoring without the "Guess and Check" method<div>Factoring can be quite difficult for those who are new to the concept. There are many ways to go about it. The guess and check way seems to be the most common, and in my mind, it is the best, especially if one wants to go further into mathematics, than Calculus 1. But for those just getting through a required algebra course, here is another way to consider, that I picked up while tutoring some time ago:</div>
<div> </div>
<div>If you have heard of factor by grouping, then this concept will make some sense to you. Let's use an example to demenstrate how to do this operation:</div>
<div> </div>
<div>Ex| x<sup>2</sup> + x - 2</div>
<div> </div>
<div>With this guess and check method, we would use (x + 1)(x - 2) or (x + 2)(x - 1). When we "foil" this out, we see that the second choice is the correct factorization. But, instead of just using these guesses, why not have a concrete way to do this.</div>
<div> </div>
<div>Let's redo the example, with another method.</div>
<div> </div>
<div>Ex| x<sup>2</sup> + x - 2</div>
<div> </div>
<div>First notice the -2, the negative shows us that the only way to this is (x - ?)(x + ?)</div>
<div>Now we look for the factors of 2. The only factors are 1,2 and 2, 1.</div>
<div>(In another case in which there are more factors, the difference or addition of these two factor must equal the middle term. That determines the correct pair of factors.</div>
<div>Now notice the middle term is positive. That means the larger of the two factors we choose must be positive also.</div>
<div>So, we can now write this as:</div>
<div> </div>
<div>x<sup>2</sup> + 2x - 1x - 2</div>
<div> </div>
<div>What we just did was replace the original "+x" with "2x - 1x". These two statements are equivalent, so this is fair game.</div>
<div> </div>
<div>Now, we group the left and right sides together. Note that when they are grouped, the negative stays with the 1, as shown below:</div>
<div> </div>
<div>(x<sup>2</sup> + 2x) + (-1x - 2)</div>
<div> </div>
<div>Now we factor out common terms from each:</div>
<div> </div>
<div>x is common on the left, and -1 is common on the right, so we factor out each of these:</div>
<div> </div>
<div>x(x + 2) - 1(x + 2)</div>
<div> </div>
<div>Now we check to see if our terms in parenthesis are equal. Since they are, we can group them into one term, and and the outer terms together. These two expressions are then multiplied:</div>
<div> </div>
<div>(x + -1)(x + 2)</div>
<div> </div>
<div>Which equals:</div>
<div> </div>
<div>(x - 1)(x + 2)</div>
<div> </div>
<div>If we foil this out, we get:</div>
<div> </div>
<div>x<sup>2</sup> + x - 2</div>
<div> </div>
<div>Which is our original problem. Therefore we have correctly factored this. And in so doing, we have learned a new way to factor that requires no guessing.</div>Mon, 10 Feb 2014 12:53:30 -06002014-02-10T12:53:30-06:00257170http://www.wyzant.com/resources/blogs/257170/all_my_students_passed_the_ny_state_regents_examsGilant P.http://www.wyzant.com/resources/users/view/77505480All my students passed the NY State Regents Exams...<div>I am happy to announce that all my students have passed the NY State Regents examinations, except one student. The subjects varied from Algebra 1, Algebra 11/Trigonometry, English, US and Global History and Living Environment. I am so proud of them. Most of these students are students who struggled quite a bit. It was a long journey but one I would do again. </div>
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<div>I am very proud of them as most of them will be graduating this year. The NY State Common Core examinations are next.</div>Wed, 05 Feb 2014 09:52:04 -06002014-02-05T09:52:04-06:00255740http://www.wyzant.com/resources/blogs/255740/high_school_geometry_for_some_why_is_this_more_challenging_than_algebra_1Bruce H.http://www.wyzant.com/resources/users/view/76725150High School Geometry: For Some, Why Is This More Challenging Than Algebra 1? <div>Several of my current Geometry students have commented on this very contrast. This has prompted me to offer a few possible reasons.</div>
<div> </div>
<div>First, Geometry requires a heavy reliance on explanations and justifications (particularly of the formal two-column proof variety) that involve stepwise, deductive reasoning. For many, this is their first exposure to this type of thought process, basically absent in Algebra 1.</div>
<div> </div>
<div>Second, a large part of Geometry involves 2-d and 3-d visualization abilities and the differences in appearance between shapes even when they are not positioned upright. Still further, for a number of students, distinguishing the characteristic properties amongst the different shapes becomes a new challenge.</div>
<div> </div>
<div>Third, in many cases Geometry entails the ability to form conjectures about observed properties of shapes, lines, line segments and angles even before the facts have been clearly established and stated. This level of abstract thinking is rarely encountered in Algebra 1.</div>
<div> </div>
<div>Implications based upon the aforementioned are significant. For those following a Science/Technology/Engineering/Math (STEM) course track and headed into Precalculus, Calculus, General Chemistry and Organic Chemistry, the ability to visualize shapes and chemical structures in 2-d and 3-d space is extremely important.</div>
<div> </div>
<div>In order to better prepare students for dealing with Geometry's rigors, many educators have believed that certain math instructional practices even going back to the elementary school level will have to be improved within the classroom. In the meantime, all students can still take steps to increase their current mental capacities and reasoning abilities by following a brain-enhancing diet (such as my "Maximum Mental Health Diet" recommendations offered at the first tutoring session). This includes daily consumption of long-chain Omega-3 essential fatty acids through fish or supplements and an abundance of colorful, antioxidant-rich plant foods.</div>Tue, 04 Feb 2014 14:04:24 -06002014-02-04T14:04:24-06:00250445http://www.wyzant.com/resources/blogs/250445/easy_way_to_remember_the_quadratic_formulaKatie H.http://www.wyzant.com/resources/users/view/80680710Easy way to remember the quadratic formula!<div>As a student, I found that I remembered information a lot easier when the information was in a song. I learned the 'quadratic formula song' in one of my math classes and have not forgotten the formula since. Several of my students have also found this song helpful (and catchy!), so I though I'd share:</div>
<div> </div>
<div>The 'Quadratic Formula Song' (sung to the lyrics of 'Pop Goes the Weasel')</div>
<div>The quadratic formula is negative b</div>
<div>plus or minus the square root</div>
<div>of b squared minus four a c</div>
<div>all over 2a!</div>
<div> </div>
<div>(Warning, this will get stuck in your head!) </div>Sun, 12 Jan 2014 00:02:49 -06002014-01-12T00:02:49-06:00249113http://www.wyzant.com/resources/blogs/249113/favorite_math_resources_k_12_gedJessica H.http://www.wyzant.com/resources/users/view/84101090Favorite Math Resources, K-12 + GED<div>Here are some of my favorite Math resources. Check back again soon, this list is always growing! I also recommend school textbooks, your local library, and used bookstores. <br /><br />As a note, college-level math textbooks are often helpful for high school math students. Why is that? Isn't that a little counter-intuitive? Yes, it would appear that way! However, many college-level math textbooks are written with the idea that many college students may not have taken a math class in a year or more, so they are written with more detailed explanations. This can be particularly helpful for high school students taking Algebra, Geometry, and Trig. I have a collection of college-level math books that I purchased at a local used bookstore. The most expensive used math book I own cost $26 used. Books that focus on standardized test prep (such as the SAT, AP, or GED prep) can be helpful for all core subjects, as they summarize key ideas more succinctly than 'normal' textbooks. These are GREAT for test review and studying as a supplemental text, whethre you're studying for that standardized test or not.</div>
<div> </div>
<div><br />(K-12) <a href="http://www.kahnacademy.org">KahnAcademy.org</a> – All math subjects, with video tutorials and practice problems (plus answers!)</div>
<div><br />(K-Algebra\Geom.) <a href="http://www.mathplayground.com">MathPlayground.com</a> – Fun math tutorials and games to help reinforce what you’ve learned!</div>
<div><br />(K-Geometry) <a href="http://www.mathisfun.com">Mathisfun.com</a> – Number Value (K/Gr. 1) to Geometry</div>
<div><br />(K-Geometry) <a href="http://www.aaamath.com">Aaamath.com</a> – Explanations, practice problems (with answers), and more!</div>
<div><br />(Gr. 2+) iPhone app – Math Help – Division, multiplication, division with remainders</div>
<div><br />(Algebra) <a href="http://www.purplemath.com">Purple Math.com</a> – All things Algebra, with detailed explanations and (a few) practice problems.</div>
<div><br />(Gr. 9-12) iPhone/Droid app- GED Math Lite – All high school math topics, with quizzes and answer explanations.<br /><br /></div>
<div>(Gr 3-12) <a href="http://www.convert-me.com/en/">Convert-Me.com</a> - Not sure how to convert kilometers to meters? Gallons to quarts? Give this site a try!</div>Tue, 31 Dec 2013 14:09:48 -06002013-12-31T14:09:48-06:00246136http://www.wyzant.com/resources/blogs/246136/i_am_in_need_of_some_studentsAshley M.http://www.wyzant.com/resources/users/view/84377950I am in need of some students. <div>Hi,</div>
<div> </div>
<div>I would be honored in having the opportunity of working with students and parents. The education and success of students are very important to me and I would love to do what I can to help. I am a math and education major with an Associate's of Arts and Teaching Degree from Lee College and I am seeking a teaching career. I live in the Baytown area and I am not able to provide my own transportation due to the fact that I have a disability which prevents me from driving, so I can only rely on public transportation and I am limited to how far I can travel. Therefor, communication is much needed. I am available until 4:30 p.m. Monday through Friday. Anyone needing a private tutor, please contact me. I would be happy to help you at any time.</div>Sat, 23 Nov 2013 19:47:33 -06002013-11-23T19:47:33-06:00245878http://www.wyzant.com/resources/blogs/245878/christmas_presentAndrew L.http://www.wyzant.com/resources/users/view/75937410Christmas Present!!<div>Hi All!</div>
<div> </div>
<div>In the spirit of giving, starting on 11/29/2013, I will be offering a few brainteasers/ trivia questions where the first 3 people to email me the correct answer will receive a free, one hour, tutoring session in any subject that I offer tutoring for (via the online platform)! That's right free! Get your thinking hats on everyone!</div>
<div> </div>
<h1><span class="greenText"><strong>Merry <span class="orange">Christmas</span>!! </strong></span></h1>
<div><span class="greenText"><strong><a href="http://www.wyzant.com/Tutors/AndrewLane">Andrew L. Profile</a></strong></span></div>
<div> </div>Thu, 21 Nov 2013 01:35:13 -06002013-11-21T01:35:13-06:00243335http://www.wyzant.com/resources/blogs/243335/math_equations_formulas_and_vocabularyKayleigh T.http://www.wyzant.com/resources/users/view/84291860Math Equations, Formulas and Vocabulary<div>
<ul>
<li>Area, Volume and Circumference equations:</li>
<li>Area of a Square</li>
</ul>
</div>
<div>A=S<sup>2</sup></div>
<div>
<ul>
<li>Area of a Triangle</li>
</ul>
</div>
<div>A=1/2bh</div>
<div>
<ul>
<li>Area of a Rectangle</li>
</ul>
</div>
<div>A=LW</div>
<div>
<ul>
<li>Right Triangle/Pythagorean Theorem</li>
</ul>
</div>
<div>a<sup>2</sup>+b<sup>2</sup>=c<sup>2</sup></div>
<div>
<ul>
<li>Area of Parallelogram</li>
</ul>
</div>
<div>A=bh</div>
<div>
<ul>
<li>Area of a Trapezoid</li>
</ul>
</div>
<div>A=1/2h(a+b)</div>
<div>
<ul>
<li>Area of a Circle</li>
</ul>
</div>
<div>A=πr<sup>2</sup></div>
<div>
<ul>
<li>Circumference of a Circle</li>
</ul>
</div>
<div>c=πd or c=2πr</div>
<div>
<ul>
<li>Volume of a Sphere</li>
</ul>
</div>
<div>V=4/3πr<sup>3</sup></div>
<div>
<ul>
<li>Surface Area of a Sphere</li>
</ul>
</div>
<div>SA=4πr<sup>2</sup></div>
<div>
<ul>
<li>Volume of a Cube</li>
</ul>
</div>
<div>V=s<sup>3</sup></div>
<div>
<ul>
<li>Volume of a Rectangular Solid</li>
</ul>
</div>
<div>V=lwh</div>
<div>
<ul>
<li>Slope of a line Equations</li>
<li>Slope-intercept form</li>
</ul>
</div>
<div>y=mx+b</div>
<div>m is the slope</div>
<div>b is the y-intercept</div>
<div>y is a y coordinate on the graph (that coincides with the line)</div>
<div>x is an x coordinate on the graph (that coincides with the line)</div>
<div>
<ul>
<li>Horizontal line</li>
</ul>
y=b</div>
<div>
<ul>
<li>Vertical line</li>
</ul>
x=a</div>
<div>
<ul>
<li>Finding slope of line containing points (x<sub>1</sub>,y<sub>1</sub>) to (x<sub>2</sub>,y<sub>2</sub>)</li>
</ul>
m=<span style="text-decoration: underline;">y<sub>2</sub>-y<sub>1</sub></span></div>
<div> x<sub>2</sub>-x<sub>1</sub></div>
<div>
<ul>
<li>Distance from one point to another</li>
</ul>
(x<sub>1</sub>,y<sub>1</sub>) to (x<sub>2</sub>,y<sub>2</sub>)<span style="font-size: 9px;"><br /></span></div>
<div>d=√(x<sub>2</sub>-x<sub>1</sub>)<sup>2</sup>+(y<sub>2</sub>-y<sub>1</sub>)<sup>2</sup></div>
<div>
<ul>
<li>Midpoint of a Line Segment</li>
</ul>
( <sup><span style="text-decoration: underline;">x1+x2 </span> , <span style="text-decoration: underline;">y1+y2</span> </sup>)</div>
<div> <sup>2 2</sup></div>
<div>
<ul>
<li>Quadratic Formula</li>
</ul>
x=<span style="text-decoration: underline;">-b±√b<sup>2</sup>-4ac</span></div>
<div> 2a</div>
<div>
<ul>
<li>Math Vocabulary</li>
</ul>
<ol>
<li>Equilateral Triangle- Three sides of equal length and 3 angles 60° each</li>
<li>Scalene Triangle- Three unequal sides and 3 unequal angles</li>
<li>Isosceles Triangle- Two sides of equal length and two base angles are equal</li>
<li>Commutative Property- a+b=b+a; ab=ba</li>
<li>Associative Property- a=(b+c)=(a+b)+c; a(b+c)=ab+ac</li>
<li>Distributive Property- a9b+c)=ab+ac</li>
<li>Additive Identity- a+0=0+a=a</li>
<li>Multiplicative Identity- a•1=1•a=a</li>
<li>Additive Inverse- -a+a=a+(-a)=0</li>
<li>Multiplicative Inverse- a•1/a=1, a≠0</li>
</ol></div>Wed, 06 Nov 2013 00:09:02 -06002013-11-06T00:09:02-06:00243067http://www.wyzant.com/resources/blogs/243067/why_i_hate_foilDevin F.http://www.wyzant.com/resources/users/view/78876240Why I Hate FOIL<div>Let's use our imagination a bit. Picture yourself in math class (Algebra I to be exact), minding your own business, having fun playing with the axioms (aka rules) of algebra, and then one day your teacher drops this bomb on you:<br /><br />"Expand (x+3)(x-1)"<br /><br />And you might be thinking, "woah now, where did come from?"<br /><br />It makes sense that this would shock you. You were just getting used to the idea of expanding 3(x-1), and you probably would have been fine with x+3(x-1), but (x+3)(x-1) is a foreign idea all together.<br /><br />Well, before you have much time to think about it on your own and discover anything interesting, your teacher will probably tell you that even though you don't know how to solve it now, there is a "super helpful", magical technique that will help you…<br />FOIL<br /><br />For those of you lucky enough never to have heard of FOIL, I will explain. FOIL stands for First Outside Inside Last and is a common mnemonic device used to confuse children about a fairly easy concept.<br /><br />If you remember the distributive property a(b+c) = ab+ac, then it might seem odd that all of a sudden we put two grouping next to each other and now we are doing something "new".<br />But is FOIL really new?<br /><br />The answer is of course no, what we are actually doing is just a short cut for the distributive property. And if you were aloud to try and solve it before being told what to do, you might actually have figured that out. For example, if we have (a+b)(c+d), we could distribute (a+b) as if it was a whole quantity. So (a+b)(c+d) = c(a+b) + d(a+b) and then we distribute again and get ac + bc + ad + bd.<br /><br />To me this seems much simpler than having to learn a mnemonic device, and remember how to draw our "rainbow lines" and remember where to put a plus and a minus, and so son and so forth. We merely follow a simple rules we already know.<br /><br />Another useful reason not to teach FOIL is because in only works for expressions similar to (a+b)(c+d). But what about expressions that look like (a+b)(c+d+e), or even (a-b+c)(d+e)(f-g-h+i+j)(k-l+p)? You can't use FOIL for these, but of course, you use the distributive property.<br /><br />So please, if you are a math teacher, the next time you have a chance to teach FOIL… don't. Spare your students the confusion and teach them what is really going on. FOIL might be quicker, but math isn't about the destination, it's about the journey.</div>Sun, 03 Nov 2013 10:08:24 -06002013-11-03T10:08:24-06:00242939http://www.wyzant.com/resources/blogs/242939/algebraic_equations_and_expressionsMichael L.http://www.wyzant.com/resources/users/view/82542570Algebraic Equations and Expressions<div>One of the major differences between algebraic equations and algebraic expressions consist of the equal sign because the equal sign consitutes for a solution that can be checked to verify that it is the solution. Expressions are meant to be simplified so common factors are important in simplifying expressions. Equations give a way to actually check the answer by subsitution for the variable while expressions are normally checked by multiplication or another type of operation.</div>Fri, 01 Nov 2013 19:40:00 -05002013-11-01T19:40:00-05:00241993http://www.wyzant.com/resources/blogs/241993/a_few_thoughts_on_why_we_learn_algebraEllen S.http://www.wyzant.com/resources/users/view/75479140A Few Thoughts on Why We Learn Algebra<div>I recently responded to a question on WyzAnt's “Answers” page from a very frustrated student asking why he should bother learning algebra. He wanted to know when he would ever need to use it in the “real world” because it was frustrating him to tears and “I'm tired of trying to find your x algebra, and I don't care y either!!!”<br /><br />Now, despite that being a pretty awesome joke, I really felt for this kid. I hear this sort of complaint a lot from students who desperately want to just throw in the towel and skip math completely. But what bothered me even more were the responses already given by three or four other tutors. They were all valid points talking about life skills that require math, such as paying bills, applying for loans, etc., or else career fields that involve math such as computer science and physics. I hear these responses a lot too, and what bothers me is that those answers are clearly not what this poor student needed to hear. When you're that frustrated about math, the last thing you want to hear is that you'll “need” to be able to do math in order to live. That might cause you to take those responses as a sign that you should not venture anywhere near computer science or physics, or ever rent an apartment or own a car. What this student needed to hear was the larger picture of why math is useful even if you never touch another x or y in your life. So here are my thoughts, in the form of an example from my own life.<br /><br />When you get right down to it, at its most basic level, algebra centers around the idea that you can add, subtract, multiply, or divide both sides of an equation by the same number and the equation will continue to be true. That very basic concept of “doing the same thing to both sides” has the implication of allowing you to rewrite the same equation a multitude of different ways without changing its value. In essence, algebra is problem solving at its most basic. You start with what you know, and by working step by step through rewriting the equation while maintaining its value, you arrive at a version of the equation that makes the unknown element clear. The whole time, the laws of algebra remind you that you're not changing what the equation means; you're just rewriting it in a way that's easier to work with and understand. This type of step-by-step problem solving has a multitude of uses in everyday life that don't involve a single number.<br /><br />Here's my example. One evening in college, I arrived back at my dorm building after a long day of classes, only to find that my wallet was not in my bag. I had no idea how long I'd been without my wallet, and even less idea where it was. On top of that, I had a small window of time in which I was supposed to go home and change out my books before heading out again, so I needed to get into the dorm NOW, which I couldn't do without my student ID card, which was – you guessed it – in my wallet. So what do I do?<br /><br />Well, I'll be honest, I began to panic slightly. But I worked through the panic and figured out my first plan: retrace my steps until I found my wallet. Fortunately, all of my classes that day had been in the same building, so I didn't have far to go. Unfortunately, my wallet was not anywhere on the path I'd taken from the dorm to the classroom, the path home, or anywhere in between. The wallet was lost.<br /><br />Having hit a dead-end on that front, I decided to set that problem aside and deal with the second issue: I still needed to get into the dorm to change out my textbooks. I figured I'd work on getting into the dorm, and perhaps once I was there more options for the lost wallet would present themselves. So instead of heading for the back door, which required an ID swipe to get in, I walked around to the front entrance and went into the lobby (a public area). I then headed over to the door that led to my wing, and killed time by pretending to read the bulletin board on the wall nearby. Soon, another student came by and swiped her card to open the door. I hurriedly slipped in behind her before the door closed, knowing that most people ignored the signs saying to not let anyone else in after you. I ran up to my room, opened the door (thankfully I still had my keys!) and there was my wallet, lying on the floor in the middle of the room.<br /><br />So what does any of this have to do with algebra? Well, compare my problem-solving strategies to the process of solving a system of equations. In my case, I had two variables: I needed to get into the dorm, and my wallet was gone. I started by trying to find my wallet – when solving a system, you start by solving one equation for one variable. I got as far as I could go on that path and eventually ended up with wallet = gone. I had to set that equation aside for a moment and deal with the other variable, just as you then switch equations in the system. I plugged “I don't have my ID” into the equation of “getting into the dorm” and solved that problem using what I knew about the building and the residents' laziness, and managed to get into my room (I solved for “I need to get into the dorm”). Once in my room, the first equation became solvable again, since my wallet turned out to be there – right where it had fallen out of my bag before I left the room that morning.<br /><br />This may sound way too coincidental, but the truth is that algebraic reasoning is incredibly important for a lot of tasks that have nothing to do with numbers. The ability to rewrite an equation while maintaining its value until the answer presents itself is at the heart of all problem-solving abilities. I often remind my students of the larger usefulness of the skills learned in math class by encouraging them to “take the numbers out of it.” What exactly are you doing in a broader sense, and how might you be able to use those skills in other situations? Give it a try – you might find that you like math more than you thought.</div>Wed, 23 Oct 2013 08:47:38 -05002013-10-23T08:47:38-05:00239146http://www.wyzant.com/resources/blogs/239146/how_to_be_successful_in_mathematicsAdrienne J.http://www.wyzant.com/resources/users/view/81955960How to be Successful in Mathematics<p><strong>How to be Successful In Mathematics</strong><br /><br />Math is a complicated subject. Students struggle with it, parents don’t feel comfortable helping with homework, and teachers find it impossible to “re-teach” every year. It is for these reasons that I feel having a good foundation in math is imperative. Students that have a great foundation feel confident and are not afraid of tackling a problem until they figure it out.</p>
<p> <br /><strong>What do students need to know to have a good foundation? </strong></p>
<p>Well, I think the most basic concepts they need to master are the concepts learned in pre-algebra. Most parents would be shocked to hear that students begin to learn these concepts as early as second grade. <br />Some are those concepts include properly using the order of operations; being able to add, subtract, multiply and divide negative numbers, fractions and decimals; and working problems with more than one variable. I encounter students “freezing” all the time when they encounter fractions, negative numbers, and variables. If they had more confidence in these concepts, they would feel comfortable trying a more complicated problem without hesitation.</p>
<p><br />My recommendation for mastering these concepts is practice, practice, practice. Identify where you are weak and do as many problems as you can to work those concepts. It is amazing how working the same kind of problem over and over again can drive a concept home.</p>
<p><br /><strong>Where can you find extra problems??</strong></p>
<p><br />Many students don’t realize that most textbooks supply extra problems in the back of their textbook. Also, there is a mixed review at the end of each chapter that combines all of the concepts of that chapter in one place. Feel free to go back to previous chapter and work problems for a review. The index is also a great tool to help you find extra problems.</p>Wed, 25 Sep 2013 20:39:42 -05002013-09-25T20:39:42-05:00237366http://www.wyzant.com/resources/blogs/237366/the_ordering_of_algebraic_operationsYingda Z.http://www.wyzant.com/resources/users/view/81959800The ordering of algebraic operations<div>When both writing down and reading the algebraic expressions, the binary operation (including addition+, subtraction-, multiply*, divide/, exponential^) follow a conventional order:<br /> <br />0) Parenthesis, including {}, [], ()<br />1) Exponent, multiply and divide<br />2) Addition and subtraction<br /> <br />The ordering is 0)>1)>2). Then there is no ordering within each group, eg multiply and divide are at the same level of priority except that 0) comes in such as a parenthesis.<br /> <br />Let's take a look at one quick example: 3+(8-2)*6.<br />First compute (8-2)=6;<br />Then compute (8-2)*6=6*6=36;<br />Finally compute 3+(8-2)*6=3+36=39.<br /> <br />Another example: 3^2+3/(5-2)<br />First compute (5-2)=3;<br />Then do 3/(5-3)=3/3=1;<br />Next compute 3^2=3*3=9;<br />Finally add 3^2+3/(5-2)=9+1=10.<br /> <br />Hope it helps!</div>Mon, 16 Sep 2013 09:22:10 -05002013-09-16T09:22:10-05:00237223http://www.wyzant.com/resources/blogs/237223/get_a_quick_start_don_t_procrastinate_and_get_behindSteven L.http://www.wyzant.com/resources/users/view/82748460Get a quick start. Don't procrastinate and get behind. <div>My worst school years were when I did not keep up because I didn't care for the subject. Get over it. If the course is required you have to take it and do well. Putting off studying and keeping up with the curriculum will only make getting ready for tests more difficult and you will not have as good understand of the subject. This can rub off on other subjects as well while you cram for exams. </div>
<div> </div>
<div>The semesters I got a jump on all subjects, especially the ones I did not think I would like, I did much better. Whether it was by reading text book ahead, ready to ask questions in class or understand the lecture and making sure my class notes were well done and I reviewed them after class to fill in gaps, it all helps build the foundation for the subject matter. Generally if I did this, by the time the semester was 60% complete, the remainder was a breeze. Made all the difference for me.</div>Sat, 14 Sep 2013 10:20:39 -05002013-09-14T10:20:39-05:00234894http://www.wyzant.com/resources/blogs/234894/will_the_power_of_x_increase_or_decrease_its_absolute_value_i_e_the_value_regardless_of_the_sign_orErik S.http://www.wyzant.com/resources/users/view/82277360Will the power of x increase or decrease its absolute value (i.e. the value regardless of the sign, + or -)?<p>Well, there are two exceptions to this question. X cannot be 0 or 1 because 0*0=0, and 1*1=1. No matter how many times you multiply 0 by itself, you will always get 0, and no matter how many times you multiply 1 by itself, you will always get 1. That's why the power of x will never change its value if x is 0 or 1. Now that we realize the two exceptions of 0 and 1 for x, x would have to be in one of two certain ranges: 0<x<1 or x>1.</p>
<p>If 0<x<1, then that would mean that x is a proper fraction when the numerator is smaller than the denominator (e.g. 5/6). Let's use the easiest fraction value for x, 1/2, and the easiest power of x, x^2. Plug in the value of x, and you will get x^2=(1/2)^2. This will multiply the fraction of 1/2 twice by itself: (1/2)*(1/2). Now, since any number times 1 is that number, (1/2)*1=1/2 so that 1/2 remains the same. So if the second term is less than 1, it will make the first term smaller than itself as (1/2)*(1/2)=1/4. Therefore, the power of x will decrease its value if 0<x<1.</p>
<p>If x>1, then x would have to be anything beyond the whole number of 1 in any of the following forms: improper fraction (e.g. 3/2 because the numerator is greater than the denominator), mixed number (e.g. 1 1/2), decimal (e.g. 1.01 with at least one non-zero digit on each side of the decimal point), or whole number (e.g. 2). Let's try using for a value of x a decimal number of 1.1 as one of those forms. Plug in the value of x, and you will get x^2=(1.1)^2. This will multiply the fraction of 1.1 twice by itself: (1.1)*(1.1). Now, since any number times 1 is that number, (1.1)*1=1.1 so that 1.1 remains the same. So if the second term is greater than 1, it will make the first term larger than itself as (1.1)*(1.1)=1.21. Therefore, the power of x will increase its value if x>1.</p>
<p>In conclusion, the power of x will either increase or decrease its value as it depends on two specific ranges: 0<x<1 or x>1. If the value of x is in the former range, then the exponent will decrease its value, and the greater the exponent, the closer to 0 x will get. Otherwise, if the value of x is in the latter range, then the exponent will increase its value, and the higher the exponent, the higher the value of x. Otherwise, the value of x will always be 0 or 1 whichever one of these values it is. Consequently, x can be any value but 0 and 1 for its power to alter its value.</p>Thu, 15 Aug 2013 19:33:47 -05002013-08-15T19:33:47-05:00234412http://www.wyzant.com/resources/blogs/234412/things_a_good_tutor_should_say_and_discuss_with_youNatalie B.http://www.wyzant.com/resources/users/view/77223300Things a good tutor should say and discuss with you:<p>As the school year ramps up again, I wanted to put out a modified version of a Memo of Understanding <a href="http://en.wikipedia.org/wiki/Memo_of_understanding">http://en.wikipedia.org/wiki/Memo_of_understanding</a> for parents and students. It seems each year in the rush to get through the first weeks of school parents and students forget the basic first good steps and then the spiral downwards occurs and then the need for obtaining a tutor and then the ‘wish for promises’ from a tutor.
Pay attention to your child’s folder or agenda book. A student is generally not able to self regulate until well into high school. Some people never quite figure it out. Be the best person you can be by helping your child check for due dates, completeness, work turned in on time. Not only will this help your child learn to create and regulate a schedule, it prevents the following types of conversations I always disliked as a teacher ("Can you just give my child one big assignment to make up for the D/F so they can pass"; "I am going to talk to the principal about this grade - you are unfair"; "I had no idea you gave homework or long term assignments - we never had that when I was a student"; "you mean my child has to catch up on all that for quarter or semester grades - they won’t be able to do anything fun this weekend or go to their practice"....). When you are following the folder, agenda book, grades - you WILL know when something is off and can immediately do something about the situation.
As a tutor, I am able to do the best work before the crisis hits (averting the crisis) as opposed to cleaning up the mess after the crisis hits. The minute the student and parents see grade slippage is the time to look for a tutor. Many schools have online grades where you can look at assignments and tests once a week and you can have your child do the same. This is called responsibility. I do not mean a one off quiz grade, I mean the two or three off which occur over a couple weeks and can effect harm on overall grades. Waiting until a student is already in a downward spiral means dealing with what the student did not understand AND keeping them up on what is current. A way to think about this is your desk after returning from a two to three week vacation. I have never met some one who did not benefit from understanding grading, most especially when grades are weighted <a href="http://www.blacksdomain.com/files/Notes/Calculating_WA.php">http://www.blacksdomain.com/files/Notes/Calculating_WA.php</a> and it is rare when I can find some one who understands how GPA is calculated normally, with AP courses, etc. <a href="http://www.back2college.com/gpa.htm">http://www.back2college.com/gpa.htm</a>
<br>Tutoring is important - and so is being a scholar athlete. It is not one or the other, it is both. Please don’t frame one as more important, rather, speak to your child’s coach and see what can be done to allow for a bit less practice time to have a tutor come in. In the long run, this is always preferable to having to pull a kid out of sports if things head south. It has been my experience, great coaches always want scholar athletes and will work with parents/students to make an accommodating schedule, even if it seems to be an awkward conversation.
<br>I can not promise to raise your child’s grade - if I could do that, I would be in a whole other line of business with snakes and oil and funny signs. What I can promise is to put in 150% with your child and you to help get them back on track. I can let you know specifically where your child is having a problem or two and zero in on that content area. I can let you know if I think this is more than the routine spring time blahs or too much vacation and too little review. I can offer to communicate with your child’s teacher(s) to see if there is something to prepare for and get ahead of.
<br>One thing I can do is help you understand how grading is done, what it means and does not mean, what a rubric is and how it is used, when to question how something is graded or to accept reality. I can work with you, the parent, on better communication with the teacher on behalf of your child. Part of being a tutor is also being a resource /conduit for obtaining information to help your child learn more in a particular subject area, find ways to make learning easier, better, cheaper.
<br>A great tutor wishes to tutor themselves right out of a job... then they can help other students. Part of doing this is helping get the right pieces in place from the beginning so a student can do well and then knows the ways to do well going forward. While failure presents opportunities for great learning experiences and stories, it rarely provides for one to feel good about what could have been.</p>Sun, 04 Aug 2013 13:33:54 -05002013-08-04T13:33:54-05:00233897http://www.wyzant.com/resources/blogs/233897/algebra_word_problems_relative_comparisonsJorge L.http://www.wyzant.com/resources/users/view/83164680Algebra Word Problems: Relative Comparisons<p><b>Solving Word Problems with Proportions and Relative Comparisons</b>
<br>These word problems are set-up where the dependent variable is not provided as is, but rather as a part of an operation. You will have to set-up each side of the equality with its own operations.
<br><u>Example 1:</u>
<br>“Shelley finished x number of her math homework problems before dinner. Had she finished 3 more, she would have finished half her math homework. Write an equation which represents the relationship between y, total problems and x, number of problems Shelley completed.”
<br>This isn’t set-up in the same way as problems presented in previous entries because there isn’t a defined rate of change right away. So, it will be set-up this way with one variable on each side of the equality. You're already given the variables to use in the problem.
<br>Proportion of completed problems = <i>proportion</i> of total problems.
<br>“3 more than completed problems” = “half her math homework” (half total problems)
<br>(x + 3) = (1/2)*(y)
<br>Or
<br>x + 3 = y/2
<br>If you’re asked in the problem to isolate – usually if the problem wants you to graph or establish the rate of change - you will isolate y by multiplying each side by 2. You get:
<br>2x + 6 = y.</p>
<p><u>Example 2:</u>
<br>“On Thursday Brenda’s schedule allows for 60 more minutes of training than any other day of the week. One round of strength routine takes 8 minutes. One round of endurance training takes 12 minutes. Write an equation which represents the time available to Brenda to spend on each type of routine on Thursday compared to any given day. Assume all other days she can train the same amount of time.”
<br>So, let’s start by doing a comparison of how much time you can spend on any day to Thursday. Thursday is “60 more than any other day” or 60 + d. d represents the amount of time she can train any other day in minutes.
<br>So, you get
<br>“Amount of Training on Thursday” = Time spent on Strength + Time spent on Endurance.
<br>60 + d = 8 minutes per session + 12 minutes per session.
<br>60 + d = 8s + 12e</p>
<p><u>Example 3:</u>
<br>“Jennifer and Alex work at a call center for a major company, but in different departments. Monday was a particularly busy day for Jennifer’s department. If Jennifer had received 10 fewer calls, she would have answered 40% more calls than Alex.”
<br>Set-up each side as:
<br>“10 fewer calls” than actual number for Jennifer = “40% more” than actual calls for Alex.
<br>Translate each side into symbols:
<br>j – 10 = a + 0.40a
<br>Note: The problem indicates 40% more calls, not 40 more calls. That's why we use 0.40a and not simply add 40 to the right side.</p>Sun, 04 Aug 2013 11:22:11 -05002013-08-04T11:22:11-05:00234231http://www.wyzant.com/resources/blogs/234231/algebra_word_problems_problems_with_provided_valuesJorge L.http://www.wyzant.com/resources/users/view/83164680Algebra Word Problems: Problems with provided values<p><b>Word Problems with Multiple Variables and Given Values</b>
<br>This type of problem will be presented such that you'll have to set-up the equation or relation between the variables. Additionally, you will be given the value of one or more variables. On all of these problems you are not asked to solve the problem, only set-up the equation.</p>
<p><u>Example 1:</u>
<br>“A weather balloon is launched from a height of 100 meters above sea level. The balloon rises at a constant rate of 27 meters per minute. Write an equation that can be used to determine the time in minutes it will take the balloon to reach a height of 2889 meters above sea level.”
<br>Start with the relationship of variables:
<br>Dependent Variable = Fixed Value + Rate of Change * Independent Variable.
<br>“Height of balloon” = “Initial height” + “27meters per minute”
h = 100m + 27m
<br>For the final step, substitute the given height of 2889:
<br>2889 = 100 + 27m.</p>
<p><u>Example 2:</u>
<br>“The perimeter of a rectangle is 40. If the measurement of the length is 3 more than the width. Write an expression that can be used to find the width of the rectangle.”
<br>In this case, you will need to remember the formula for perimeter of a rectangle:
<br>p = 2w + 2l (the same as the sum of its sides: w + w + l + l ).
<br>You will have to substitute the phrase “3 more than the width” in place of length. This expression translates as w + 3.
<br>p = 2w + 2( w + 3).
<br>NOTE: You have to place parenthesis around w + 3 because you’re multiplying 2 by the whole measurement of length. 2w + 3 means “3 more than twice the width”.
<br>Now, you can substitute the known perimeter and you get the final form:
<br>40 = 2w + 2(w + 3)</p>
<p><u>Example 3:</u>
<br>“Brandon has a budget of $156 to spend on clothes. The shirts he wants to buy are on sale for $24 each, and the pair of pants he wants to buy is $36. Write an expression to help determine the number of shirts Brandon can purchase on his budget.”
<br>Dependent Variable = Rate of Change * Variable + Rate of Change * Variable.
<br>“Total cost”= “24 per shirt” + “36 per pair of pants”
<br>c = 24s + 36p
<br>Because the problem indicates there is only one pair of pants, you substitute and finalize as such:
<br>c = 24s + 36
<br>In budget type problems, when setting up the equations the budget is treated the same way as the total cost because the total cost cannot be more than the budget. Because of this relationship, these problems will be stated as an inequality.
<br>Total cost < budget or Budget > Total Cost
<br>So, substitute the inequality in place of cost and you get:
<br>156 > 24s + 36</p>
<p><u>Example 4:</u>
<br>“At Turtle South East Book Shop, DVDs on sale cost $8 each, while Blu-Rays cost $14 each. Write an inequality which best describes how many DVDs and Blu-Rays can be purchased for $78 or less.”
<br>Start with relationship of total cost to DVDs and Blu-Rays purchased.
<br>Total Cost = Cost of DVDs + Cost of Blu-Rays
<br>“Total Cost” = “$8 per DVD” + “$14 per Blu-Ray”
<br>c = 8d + 14b.
<br>And finalize by setting up the inequality Budget > Cost.
<br>78 > 8d + 14b</p>Sun, 04 Aug 2013 11:04:44 -05002013-08-04T11:04:44-05:00234232http://www.wyzant.com/resources/blogs/234232/algebra_word_problems_rate_of_changeJorge L.http://www.wyzant.com/resources/users/view/83164680Algebra Word Problems: Rate of Change<p><b>Writing Expressions Involving Rate of Change</b>
<br>These real-world problems can be best translated when broken down into their components (variables and operations). When you see the words “is” or “are”, this is the points where you set-up the equality. Whenever you see the word “per”, “each” the implication is a multiplication. This indicates the rate of change between the variables.
<br>The general format for these problems is:
<br>Dependent Variable = Fixed Value + Rate of Change * Independent Variable.
<br>The fixed value is generally a fixed value which does not change. Most commonly, it will be the initial value in a situation.</p>
<p><u>Example 1:</u>
<br>“Mark is purchasing a new computer. The cost of the computer is $2400 after tax. He will make monthly payments of $150. Write an equation which describes the balance on the account after any given number of months”
<br>Variables present: balance and number of months.
<br>The rate of change in this case is the $150 per month. The word “per” is the indicator of the rate of change.
<br>The relationship between the variables is:
<br>Dependent Variable = Fixed Value + Rate of Change * Independent Variable.
<br>“Balance” = “Cost of Computer” – “150” per “month”
<br>b = 2400 – 150*m.
<br>Because the balance on the account is going to be lower as the time passes, you will subtract from the initial value.</p>
<p><u>Example 2:</u>
<br>“Mr. Fellow bought a refrigerator that cost $1200 including tax. The cost of electricity to run the refrigerator is estimated at $63 per year. Write an equation which represents the total cost of operation”.
<br>Relationship variables are: total cost of operation and number of years.
<br>Dependent Variable = Fixed Value + Rate of Change * Independent Variable.
<br>Total cost of operation = initial cost + “cost” per “year”
<br>t = 1200 + 63*y</p>
<p><u>Example 3:</u>
<br>“Vicki works as a sales associate in a department store. She earns $6 per hour, plus a commission of 3% on her sales. Write an equation which describes her total earnings”.
<br>Aside from numbers, the relationship can be described as “Total earnings depend on sales and number of hours worked”. There are three variables. In this particular case, each variable carries its own rate of change and there is no fixed value as her earnings start at $0.
<br>Setting-up the relationship:
<br>Dependent Variable = Rate*Independent Variable + Rate*Independent Variable
<br>“Total Earnings” = “$6” per “Hour” + “3%” per “sales (in dollars).”
<br>e = 6*h + 0.03*s.</p>
<p><u>Example 4:</u>
<br>“Passengers on a commercial flight are able to make in-flight calls using the built-in telephone system. The calls cost $3 to connect plus $1.85 each minutes. Write an equation the represents the total cost t, to make a call which lasts n number of minutes.”
<br>Dependent Variable = Fixed Value + Rate of Change * Independent Variable.
<br>“Total Cost” = “Cost to connect” + “cost” “per” “minute”
<br>t = 3 + 1.85*m</p>Sun, 04 Aug 2013 09:43:43 -05002013-08-04T09:43:43-05:00