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Study Guide, Intermediate Algebra

by: Liyah Notetaker

Study Guide, Intermediate Algebra MA 0103

Marketplace > Mississippi State University > Math > MA 0103 > Study Guide Intermediate Algebra
Liyah Notetaker

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This Study Guide covers almost everything we went over these past few weeks! Enjoy!
Intermediate Algebra
Jonathan Eaton
Study Guide
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This 8 page Study Guide was uploaded by Liyah Notetaker on Wednesday February 3, 2016. The Study Guide belongs to MA 0103 at Mississippi State University taught by Jonathan Eaton in Fall 2016. Since its upload, it has received 79 views. For similar materials see Intermediate Algebra in Math at Mississippi State University.


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Date Created: 02/03/16
Aliyah Alexander  Intermediate Algebra Study Guide Test 1 This STUDY GUIDE covers what all we have learned up to now! This will all be covered  on the test! Enjoy! Adding/ Subtracting Real Numbers Same Sign: Keep the sign, Combine the numbers Different signs: Take the negative off the numbers, use sign of the larger value Multiplying/ Dividing Real Numbers Same Sign: The answer is positive. Different Sign: The answer is negative. 1. -3 + 10 = 7 2. -11 + (-15) = -26 3. 3 – (-9) = 12 4. 3 ∙ -9 = -27 5. ( -5)(-4) = 20 6. -1/4 ∙ -2/9 = 2/36 simplified to 1/18 7. 4 ÷ -2 = -2 8. -16 ÷ -4 = 4 9. 0 ÷ 7 = 0 10. -7 ÷ 0 = no solution Absolute Value ­ Measures the distance that a number is from 0 ­ Measurements are always positive Solving a Linear Equation in One Variable Step 1: Simplify each side separately.   Clear any parentheses, fractions, or decimals  Combine like terms Step 2: Isolate the variable terms on one side.  Use the addition property to put all terms with a variable on one and side and the  other numbers on the other side. Step 3: Isolate the variable.  Use the multiplication property of equality to obtain an equation that has just the  variable with coefficient 1 on one side. Step 4: Check.   Substitute the variable with its answer in the original equation. The statement  should be true. If not, rework the problem. Solve the equation 8(x−3) − (6−2x) = 2(x + 2) −5 (5−x) We begin by multiplying out the brackets, taking care, in particular, with any minus signs. 8x−24−6 + 2x = 2x + 4−25 + 5x Each side can be tidied up by collecting the x terms and the numbers together.  10x−30 = 7x−21 Now take 7x from each side, and then add 30 to each side:  3x−30 = −21 3x = 9 x = 3 And again you should take the solution (x = 3), substitute it back into the original equation  to check that we have got the correct answer. On the left: 8(x−3) − (6−2x) = 8(3−3) − (6−2 (3)) = 0−0 = 0. On the right:  2(x + 2) −5 (5−x) = 2(3 + 2) −5 (5−3) = 10−10 = 0.  So both sides equal zero. The equation balances and so x = 3 is the solution.   Solve 0.2x + 0.9 = 0.3 – 0.1x This equation solves just like all the other linear equations. It just looks worse because of  the decimals. But that's easy to fix: however many decimal places I have, I can multiply by  "1" followed by that number of zeroes. In this case, I'll multiply through by 10: 10(0.2x) + 10(0.9) = 10(0.3) – 10(0.1x)  2x + 9 = 3 – 1x  Then I solve as usual: 2x + 1x + 9 – 9 = 3 – 9 – 1x + 1x  3x = –6  x = –2  If one of the decimals had had two decimal places, then I'd have multiplied through by 100; for three, I'd have multiplied through by 1000.  Solve   To simplify my computations for equations with fractions, I can first multiply through by the  common denominator. For this equation, the common denominator is 12:                   3x + 12 = 2x + 6    3x – 2x + 12 – 12 = 2x – 2x + 6 – 12                              x = –6 Solving for a Specified Variable Step 1: Clear any parentheses and fractions.  If the equation contains parentheses, clear them by using the distributive property.  Clear them by multiplying both sides by the LCD. (Least common denominator) Step 2: Isolate all terms with the specified variable.  Transform so that all terms containing the specified variable are on one side of the  equation and all terms without that variable are on the other side. Step 3: Isolate the specified variable.  Divide each side by the factor that is the coefficient of the specified variable. If you need both variables: Just plug W in the solution or vice versa.  You have $50,000 to invest, and two funds that you'd like to invest in. The You­ Risk­It Fund (Fund Y) yields 14% interest. The Extra­Dull Fund (Fund X) yields  6% interest. Because of college financial­aid implications, you don't think you can  afford to earn more than $4,500 in interest income this year. How much should you  put in each fund?" The problem here comes from the fact that I'm splitting that $50,000 in principal into two  smaller amounts. Here's how to handle this:   I P r t Fund X ? ? 0.06 1 Fund Y ? ? 0.14 1 total 4,500 50,000 ­­­ ­­­ How do I fill in for those question marks? I'll start with the principal P. Let's say that I put "x" dollars into Fund X, and "y" dollars into Fund Y. Then x + y = 50,000. This doesn't  help much, since I only know how to solve equations in one variable. But then I notice that  I can solve x + y = 50,000 to get y = $50,000 – x. THIS TECHNIQUE IS IMPORTANT! The amount in Fund Y is (the total) less (what  we've already accounted for in Fund X), or 50,000 – x. You will need this technique, this  "how much is left" construction, in the future, so make sure you understand it now.   I P r t Fund X ? x 0.06 1 Fund Y ? 50,000 – x 0.14 1 total 4,500 50,000 ­­­ ­­­ Now I will show you why I set up the table like this. By organizing the columns according  to the interest formula, I can now multiply across (right to left) and fill in the "interest"  column.   I P r t Fund X 0.06x x 0.06 1 Fund Y 0.14(50,000 – x) 50,000 – x 0.14 1 total 4,500 50,000 ­­­ ­­­ Since the interest from Fund X and the interest from Fund Y will add up to $4,500, I can  add down the "interest" column, and set this sum equal to the given total interest: 0.06x + 0.14(50,000 – x) = 4,500  0.06x + 7,000 – 0.14x = 4,500  7,000 – 0.08x = 4,500  –0.08x = –2,500  x = 31,250  Then y = 50,000 – 31,250 = 18,750. I should put $31,250 into Fund X, and $18,750 into Fund Y. You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and  a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution,  you decide to mix 10% solution with 30% solution, to make your own 15% solution. You need  10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use? Let x stand for the number of liters of 10% solution, and let y stand for the number of liters of  30% solution. (The labeling of variables is, in this case, very important, because "x" and "y" are  not at all suggestive of what they stand for. If we don't label, we won't be able to interpret our  answer in the end.) For mixture problems, it is often very helpful to do a grid:   liters sol'n percent acid total liters acid 10% sol'n x 0.10 0.10x 30% sol'n y 0.30 0.30y mixture x + y = 10 0.15 (0.15)(10) = 1.5 Since x + y = 10, then x = 10 – y. Using this, we can substitute for x in our grid, and eliminate  one of the variables:       liters sol'n percent acid liters acid 10% sol'n 10 – y 0.10 0.10(10 – y) 30% sol'n y 0.30 0.30y mixture x + y = 10 0.15 (0.15)(10) = 1.5 When the problem is set up like this, you can usually use the last column to write your equation:  The liters of acid from the 10% solution, plus the liters of acid in the 30% solution, add up to the  liters of acid in the 15% solution. Then: 0.10(10 – y) + 0.30y = 1.5  1 – 0.10y + 0.30y = 1.5  1 + 0.20y = 1.5  0.20y = 0.5  0.5 y =  / 0.20 2.5  Then we need 2.5 liters of the 30% solution, and x = 10 – y = 10 – 2.5 = 7.5 liters of the 10%  solution. (If you think about it, this makes sense. Fifteen percent is closer to 10% than to 30%,  so we ought to need more 10% solution in our mix.) Linear Equations in One Variable An inequality consists of algebraic expressions related by one of the following symbols: < “is less than” ≤ “is less than or equal to  “ is greater than” >   ≥ “is greater than or equal to We solve an inequality by finding all real number solutions for it. To write intervals, we use INTERVAL NOTATION, which includes the infinity symbols, ∞  or ­  ∞ .  Remember the following important concepts regarding interval notation.  A parentheses indicates that an endpoint is NOT included.  A square bracket indicates that an endpoint is included.   A parenthesis is always used next to an infinity symbol.  The set of real numbers is written in interval notation.  This notation is my favorite for intervals.  It's just a lot simpler!  Let's look at the intervals we did with the set­builder notation:      Let's start with the first one:   This is what it means;   So, we write it like this:  Use  [ or ] for closed dots   Use  ( or ) for open dots  


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