PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY

MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive Properties 8 4. Properties of Frctions 9 5. Properties of Exponents 11 6. Properties of Rdicls 13 7. The Slope nd Eqution of Line 14 1. ACT Compss Prctice Tests If you hve Internet ccess, you cn ccess the online resources below through the pdf file by simply clicking on the links below.you cn use these resources to prctice with smple ACT Compss tests online or wtch video tutorils on Google Video. If you hve only printed copy of the pdf file, you cn still find these Internet resources by using the provided web links. (1) CUNY Compss Prctice Tests from Hostos Community College http://www.hostos.cuny.edu/o/compss/prelgebr.htm (2) Kentucky Erly Mth Testing Progrm Prctice Tests https://www.mthclss.org/wqs/k.sp?stte=1 (3) Google Video Tutoril on Order of Opertions http://video.google.com/googleplyer.swf?docid=-3581910795500993427 (4) Google Video Pre-Algebr Tutoril http://video.google.com/videoply?docid=-2898932824775207461 (5) Google Video Tutoril on Solving Equtions http://video.google.com/videoply?docid=4755123812601857335 Dte: November 2009. 1

2 PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY 2. Common Mistkes Common Mistke 1. A surprisingly common mistke is to incorrectly copy the problem in your exm booklet. Mke sure you re working on the correct problem! Common Mistke 2. Alwys put prenthesis round negtive number, especilly when you hve to multiply it by nother number, s in this cse: 5 + 2 ( 5) = 5 10 = 5 Never drop the prenthesis round the negtive number becuse you will forget tht you hve multipliction nd you will get this insted: 5 + 2 5 = 2 which is wrong nd hs nothing to do with the originl problem. Common Mistke 3. Be very creful with the order of opertions. The correct order of opertions is given below: (1) First do the opertions inside the Prenthesis. (2) Then tke cre of Exponents, (3) Multipliction, Division, (4) Addition, Subtrction. Consider s n exmple the lgebric expression 2 + 3 ( 9). There re severl mistkes you cn mke. Firstly, if you don t put the prenthesis round the negtive number, you will get: 2 + 3 9 = 4, which is wrong. Secondly, you cn get the order of opertion wrong: 2 + 3 ( 9) = 5 ( 9) = 45 Wrong! fter first dding 2 nd 3 nd then multiplying the result by ( 9). This is wrong, becuse multipliction hs priority, so one should first multiply 3 nd ( 9) to get 27 nd only then dd 2 to the result. So, the correct thing to do is the following: 2 + 3 ( 9) = 2 + ( 27) = 2 27 = 25 Correct!

MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES 3 It would be different if we hve the following expression (2+3) ( 9). The difference is tht 2 nd 3 re now inside prenthesis, so we would hve to do the opertion inside the prenthesis first nd then multiply: (2 + 3) ( 9) = 5 ( 9) = 45 Correct! Common Mistke 4. A few common mistkes re relted to the properties of exponents. For exmple, note tht 2 3 2 (2 3) 2 becuse tking the exponent hs priority over multipliction. So, if one wnts to clculte 2 3 2 then one should tke the exponent first 3 2 = 9 nd then multiply the result by 2 to get 18, tht is 2 3 2 = 2 9 = 18. While for (2 3) 2, we first do the multipliction inside the prenthesis to get 6, which we then squre: (2 3) 2 = 6 2 = 6 6 = 36 Another common mistke relted to exponents is to write It s lso wrong to write insted of using the correct property 3 2 3 5 = 3 10 Wrong! 3 2 3 5 = 9 7 Wrong! 3 2 3 5 = 3 2+5 = 3 7 Correct! Keep in mind tht the generl property reds m n = m+n Correct! Finlly, if we hve to tke the power of power, it is wrong to write (x 2 ) 5 = x 7 Wrong! The correct ppliction of the power property reds (x 2 ) 5 = x 10 Correct! Remember the generl property hs the form: (x m ) n = x m n Correct!

4 PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Exmple 1. Consider now the following exmple. Evlute the expression: 12 2 3 2 3 (2 5) 3 One common mistke is to write the second term in this difference s 3 (2 5) 3 = (6 15) 3 Wrong! This is wrong becuse we re distributing 3 inside the prenthesis pretending tht we hve (2 5) nd ignoring the fct tht we ctully hve this difference rised to the power of 3. To do it correctly, we need to do the opertion inside the prenthesis first nd then rise the result to 3rd power: 3 (2 5) 3 = 3 ( 3) 3 = 3 ( 27) = 3 27 = 81 Correct! nd only then multiply the result by 3. In our cse, we hve ( 3) 3 = ( 3) ( 3) ( 3) = ( 1) 3 3 3 = 27 Remember tht negtive number rised to n odd power must be negtive nd negtive number rised to n even power must be positive (negtive) odd = negtive (negtive) even = positive For exmple, ( 1) 2 = 1, ( 1) 3 = 1, ( 1) 4 = 1, ( 1) 5 = 1. Now, let s go bck to the originl exmple. In the first term, we must tke cre of the exponent in the numertor first, so write 2 3 = 2 2 2 = 8 nd t the sme time simplify the second term s we did erlier: 12 2 3 2 3 (2 5) 3 = 12 8 2 3 ( 3) 3 = 4 2 3 ( 27) Note here tht 4 2 = 4 2 = 2 (don t forget the minus sign): 2 3 ( 27) = 2 + ( 3) ( 27) = 2 + 3 27 = 2 + 81 = 79

MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES 5 We hve here the product of two negtive numbers 3 nd 27, which gives us positive number ( 3) ( 27) = 3 27 = 81. Finlly, if you re confused bout the sum 2 + 81, note tht this is relly the sme s the difference 81 2, we simply tke 2 wy from 81 to get 79. Common Mistke 5. Remember tht the frction A mens tht we divide A by B, B i.e. we hve A B. For tht reson, we cn express division by B in terms of multipliction by the reciprocl of B, which is 1, nmely B A B = A 1 B = A B Consider the division when A = 15b3 nd B = 5b 2, then we hve 2 ( ) 15b 3 (5b 2 ) 2 Sometimes, students ttempt to use the division rule bove but since they cnnot quite remember it, they would write something like this ( ) ( ) 15b 3 15b (5b 2 3 ) = 5b2 Wrong! 2 2 1 This is wrong, of course, becuse the division is replced by multipliction but the reciprocl of 5b 2 is not tken. Insted, 5b 2 is divided by 1, which does not chnge nything since ny number divided by 1 is the number itself, tht is 5b2 = 5b 2. This 1 wy, the division is simply replced by multipliction while nothing else chnges nd this is wrong. The correct thing to do is to tke the reciprocl of 5b 2 when replcing division by multipliction, nmely: ( ) 15b 3 (5b 2 ) = 15b3 2 2 1 5b = 15b3 2 10b = 3b 2 2 Correct! Let s recll the rule for multiplying two frctions tht is used bove. We multiply the nomintors nd the denomintors of both frctions b c d = c b d

6 PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY In the exmple bove, we reduce the frction 15 = 5 3 = 3 by cnceling out the 10 5 2 2 common fctor of 5. Remember tht if we hve the sme number bove nd below the br of frction then we cn cncel this number: b c = b c becuse the br of the frction represents division. The property of exponents, used bove to get the finl result, is b m b n = bm n which we pply to our exmple to conclude tht b 3 b 2 = b3 2 = b 1 = b Common Mistke 6. One of the most common mistkes when deling with frctions is the following rule tht students invent to dd unlike frctions: b + c d = + c b + d Wrong! It is very esy to see tht this rule is not correct by checking simple exmple. Tke = 2, b = 1, c = 3 nd d = 1 nd if this rule is correct we should get true sttement: 2 1 + 3 1? = 2 + 3 1 + 1 2 + 3? = 5 2 which is clerly flse sttement, since 5 5, so the rule cnnot be correct. 2 The correct rule we get by cross-multiplying numertors by denomintors nd the sum of the two products gives us the new numertor, while the new denomintor is just the product of the two denomintors: b + c d = d + c b b d

MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES 7 Of course, given specific numbers, one cn lso look for the LCD (lest common denomintor) but in generl mny students find the LCD concept more difficult. For exmple, let s dd the two frctions using the correct rule: 5 6 + 11 12 = 5 12 + 11 6 6 12 = 60 + 66 6 12 = 126 6 12 = 6 21 6 12 = 21 12 = 3 7 3 4 = 7 4 Alterntively, it is esy, in this cse, to find the LCD, which is 12. The next step is to write the first frction s n equivlent frction hving denomintor of 12 nd then we cn esily dd the like frctions. Tht s why we multiply by 2 the numertor nd denomintor of the first frction: 5 6 + 11 12 = 5 2 6 2 + 11 12 = 10 12 + 11 12 = 10 + 11 12 = 21 12 = 7 4 Remember the rule for dding like frctions (with the sme denomintors): b + c b = + c b which we use bove to dd the like frctions: 10 12 + 11 12 = 10 + 11 12 Common Mistke 7. Some common mistkes relted to rdicls re writing: x 16 = x 4 or x 9 = x 3 Wrong! We cn check esily if our guess is correct by simply using the definition of squre root, which in the first cse would men tht if we tke the squre of our guess x 4, we should get wht is inside the rdicl: (x 4 ) 2 should be equl to x 16. However, (x 4 ) 2 = x 4 2 = x 8 x 16, so our guess x 4 cnnot be correct. I cn only guess tht the logic tht leds to the wrong clims bove goes long these lines x 16 = x 16 = x 4 or x 9 = x 9 = x 3 Wrong! The correct rule to pply in the cse of n even power under the rdicl sign is: x 16 = x 16 2 = x 8 Correct!

8 PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY It is esy to see tht x 8 is the correct nswer becuse if we squre it, we get (x 8 ) 2 = x 8 2 = x 16 nd x 16 is wht s inside the rdicl sign. In the second cse, when we don t hve n even power, 16 is n even power but 9 is not, we need to split the odd power in order to get n even power: x9 = x 8 x = x 8 x = x 8 2 x = x4 x Correct! Here for, b > 0, we used the product rule for rdicls b = b nd we cn generlize the rule we used bove for ny positive even power under the rdicl sign, like 16 or 8 x even = x even 2 3. Distributive Properties Let, b, c, d, e be ny numbers, positive (negtive) or terms contining vribles. If we hve to multiply two fctors, we must multiply every term in the first fctor by every term in the second fctor. It is useful to drw rrows indicting ll possible products. The following distributive properties re often used: (b + c) = b + c ( + b) (c + d) = c + d + b c + b d ( + b) (c + d + e) = c + d + e + b c + b d + b e Exmple 2. Multiply nd combine like terms in the following exmples: (1) 2x (4x 2 + 3) = 2x 4x 2 + 2x 3 = 8x 3 + 6x (2) 2 (4 3) = 2 4 2 ( 3) = 8 + 6 = 2 (3) (x y) (x + y) = x x + x y y x y y = x 2 y 2 (4) (x 4) (x 2 + 2x 2) = x x 2 + x 2x + x ( 2) 4 x 2 4 2x 4 ( 2) = = x 3 + }{{} 2x 2 2x 4x2 }{{} 8x + 8 = x3 2x 2 10x + 8

MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES 9 4. Properties of Frctions b c b = c Exmple 3. This pplies to numericl s well s lgebric frctions. Remember tht frction represents division, so if we hve the sme term (b in the formul) bove nd below the division line, being the br of the frction, then we cn cncel this term, s we re relly dividing the term by itself. 5 7 7 2 = 5 2 (x 1) (x+2) = (x 1) (x+2) (x+5) (x+5) b = b = b Exmple 4. We cn move the minus sign from the denomintor to the numertor or we cn plce it in front of the frction. If both the denomintor nd the numertor re negtive, we cn cncel out the minus. 5 7 = 5 7 = 5 7 5 7 = 5 7 b c d = c b d Exmple 5. The product of two frctions is frction whose numertor is the product of the two numertors ( c bove) nd whose denomintor is the product of the two denomintors (b d bove). 5 2 = 10 7 3 21 (x 4) (x 2) = (x 4) (x 2) (x+1) 3 3(x+1) b c d = b d c Exmple 6. We divide one frction (the divident) by nother (the divisor) by multiplying the divident by the reciprocl of the divisor.

10 PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Remember, the reciprocl of c d is d c 5 2 = 5 3 = 15 7 3 7 2 14 (x 3) (x+5) (x 2) (x 1) = (x 3) (x+5) (x 1) (x 2) = (x 3) (x 1) (x+5) (x 2) d + b d = + b d Exmple 7. We dd two like frctions (hving the sme denomintors) by dding the numertors nd keeping the common denomintor. 5 7 + 4 7 = 9 7 5 x + 2 x = 7 x d b d = b d Exmple 8. We subtrct two like frctions (hving the sme denomintors) by subtrcting the corresponding numertors nd keeping the common denomintor. 5 7 4 7 = 1 7 5 x 2 x = 3 x b + c d = d + c b b d Exmple 9. We dd two unlike frctions hving different denomintors b d, by trnsforming them first into equivlent like frctions, hving the sme denomintors: 5 + 2 = 5 3 + 2 7 = 15+14 = 29 7 3 7 3 3 7 21 21 2 + 5 = 2 x + 5 = 2x+5 x x x x 2 + 5 = 2 + 5 x = 2 + 5 x = 2+5x x 2 x x 2 x x x 2 x 2 x 2 b c d = d c b b d

MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES 11 Exmple 10. We subtrct two unlike frctions hving different denomintors b d, by trnsforming them into equivlent like frctions, hving the sme denomintors: 5 7 2 3 = 5 3 7 3 2 7 3 7 = 15 14 21 = 1 21 2 5 x = 2 x x 5 x = 2x 5 x 2 5 = 2 5 x = 2 5 x x 2 x x 2 x x x 2 x 2 = 2 5x x 2 5. Properties of Exponents x n = n { }} { x x x Exmple 11. Rising number (or n lgebric expression) to some positive integer power is the sme s multiplying this number (or n lgebric expression) by itself s mny times s the positive integer power. ( 2) 3 = ( 2)( 2)( 2) = 8 (x 2) 3 = (x 2)(x 2)(x 2) ( 2) 4 = ( 2)( 2)( 2)( 2) = 16 x m x n = x m+n Exmple 12. Observe tht ll powers in the formul bove hve the sme bse x, which could be number or more generl lgebric expression. To multiply two powers hving the sme bse (x bove), we dd the exponents nd keep the bse. 2 3 2 5 = 2 3+5 = 2 8 the bse here is 2 (x 2) 3 (x 2) 4 = (x 2) 3+4 = (x 2) 7 the bse here is (x 2) 4 5 = 4+5 = 9 the bse here is 2 4 3 5 (2 3) 4+5 bses re different, 2 nd 3 (x m ) n = x m n Exmple 13. The power of power rule leds us to multipliction of exponents.

12 PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY (2 3 ) 5 = 2 3 5 = 2 15 ((x 2) 3 ) 4 = (x 2) 3 4 = (x 2) 12 ( 4 ) 5 = 4 5 = 20 (2 4 ) 5 2 4+5 x m x n = xm n Exmple 14. The quotient of two powers hving the sme bse (x bove) nd ny exponents (m nd n bove) is the bse to the power tht is the difference of the two exponents (m n bove), where the exponent tht is below the br of the frction (n bove) is subtrcted from the exponent bove the br (m bove). 25 2 3 = 2 5 3 = 2 2 = 2 2 = 4 bse is 2 x5 x 3 = x 5 3 = x 2 bse is x 104 10 2 = 10 4 ( 2) = 10 4+2 = 10 6 bse is 10; used in Scientific Nottion 10 4 10 3 = 10 4 ( 3) = 10 4+3 = 10 1 = 1 10 used in Scientific Nottion x n = 1 x n, x0 = 1 Exmple 15. Used to convert negtive exponent into positive one. 2 3 = 1 2 3 10 4 = 1 10 4 (x y) n = x n y n Exmple 16. The power of product is the product of the powers. (2 5) 3 = 2 3 5 3 (5x) 3 = 5 3 x 3 (2xy 2 ) 4 = 2 4 x 4 (y 2 ) 4 = 16x 4 y 8

MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES 13 ( ) n x = xn y y n Exmple 17. The power of quotient is the quotient of the powers. ( ) 2 2 5 = 2 2 5 2 ( ) x 3 5 = x 3 5 3 6. Properties of Rdicls = ( ) 2 =, > 0 Exmple 18. Tking the squre of squre root leves us with the number inside the rdicl. We undo the squre root by tking the squre. 5 5 = 5 ( x) 2 = x, x > 0 2 =, > 0 Exmple 19. We cn undo the squre root by tking the squre inside the rdicl. 5 2 = 5 x 2 = x, x > 0 b = b, > 0, b > 0 Exmple 20. The squre root of the product is the product of the squre roots. 25 16 = 25 16 = 5 4 = 20 x 2 y = x 2 y = x y 49 x y 2 z 2 = 49 x y 2 z 2 = 7yz x

14 PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY b = b, > 0, b > 0 Exmple 21. The squre root of the quotient is the quotient of the squre roots. 5 2 = 5 2 36 36 = 5 6 x 2 = x 2 64 64 = x 8 2n = 2n 2 = n or even = even 2 Exmple 22. If we hve n even exponent inside the rdicl, 2n bove, we cn undo the rdicl by tking hlf of the even exponent, 2n 2 = n. 5 14 = 5 14 2 = 5 7 25x 16 y 10 = 25 x 16 2 y 10 2 = 5x 8 y 5 x 8 y 17 = x 8 2 y 16 y = x 4 y 16 2 y = x 4 y 8 y 7. The Slope nd Eqution of Line The slope of line is number determined by the coordintes of ny two points on the line. If P (x 1, y 1 ) nd Q(x 2, y 2 ) re ny two points on the line, given by their coordintes, then the slope is the number m = y 2 y 1 x 2 x 1 = y x Note tht y, the difference in the y-coordintes, is on the top, while x, the difference in the x-coordintes, is on the bottom. It is common mistke to use x y y insted of x to compute the slope. Notice tht we chose bove to strt with the second point Q nd tht is why both differences begin with the coordintes of the second point (y 2 y 1 ) nd (x 2 x 1 ). One cn strt insted with the first point P, in which cse both differences should strt with the coordintes of the first point, nmely (y 1 y 2 ) nd (x 1 x 2 ), which gives the sme slope s bove

MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES 15 m = y 1 y 2 x 1 x 2 = (y 2 y 1 ) (x 2 x 1 ) = y 2 y 1 x 2 x 1 Exmple 23. Find the slope of the line pssing through the points P ( 1, 2) nd Q( 4, 5). If we choose to strt with the first point P, then both differences should strt with the coordintes of the first point, with y difference on the top Slope = m = 2 ( 5) 1 ( 4) = 2 + 5 1 + 4 = 3 3 = 1 If we choose to strt with the second point Q, then both differences should strt with the coordintes of the second point, with y difference on the top Slope = m = 5 ( 2) 4 ( 1) = 5 + 2 4 + 1 = 3 3 = 1 Exmple 24. The following results re useful to remember: Any horizontl line hs slope zero. For verticl lines the slope is not defined. If line hs positive slope, the line rises from left to right. If line hs negtive slope, the line flls down from left to right. The eqution of line pssing through two given points P (x 1, y 1 ) nd Q(x 2, y 2 ) is given in terms of the slope m nd the coordintes of one of the points: y = y 1 + m(x x 1 ) Note tht y nd x re vribles, representing the coordintes of n rbitrry point on the line. Here, we chose to use the coordintes of the first point P, nmely the given numbers x 1 nd y 1 but one cn lso use the coordintes of the second point Q nd still get the sme eqution for the line y = y 2 + m(x x 2 )

16 PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Exmple 25. Tke the points bove P ( 1, 2) nd Q( 4, 5), for which we computed the slope m = 1. The eqution of the line pssing through these two points, using the coordintes of the first point, is then y = 2 + 1(x ( 1)) = 2 + x + 1 = x 1 We should get the sme eqution if we use the coordintes of the second point insted y = 5 + 1(x ( 4)) = 5 + x + 4 = x 1 Exmple 26. Let the eqution of line be given by 3x + 2y = 5 How cn we find the slope of the line from this eqution? We need to solve this eqution for y in terms of x: 2y = 5 3x subtrct from both sides 3x y = 5 3x 2 then divide both sides by 2 y = 5 2 3 2 x this is wht we need to find the slope Once we hve the eqution in the form (for some numbers m nd b): y = mx + b slope = m nd y-intercept = b the slope is simply the number m (including the sign) in front of the vrible x, while the number b is the y-intercept. In our cse, The slope is the signed number in front of the vrible x, nmely slope = 3 2 nd the y-intercept = 5 2 A common mistke is to include the vrible x in the nswer for the slope. Remember tht the slope is number. Another common mistke is to forget the sign nd write 3 for the slope, insted of 3. Remember tht the y-intercept is the y-coordinte of 2 2 the point of intersection of the line nd the y-xis, when x = 0. E-mil ddress: bkostdinov@citytech.cuny.edu