SNAP Centre Workshop. Basic Algebraic Manipulation

Transcription

1 SNAP Cetre Workshop Basic Algebraic Maipulatio 8

2 Simplifyig Algebraic Expressios Whe a expressio is writte i the most compact maer possible, it is cosidered to be simplified. Not Simplified: x(x + 4x) Simplified: 5x 2 Substitutio Oe of the most straightforward forms of simplificatio arises whe a expressio cotaiig variables is preseted, accompaied by the variables umerical values. Example 1 m + ( 2) 2 m = 3, = 5 By substitutig the variables values ito the expressio ad evaluatig, the expressio ca be fully simplified. (3) + ((5) 2) 2 = 3 + (3) 2 = = 12 Note that after substitutio, we follow the stadard order of operatios. Groupig Like Terms Whe give a expressio cotaiig variables whose values are ukow, it may still be possible to simplify the expressio by groupig like terms. Like terms are terms whose variables or combiatio of variables match exactly. Whe beig added or subtracted i a expressio, we ca group like terms simply by addig or subtractig the like terms coefficiets. Example 2 2p + 3p By recogizig both terms have the variable p i commo, their coefficiets ca be added to simplify the expressio. = (2 + 3)p = 5p 9

3 The same priciples apply whe the variable compoet of the like terms is composed of more tha oe variable. We ca add the terms coefficiets as log as the variable compoets of the terms match exactly. Example 3 8xy 2 17xy 2 = (8 17)xy 2 = 9xy 2 It is also possible for a expressio to cotai more tha oe set of like terms. To simplify, we ca group all matchig like terms together. Example 4 8q + 3q 2 7r + 2q + 2r Although it is ot ecessary, we ca rearrage the equatio to help idetify ad orgaize like terms. = 3q 2 + 8q + 2q 7r + 2r = 3q 2 + (8 + 2)q + ( 7 + 2)r = 3q q 5r The q 2 ad q terms do ot group sice they differ by a expoet ad, as such, we do ot cosider them like terms. Also, otice that the egative sig associated with the 7r stayed with its coefficiet after we grouped it. If we had kept it outside the paretheses durig groupig, the fial expressio would have icorrectly cotaied the term 9r. Distributio Aother way we ca simplify certai algebraic expressios is by distributio. Algebraic expressios have a distributive property whereby the term i frot of a set of paretheses ca be multiplied through the terms cotaied withi the paretheses. Example 5 3(x + 2) = (3)(x) + (3)(2) = 3x + 6 The term beig distributed does ot eed to be a costat. It ca be a combiatio of costats ad variables. Additioally, if there is a egative sig associated with the term beig distributed, it must be distributed through the paretheses alog with the term. Example 6 3y(8x 2y) = ( 3y)(8x) + ( 3y)( 2y) = 24xy + 6y 2 10

4 As metioed, the egative sig is distributed with the 3y term. Whe multiplied by the 2y term, the result is a positive term, which ca be see i the fial simplified expressio as 6y 2. Simplifyig a expressio will ofte ivolve distributio, followed by groupig of like terms, ad fially substitutio. I these cases, it is importat to follow the correct order of operatios. Also, it ca be useful to break the problem dow ito smaller problems to make it more maageable. Example 7 3x 3 x(x 2 + x(3y 2y + 2x)) Simplify the equatio, give the followig coditios: If x = 4, y = 3 If x = 2, y = 5 If x = 3, y = 1 We could substitute the give x ad y values ito the expressio immediately, however, sice three differet sets of values are give, it is a better idea to perform distributio ad like term groupig simplificatios first. We ca begi by groupig the like terms i the iermost set of paretheses. Here, simplifyig the expressio 3y 2y + 2x ca be see as a smaller simplificatio problem withi the larger overall problem. 3x 3 x(x 2 + x(3y 2y + 2x)) = 3x 3 x(x 2 + x((3 2)y + 2x)) = 3x 3 x(x 2 + x(y + 2x)) Next, we distribute the x term across the expressio i the iermost paretheses, the group the resultig like terms. = 3x 3 x(x 2 + xy + 2x 2 ) = 3x 3 x((1 + 2)x 2 + xy) = 3x 3 x(3x 2 + xy) Distribute the x term across the remaiig set of paretheses, ad fiish by groupig the resultig like terms. = 3x 3 3x 3 (x 2 )y = (3 3)x 3 (x 2 )y = (0)x 3 (x 2 )y = (x 2 )y Fially, substitute the give x ad y values ito the simplified variable expressio. (x 2 )y If x = 4, y = 3 11

5 = (4 2 )(3) = (16)(3) = 48 (x 2 )y If x = 2, y = 5 = (( 2) 2 )(5) = (4)(5) = 20 (x 2 )y If x = 3, y = 1 = (3 2 )( 1) = (9)( 1) = 9 By groupig like terms ad distributig first, our substitutio step was made cosiderably easier. We are ot limited to distributig sigle terms across paretheses. It ca also be doe with polyomials. Example 8 (3x 2y)(7x + 4y) We begi by distributig the first term i our first expressio through the etire secod expressio, the distribute our secod term i our first expressio through the etire secod expressio. We fiish by groupig all like terms. = 3x(7x + 4y) 2y(7x + 4y) = 21x xy 14xy 8y 2 = 21x 2 2xy 8y 2 Example 9 (3x 2 + 2x + 1)(x + 5) Agai, we distribute each of the terms i our first expressio across our etire secod expressio. Now, we group all like terms. = 3x 2 (x + 5) + 2x(x + 5) + 1(x + 5) = 3x x 2 + 2x x + x + 5 = 3x x x

6 Solvig Oe Equatio, Oe Ukow Algebraic Equatios Beyod simplifyig expressios, we ca also use algebraic maipulatio to solve equatios. It is ot uusual to be preseted with a equatio cotaiig some ukow variable whose value ca be determied usig algebraic maipulatio. Whe give a sigle equatio that cotais a sigle ukow variable, we ca determie the value(s) that could be substituted ito the equatio i order to make it a true statemet. This value, or set of values, is called the solutio. Oe way to check for solutios is to substitute a give value ito a equatio. Example 10 5x + 2 = 4 + 7x Substitute to see if x = 3 is a solutio. 5(3) + 2 = 4 + 7(3) = = 17 Sice the statemet 17 = 17 is true, x = 3 is a solutio. Example 11 5x + 2 = 4 + 7x Substitute to see if x = 2 is a solutio. 5(2) + 2 = 4 + 7(2) = = 10 Sice the statemet 12 = 10 is false, we kow that x = 2 is ot a solutio. I order to avoid checkig umbers oe at a time to fid solutios, we ca use algebraic maipulatio to isolate the variable i a give equatio, effectively givig the solutio. Example 12 5x + 2 = 4 + 7x The first step i isolatig the variable x i the above equatio is groupig all of the terms cotaiig x o oe side of the equatio, ad all of the costats o the other. This ca be accomplished by performig the same operatios o both sides of the equatio. 5x + 2 7x 2 = 4 + 7x 7x 2 5x 7x = 4 2 To elimiate 7x from the right side, it was subtracted from both the right ad the left sides. Similarly, to elimiate 2 from the left side, we subtracted it from both the right ad the left sides. Next, it is ecessary to group the like terms o the left side of the equatio. (5 7)x = 6 2x = 6 13

7 Fially, divide both sides by the ukow variable s coefficiet, makig sure to iclude the egative sig. 2x 2 = 6 2 x = 3 The resultig equatio idicates to us that x = 3 is ot just a solutio to the give problem (as was already show i the previous example to be true), but the oly solutio, elimiatig the eed to substitute ay other values of x ito the equatio. Sometimes we eed to simplify a expressio withi a equatio before fidig a solutio. Example 13 3(z + 7) z = 14 2z First, the 3 is distributed across the paretheses o the left had side of the equatio. 3z + 21 z = 14 2z Next, group ay like terms o both sides of the equatio. 2z + 21 = 2z + 14 Oce we have grouped all like terms, we ca move the terms cotaiig z to either side of the equatio, ad all costats to the other. 2z z 14 = 2z z 14 7 = 4z Fially, we divide both sides by the variable term s coefficiet = 4z 4 = z The solutio for this particular equatio is a fractio, ad uless specified otherwise it is perfectly acceptable to leave it as oe. Special Cases There are two special cases that we ca ecouter whe solvig algebraic equatios, both of which occur whe all variable terms cacel to 0. Oe possible outcome is that a true statemet remais oce the variable terms cacel. Example 14 2p 6 2(p 3) + 4 = 4 2p 6 2p = 4 4 = 4 14

8 After we simplified the left side of the equatio, all terms cotaiig the variable p cacel, leavig the true statemet 4 = 4. There is o real umber p ca be that would make this statemet false, therefore the solutio is all real umbers. The secod possible outcome is that a false statemet remais oce all variable terms cacel. Example 15 3(2t + 4) = 2 (3t + 3) + 4 6t + 12 = 6t t + 12 = 6t + 10 After simplifyig both sides of the equatio, the ext step should be to move all terms cotaiig t to oe side of the equatio. Attemptig to do this, however, results i all terms cotaiig t cacellig. 6t t = 6t t 12 = 10 After simplificatio, the false statemet 12 = 10 remais. There is o real umber t that would result i this statemet beig true, idicatig there is o solutio. Isolatig Variables Whe preseted with a equatio that has more tha oe ukow term, it is sometimes ecessary to be able to isolate oe of the variables ad express it i terms of the other variables. Methods similar to those used i solvig equatios with oe ukow are used, however, the ability to judge whe to ad whe ot to simplify a give term is also eeded. Example 16 3p 2q + 4 = 5q + p 3 Express p i terms of q. The goal here is to move all costats ad terms cotaiig q to oe side of the equatio, ad isolate p o the other side of the equatio. Simplifyig both sides of the equatio seems like a good place to start, however, distributig the 3 i the deomiator o the right side of the equatio would result i the eed to work with fractios. Istead, otice that multiplyig both sides of the equatio by 3 results i whole umbers throughout the etire equatio. 3(3p 2q + 4) = 3(5q + p) 9p 6q + 12 = 5q + p 3 Next, perform the appropriate additio ad subtractio operatios to gather all p terms o the left, ad all costats ad q terms o the right, ad the group ay like terms. 9p 6q + 12 p + 6q 12 = 5q + p p + 6q 12 9p p = 5q + 6q 12 15

9 8p = 11q 12 Fially, divide both sides of the equatio by 8 to isolate p completely. 8p 8 p = = 11q q 12 8 The ext example illustrates how a expressio iside paretheses ca ad usually should be treated i the same way a sigle term would be treated as log as oe of the terms iside the paretheses cotai the variable that is beig isolated. Example 17 y + 2 = 2x(y + 3) Express x i terms of y. Normally, our first step i this problem would be to distribute 2x across the paretheses o the right side of the equatio, however, oce this term is distributed, isolatig x would be far more difficult. Istead, both sides of the equatio will be divided by the term i paretheses. y + 2 y + 3 y + 2 y + 3 = 2x(y + 3) y + 3 = 2x y + 3 y + 3 simplifies to 1, leavig 2x o the right side of the equatio. As i previous examples, the oly step left at this poit is to divide by the x term s costat coefficiet. y +2 2(y + 3) = x OR y + 2 2y + 6 = x Now x is beig expressed i terms of y. Note that the 2 i the deomiator o the left side of the equatio ca be distributed, or remai outside the paretheses. Ofte, there is more tha oe way to maipulate a equatio algebraically. The followig example ca be solved usig two equally valid methods. Example 18 4m + 4 = 2m Express m i terms of. The first method ivolves distributig the i the deomiator o the left side of the equatio across the terms i paretheses. 4m + 4 = 2m 4m + 4 = 2m 16

10 Next, we will add 4m to both sides of the equatio, group like terms, the divide both sides of the equatio by the m term s costat coefficiet to isolate m completely. 4m m = 2m + 4m 4 = (2 + 4)m 4 = 6m 4 6 = 6m = m = m The secod method begis by multiplyig both sides of the equatio by. Example 19 4m + 4 = 2m Express m i terms of. ( 4m + 4) = (2m) 4m + 4 = 2m Next, all terms cotaiig m are moved to oe side of the equatio, after which like terms are grouped. 4m m = 2m + 4m 4 = (2 + 4)m 4 = 6m Fially, we divide both sides of the equatio by the costat coefficiet ad the variable coefficiet associated with the m term, ad simplify. 4 6 = 6m = m 17