Fundamental Concepts of Algebra - tentotwelvemath · Fundamental Concepts of Algebra Suppose we...
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Fundamental Concepts of Algebra Suppose we have the equality 15 = 10 + 5 Usually we have something that looks like 15 = + 5 and our job is to solve it. As long as we perform the same operation to both sides of our equation, we preserve equality. For our example, will always be able to find ‘10’ as the value of . If we inadvertently do something different to each side we will lose ‘10’ as the answer and will end up with some other number that doesn’t make sense in the original equation. 15 = 10 + 5; not (anything else) +5. Legitimate processes preserve the original equality. Here are some examples of legitimate processes. For each process, make sure that ‘10’ makes sense as ‘x’ in the final line. Add the same thing to both sides, for example, 7: 15 = + 5 15 + 7 = + 5 + 7 Simplifies to: 22 = + 12 Take the same thing from both sides for example, 3: 15 = + 5 15 − 3 = + 5 − 3 Simplifies to: 12 = + 2
Transcript of Fundamental Concepts of Algebra - tentotwelvemath · Fundamental Concepts of Algebra Suppose we...
FundamentalConceptsofAlgebraSupposewehavetheequality
15 = 10+ 5Usuallywehavesomethingthatlookslike
15 = 𝑥 + 5andourjobistosolveit.Aslongasweperformthesameoperationtobothsidesofourequation,wepreserveequality.Forourexample,willalwaysbeabletofind‘10’asthevalueof𝑥.Ifweinadvertentlydosomethingdifferenttoeachsidewewilllose‘10’astheanswerandwillendupwithsomeothernumberthatdoesn’tmakesenseintheoriginalequation.15 = 10+ 5;not(anythingelse)+5.Legitimateprocessespreservetheoriginalequality.Herearesomeexamplesoflegitimateprocesses.Foreachprocess,makesurethat‘10’makessenseas‘x’inthefinalline.Addthesamethingtobothsides,forexample,7:
15 = 𝑥 + 5
15+ 7 = 𝑥 + 5+ 7
Simplifiesto:
22 = 𝑥 + 12
Takethesamethingfrombothsidesforexample,3:
15 = 𝑥 + 5
15− 3 = 𝑥 + 5− 3
Simplifiesto:
12 = 𝑥 + 2
Multiplybothsidesbythesamenumber,forexample2:
15 = 𝑥 + 5
2 15 = 2(𝑥 + 5)
Expandsto:
30 = 2𝑥 + 10
Noticethatalltermsonbothsidesneedtobemultipliedby2.
Dividebothsidesbythesamenumber,forexample,5:
15 = 𝑥 + 515 (15) =
15 (𝑥 + 5)
Expandsto:
3 =𝑥5 + 1
Thesecondlinecouldalsobewrittenas:155 =
𝑥 + 55
Whichsimplifiesto:
3 =𝑥5 +
55
Whichinturnsimplifiesto:
3 =𝑥5 + 1
Raiseasanexponentwiththesamebase:
15 = 𝑥 + 5
2!" = 2(!!!)
Takealogarithmwiththesamebase
15 = 𝑥 + 5
log! 15 = log!(𝑥 + 5)
Applyanyotherfunction:
15 = 𝑥 + 5
𝑓 15 = 𝑓(𝑥 + 5)
N.B.Formostfunctions,suchas‘squaring’,𝑓 𝑥 + 5 ≠ 𝑓 𝑥 + 𝑓 5 . Eg,15! = 225,
but10! + 5! = 125,not225.Functionsdonotgenerallydistribute.Whatis
illustratedhereisthatifyouputthesamevalueintoafunction,yougetthesame
valueout.Withoutthedetailsofthefunction𝑓(eg,𝑓(𝑥) = 𝑥!),wecannotsimplify
therighthandside.
Sometimes,weworkonlyoneside.Thatiswhenwewritetheexpressiononone
sideinanequivalentformat.Forexample:
2𝑥 + 5 ! − 𝑥 + 3 ! = 456
2𝑥 + 5 2𝑥 + 5 − 𝑥 + 3 𝑥 + 3 = 456
4𝑥! + 10𝑥 + 10𝑥 + 25− 𝑥! + 3𝑥 + 3𝑥 + 9 = 456
4𝑥! − 𝑥! + 20𝑥 − 6𝑥 + 25− 9 = 456
3𝑥! + 14𝑥 + 16 = 456
Inthisexample,wemanipulatethelefthandsideonly.Ifyouknowhowtosolvea
quadratic,youwouldprobablynowsubtract456frombothsides,andifyou’re
havingagooddayyoumightfactorthetrinomial.Ifyourobservationissharp,you
might’venoticedthat‘10’isoneofthetwosolutionsforthisequation.
