Understanding Algebra?
611607
Usually these types of problems would have -4x where you have -4.  Assuming that's the case...\n\nMultiply the top row by 4.  That comes out 4x + 12y = 68\n\nAdd this to the bottom row.  That is 0x + 14y = 84\n\nor just 14y = 84\n\nDivide both sides by 14 and get y = 6\n\nPlug this in to either equation to get x:\nx + 3(6) = 17\nx + 18 = 17\nx = -1\n\nCheck in other equation: -4(-1) + 2(6) = 16\n4 + 12 = 16 check\n\nHope that helps
805935
Please don't take this as an attack or insult, it is meant to be helpful and comes from a professional Electrical Engineer:\n\nThe questions you are asking are all very basic algebra problems.  A solid grasp of college-level first and second semester algebra is an absolute requirement when it comes to understanding electrical systems.  These math fundamentals are used everywhere in describing, calculating the performance of, installing, and troubleshooting electrical circuits.  Without a solid foundation in mathematics, nothing within the electrical field is going to make any sense at all.\n\nThat said, here is how to do these problems:\n\n(1) Es=E1+E2+E3 solve for E2\n\nThis is calculation of the voltage across one element in a series circuit composed of 3 elements.  The total voltage across the series circuit (Es) is equal to the sum of the voltages across each of the elements of the circuit (E1, E2, E3).  So the voltage across any one element is equal to the total voltage minus the sum of the OTHER elements:\nE2 = Es - (E1+E3)\n\n\n(2) R1=I2R2 divided by I2 solve for R\n[double check that you have copied this right -- I think you mean "solve for R2" since it is already solved for R1]:\n(I2*R2)/I2, the I2's cancel out so you are left with R2.  R1 = R2.\n\n[you may have meant "R1=I2R2 divided by I1 solve for R2" which would be a formula for calculating an unknown resistance in a parallel circuit based on a the other known resistance and the respective currents]:\nR1 = (I2 * R2) / I1\nMultiply both sides by I1:\nR1 * I1 = I2 * R2\nDivide both sides by I2\n(R1 * I1) / I2 = (I2 * R2) / I2\n(R1 * I1) / I2 = R2\nSo, R2 = R1I1 divided by I2\n\n\n(3) R1 divided by Rx = R2 divided by R3 solve for Rx\nInvert both sides of the equation:\n1/(R1/Rx) = 1/(R2/R3)\nRx/R1 = R3/R2\nMultiply both sides by R1:\nR1*(Rx/R1) = R1*(R3/R2)\nRx = R1R3 divided by R2\n\n(4) y=mx+b solve for b\nSubtract mx from both sides of the equation:\ny - mx = mx + b -mx\ny - mx = b\nso, b = y minus mx\n\n(5) I2=I1R1divided by R2 solve for I1\nI2 = (I1*R1)/R2\nMultiply both sides by R2:\nR2 * I2 = (R2*(I1*R1))/R2\nR2I2 = I1R1 because the R2's cancel out\nNow divide both sides by R1:\n(R2I2)/R1 = (I1R1)/R1\n(R2I2)/R1 = I1\nSo, I1 = R2I2 divided by R1
73629
"The Complete Idiot's Guide to Algebra" \n\nRotten title - great resource
605993
algebra is a broad topic. try to go back to ur notes and look for the formulas that your teacher has given to you. better to stick on what is in your notes and textbooks. try this also, ask someome who has taken this subject before...his information might help you...good luck.
145326
i'm good at math, but what exactly is the quustion??? there isn't any... what is it, a equation?\nany way you can post it to sciences and mathematics > mathematics, im sure someone there can help you...\n\n>>>ok one minute i'm working on it...\n\n>>>there you go:\nlets do the first: (x-3)(x-1)(x+2)\nfirst we open the last two parentheses as in usual opening, and the (x-3) remains as it is. we get (x-3)[x^2+2x-x-2]\nequals (x-3)(x^2+x-2) then we open the two parentheses left. first we multiply by the first member (x) then the second (-3). then we get:\nx*x^2+x*x+x*(-2)-3*x^2-3*x-3*(-2)\nx^3 + x^2 - 2x -3x^2 -3x + 6\nand finally:\nx^3 -2x^2 -5x + 6\n\nthe second is the same way, i'm doing it, good luck :)\n\n>>>the second: (x-5)(x+1)(x-2)\nfirst we open the last two parentheses as in usual opening, and the (x-5) remains as it is. we get (x-5)[x^2-2x+x-2]\nequals (x-5)(x^2-x-2) then we open the two parentheses left. first we multiply by the first member (x) then the second (-5). then we get:\nx*x^2+x*(-x)+x*(-2)-5*x^2-5*(-x)-5*(-2)\nx^3 -x^2 -2x -5x^2 +5x +10\nand finally:\nx^3 -6x^2 +3x + 10\n\ni hope i helped...\n\n>>>oh i leave the first one to the last because i have to leave one of the three, and it really doesn't matter which one i leave. for example on the first i left the first parenthese (x-3)[(x-1)(x+2)] but i could leave the last [(x-3)(x-1)](x+2) or the middle (x-1)[(x-3)(x+2)] and i'd still get the same answer. i have to leave one of the three bcuz i can't open them all together, it's complicated and hard, and by opening two ane leave one it becomes much easier.
716170
since 2 is a root you take (x-2)*(x-2) and -3 is one also so you multiply (x+3) now you know it is a fifth degree so you need two more so you can take (4x^2+1) and multiply\n(x-2)(x-2)(x+3)(4X^2+1) you do the algebra
857303
Hmm. It could just be something like forgetting a parenthesis. Let's go through it step by step:\n\nThe quadratic formula is:\n\n[ -b +/- sqrt(b^2-4ac) ] / 2a\n\nBecause of the +/-, we will have two equations:\n\n[ -b + sqrt(b^2-4ac) ] / 2a\nand\n[ -b - sqrt(b^2-4ac) ] / 2a\n\nThe equation is x^2 + 3x - 84 = 0\na = 1\nb = 3\nc = (-84)\n\nSo we have:\n[ -3 + sqrt(3^2-4(1)(-84)) ] / 2(1)\nand\n[ -3 - sqrt(3^2-4(1)(-84)) ] / 2(1)\n\n[ -3 + sqrt(345) ] / 2\nand\n[ -3 - sqrt(345) ] / 2\n\n7.78709\nand\n-10.7871
817481
We can use faux-algebra to solve these. You probably haven't taken algebra yet. With algebra these would be extremely easy to solve. It's why algebra was invented.\n\na. 16 x ___ has to be -42, because if we add 51 to it we get 9. Thus, ____ has to be -42/16, or -21/8.\n\nb. 2 x ____ has to be 22 - 9 x _____, because if we subtract 8 from it we get 14 - 9 x ____. And 22 has to be 11 x ____, because if we subtract 9 x ____ from it we get 2 x ____. Thus, ____ has to be 22/11, or 2.\n\nc. 9 x ___ has to be 3 x ____ + 3, because if we add 2 to it we get 3 x ___ + 5.  And 3 has to be 6 x ____, becuase if we add 3 x ____ to it we get 9 x ____. Thus, ____ has to be 3/6, or 1/2.\n\nWith algebra, no such thinking would be required. You just plug in the formulas and grind away. You'll see when you study algebra.
794610
You have to do the equations in the [ ] first.\nThen you have to do multiplication before addition.\nHere's the order of how you should do math like this:\nP - Parenthesis\nE - Exponents (a number to the something power)\nM - Multiplication\nD - Division\nA - Addition\nS - Subtraction\n\nSo you do [2+6(-3)] first\n[2-18]\n[-16]\nNow it's 15-3[-16]\n15+48\nAnswer: 63
560506
that works... so does this:\n\n-2 x -1 x 1 x 2 x 3 = 12\n\n-2*-1=2\n2*1=2\n2*2=4\n4*3=12
676624
Algebra (from Arabic: الجبر, al-ğabr) is a branch of mathematics concerning the study of structure, relation and quantity. Elementary algebra is often taught in high school and gives an introduction to the basic ideas of algebra: studying what happens when numbers are added or multiplied, and how to make polynomials and find their roots.\n\nAlgebra is much broader than elementary algebra and can be generalized. Rather than working directly with numbers, one can work with symbols or elements of some set. Addition and multiplication are viewed as general operations, and their precise definitions lead to structures such as groups, rings and fields.\n\nMathematics can be defined as the logically rigorous study of topics such as quantity, structure, space, and change. Another view, held by many mathematicians, is that mathematics is the body of knowledge justified by deductive reasoning, starting from axioms and definitions.\n\nMathematics is used throughout the world in fields such as science, engineering, medicine, and economics. These fields both inspire and make use of new discoveries in mathematics. New mathematics is also created for its own sake, without any particular application in view.\n\nThe word "mathematics" comes from the Greek μάθημα (máthēma) meaning science, knowledge, or learning, and μαθηματικός (mathēmatikós), meaning fond of learning. It is often abbreviated math in the U.S. and Canada and maths in Britain and many Commonwealth countries.\n\nThe evolution of mathematics might be seen to be an ever-increasing series of abstractions, or alternatively an expansion of subject matter. The first abstraction was probably that of numbers. The realization that two apples and two oranges have something in common, namely that they fill the hands of exactly one person, was a breakthrough in human thought. In addition to recognizing how to count concrete objects, prehistoric peoples also recognized how to count abstract quantities, like time -- days, seasons, years. Arithmetic (e.g., addition, subtraction, multiplication and division), naturally followed. Monolithic monuments testify to knowledge of geometry.\n\nFurther steps need writing or some other system for recording numbers such as tallies or the knotted strings called khipu used by the Inca empire to store numerical data. Numeral systems have been many and diverse.\n\nFrom the beginnings of recorded history, the major disciplines within mathematics arose out of the need to do calculations on taxation and commerce, to understand the relationships among numbers, to measure land, and to predict astronomical events. These needs can be roughly related to the broad subdivision of mathematics, into the studies of quantity, structure, space, and change.\n\nMathematics since has been much extended, and there has been a fruitful interaction between mathematics and science, to the benefit of both.\n\nMathematical discoveries have been made throughout history and continue to be made today. According to Mikhail B. Sevryuk, in the January 2006 issue of the Bulletin of the American Mathematical Society, "The number of papers and books included in the Mathematical Reviews database since 1940 (the first year of operation of MR) is now more than 1.9 million, and more than 75 thousand items are added to the database each year. The overwhelming majority of works in this ocean contain new mathematical theorems and their proof."\n\nArithmetic or arithmetics (from the Greek word αριθμός = number) in common usage is a branch of (or the forerunner of) mathematics which records elementary properties of certain operations on numbers, though professional mathematicians often treat arithmetic as a synonym for number theory. It is the oldest and simplest branch of mathematics, used widely by almost everyone from simple daily counting to more advanced science and business.\n\nGeometry (Greek γεωμετρία; geo = earth, metria = measure)
705864
Alright, don't listen to the above post...\n\nGrade 12 here, answering questions all day for mere fun.\n\n12)\ny - 3 = -1/3 (x + 4)\n\nEXPLAINED\n\nadd 3 to both sides\ny = -1/3 (x + 4) + 3\n\nexpand 1/3(x + 4) \n= x * 1/3 + 4 * -1/3\n= x/3 + 4/3\ny = -1/3x - 4/3 + 3\n\nTo add -4/3 and 3, use 3 = 9/3\ny = -1/3x - 4/3 + 9/3\ny = -1/3x + 5/3\n\nTrue, my bad. I originally added 4/3 instread of -4/3\n\n\n16)\n(-2, -1) (4, -4)\n\nm = Dy / Dx\nm = ((-1) - (-4)) / ((-2) - (4))\nm = 3 / -6\nm = -1/2\n\ny = mx + b\n(-1) = -1/2(-2) + b\n-1 = 1 + b\n-1 - 1 = b\n-2 = b\n\ny = -1/2x - 2\n\nCHECK\ny = -1/2(4) - 2\ny = -2 - 2\ny = -4\nTherefore it works.\n\n18)\n(0, 0) (2, 3)\n\nm = Dy / Dx\nm = ((3) - (0)) / ((2) - (0))\nm = 3/2\n\nb = 0\n\ny = mx + b\ny = (3/2)x + 0\ny = (3/2)x\n\nNote, the lower post copied my answers
1098902
Try these websites: http://www.algebrahelp.com/lessons/\nhttp://www.sparknotes.com/math/#algebra1\n\nI hope I helped. Good Luck!
57352
http://www.math.com/homeworkhelp/Algebra.html\n\nhttp://www.mathsisfun.com/number_algebra.html\n\nhttp://mathforum.org/library/drmath/sets/high_alg_toc.html\n\nhttp://www.tellmehowto.net/answers/maths.html\n\nTheres a couple, First two are aimed more towards algebra; Hope you find what you need!
584912
http://www.algebrahelp.com/ I found this site very useful perhaps you should check it out http://www.algebrahelp.com/
926781
Remember this phrase and you'll be fine!\nPlease = Parenthesis\nExcuse = Exponents\nMy = Multiplication\nDear = Division\nAunt = Addition\nSally = Subtraction\n\nNow let's work on the problem.\n\nWe can see that Parenthesis come first, however, in this problem, they terms in parenthesis are unlike terms, therefore, we move on.\n\nThere are no exponents, so we move on again!\n\nMultiplication.  Ignore the 28 for a moment and concentrate on \n-6(2x+4).\n-6 times 2x is -12x and -6 times 4 is -24. Now we have this problem:\n\n28-12x-24=124\n\nLet's combine like terms.\n\n28-24= 4\n\nNow we have 4-12x=124.\n\nsubtract 4 from each side.\n\n-12x=120\n\nnow divide each side by -12\n\n-12x/-12= x and 120/-12= -10\n\nso x= -10\n\nHope that helped!!
795656
The first thing you should remember is the phrase PEMDAS.  This is shorthand for parenthesis, exponents, multiplication/division, and addition/subtraction.  This is the order of operations in math.  You should solve all equations in this order.  So for the first one, you should first distribute (multiply) -7 with (3-3x).  After that combine like terms, i got 25x -30; you can also factor that to be 5(5x-6).  The same steps apply for the rest of the problems.  I hope it helps.