linear equations

Whatz ur doubt?give an example for that.so that i help you to solve it

* Solve 11 + 3x – 7 = 6x + 5 – 3x First, combine like terms; then solve: '11 + 3x – 7 = 6x + 5 – 3x' yields '4 = 5' Then the "solution" is "no solution". When you try to solve an equation, you are making the assumption that there is a solution. When you end up with nonsense (like the "4 = 5" above), this says that your initial assumption (that there was a solution) was wrong, so there is no solution. Since "4 = 5" is utterly false, and there is no value of x that could ever make it true, then this equation has no solution. This is different from the previous problem, where there was a value of x that would work. Don't confuse these two very different situations: "the solution exists and has the value of zero" is not the same as "no solution value exists at all". Copyright © Elizabeth Stapel 2000-2006 All Rights Reserved And don't confuse the "no solution" type of equation above with the following type: * Solve 6x + 5 – 2x = 4 + 4x + 1 First, I'll combine like terms; then I'll solve: 5 + 4x = 5 + 4x, subtract 4x from either side to get 5 = 5 Is there any value of x that would make the above statement false? Isn't 5 always going to equal 5? In fact, since there is no "x" in the solution, the value of x is irrelevant: x can be anything I want. So the solution is "all x". This solution could also be stated as "all real numbers". And if you solve by subtracting 5 from both sides to get "4x = 4x", another trivially-true statement, or subtract 4x and 5 from both sides to get "0 = 0", the solution would still be the same: "all x". * Solve 9 = 3(5x – 2) First, I have to multiply through the parentheses on the right. Then I can solve: 1 = x Then the solution is x = 1. * Solve 6x – (3x + 8) = 16 Be careful with taking negatives through parentheses. Simplify on the left-hand side first; then solve: x = 8 Then the solution is x = 8. * Solve 7(5x – 2) = 6(6x – 1) I have to be sure to take the 7 and the 6 all the way through their respective parentheses. -8 = x Then the solution is x = –8. For this type of problem, take your time and write out all of your steps. Don't try to do everything in your head. * Solve 13 – (2x + 2) = 2(x + 2) + 3x Multiply through the parentheses (a minus sign on the left, and a two on the right), combine like terms, simplify, and solve: 1 = x Then the solution is x = 1. You can always check your work on equation-solving problems. The point of a solution is that it's the x-value that makes the equation true. To check your answer, plug in your solution, and make sure that the equation works. For instance, the first problem I did was "x + 6 = –3", and I got a solution of "x = –9". Here's the check: x + 6 = –3 (–9) + 6 = –3 –3 = –3 So the solution "checks". This ability to check your answers can come in handy on tests. Once you've completed all the questions, go back and plug in your solutions. If the solution "checks", then you know you got that question right. If it doesn't, you have the chance to correct your mistake before you hand in the test!

Most linear equations require more than one step for their solution. For instance: * Solve 7x + 2 = –54 I need to undo the "times seven" and the "plus two". There is no rule about which "undo" I should do first. However, if I first divide through by 7, I'm going to have fractions. Personally, I prefer to avoid fractions if possible, so I almost always do the plus/minus before the times/divide: x = -8 Then the solution is x = –8. By the way, formatting your homework and showing your work in the manner I have done is fairly universally acceptable. Your teacher should be happy to see your homework looking like these examples. It is also a good idea to clearly rewrite your final answer at the end of each problem. Don't expect your grader to dig through your work to figure out what you meant the answer to be. * Solve –5x – 7 = 108 x = -23 Then the solution is x = –23. * Solve 3x – 9 = 33 Copyright © Elizabeth Stapel 2000-2006 All Rights Reserved x = 14 Then the solution is x = 14. * Solve 5x + 7x = 72 First, I need to combine like terms on the left; then I can solve: x = 6 Then the solution is x = 6. * Solve 4x – 6 = 6x I need to move all the x's over to one side or the other. To avoid negative coefficients on my variables, I usually move the smaller x; in this case, I'll subtract the 4x over to the other side: -3 = x Then the solution is x = –3. In the above exercise, note that it is perfectly okay to have the "x=" be on the right. The variable is not required to be on the left; we're just used to seeing it there. It's alright if your solution works out with the variable on the right. However, I've heard of some instructors who insist that, in the final answer, the variable be placed on the left-hand side. If you're in doubt about your instructor's preferred formatting, ask. * Solve 8x – 1 = 23 – 4x x = 2 Then the solution is x = 2. * Solve 5 + 4x – 7 = 4x – 2 – x Before I can solve, I need to combine like terms: x = 0 Then the solution is x = 0. In the above problem, note that it is perfectly okay for x to be zero. Zero is a valid solution. Do not say "no solution"; this equation does indeed have a solution, the solution of x = 0.

"Linear" equations are equations with just a plain old variable like "x", rather than something more complicated like x2 or x/y or square roots or such. Linear equations are the simplest equations that you'll deal with. You've probably already solved linear equations; you just didn't know it. Back in your early years, when you were learning addition, your teacher probably gave you worksheets to complete that had problems like the following: Fill in the box: 'box' + 3 = 5 Once you'd learned your addition facts well enough, you knew that you had to put a "2" in the box. Solving equations works in much the same way, but now you have to figure out what goes into the x, instead of what goes into the box. However, since you're older now, the equations can be much more complicated, and therefore the methods you'll use to solve the equations will be a bit more advanced. In general, to solve an equation for a given variable, you need to "undo" whatever has been done to the variable. For instance: Copyright © Elizabeth Stapel 2000-2006 All Rights Reserved * Solve x + 6 = –3 I want to get x by itself; that is, I want to get "x =" on one side, and whatever numbers on the other side. Since I want just x on the one side, this means that I don't like the "plus six" that's with the x. Since the 6 is added to the x, I need to subtract to get rid of it. That is, I will need to subtract a 6, in order to "undo" having added a 6. This brings up an important point: No matter what kind of equation you're dealing with -- linear or otherwise -- whatever you do to the one side, you must do the exact same thing to the other side! Equations are like little children: You have to be totally, totally fair! Probably the best way to keep track of this subtraction of 6 from both sides is to format your work this way: animation: solveing x + 6 = –3 to get x = –9 What you see here is that I've subtracted 6 from both sides, drawn an "equals" bar underneath both sides, and added down: x plus nothing is x, 6 minus 6 is zero, and –3 plus –6 is –9. Then the solution is x = –9. The same "undo" procedure works for subtraction: * Solve x – 3 = –5 Since I want to get x by itself, I don't want the "–3" that's with the variable. The opposite of subtraction is addition, so I'll undo the –3 by adding 3 to both sides, and then adding down: x - 3 + 3 = -5 + 3 Then the solution is x = –2. The "undo" of multiplication is division. If something is multiplied on x, you undo it by dividing both sides (that is, dividing each term on both sides) by whatever is multiplied on the x: * Solve 2x = 5 Since the x is multiplied by 2, I need to divide both sides by 2: 2x/2 = 5/2 x = 2.5 Then the solution is x = 5/2 or x = 2.5. Usually the fractional form is the preferred form for your answers, rather than the decimal form. That is, usually texts (and teachers) will prefer the " 5/2" answer, rather than the "2.5" answer. If in doubt, check with your instructor. The "undo" of division is multiplication: * Solve x/5 = –6 Since the x is divided by 5, I'll want to multiply both sides by 5: solving animation: x = -30 Then the solution is x = –30. There is one sort of "special case": When x is multiplied by a fraction, you can divide both sides by that fraction (since division is the "undo" of multiplication), but remember that, to divide by a fraction, you flip-n-multiply. So it's simplest to just multiply both sides of the equation by the flip ("reciprocal") of the fraction. For example: * Solve 3/5 x = 10 Since x is multiplied by 3/5, I'll want to multiply both sides by 5/3, to cancel off the fraction on the x. Many students find it helpful to also turn the 10 into a fraction, by putting it over 1. That is, 10 = 10/1: (5/3)(3/5)x = (10/1)(5/3) x = 50/3 Then the solution is x = 50/3.