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Non HW or CW maths question

  1. Aug 7, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm refreshing my maths for the restart of uni and since I have problems finding 'x' equations, I'm stumped with this fairly easy question:

    Now, because I had difficulty solving it I used an online calculator to solve and show me the steps so I can improve for next time. From looking through the steps - it introduces a '-1' to make it:


    This caught me out and I'm having trouble finding out where the -1 came from. I believe I have an inkling but would like to know the reason for the -1.

    Many thanks.
  2. jcsd
  3. Aug 7, 2013 #2
    -(a) = -1(a), if that helps. So for -(x+2), multiply x and 2 by -1 so that the left hand side of your equation equals 5 - x - 2, or 3 - x (ie: because 5 - 2 = 3).

    Next, you want all the x's on one side of the equal sign, and all the numbers on the other. So, if you add x to both sides you obtain 3 = 6x.

    Next, divide both sides by 6 to obtain 3/6 = x which is 1/2 = x.

    You can now plug this result into the equation to see if it works: 5 - (0.5 + 2) = 2.5, which is correct.
  4. Aug 7, 2013 #3
    That was the inkling I had. Unfortunately, not fully knowing the rules of algebra and the fact that the rules can change from one equation to the next make it difficult for me though I have had some success with 5x[-4 + 10(x - y)] + 7x, for example. Which is something I was never that confident attempting.
  5. Aug 7, 2013 #4

    Ray Vickson

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    No, no, no: the rules do NOT change from one example to the next. I hope you mean that which rule you apply may change from one example to the next---that can certainly be true. But the rules themselves do not change.

    It may help to see some basic rules laid out; here are the most commonly-used ones.
    1. a+b = b+a
    2. a*b = b*a
    3. a+(b+c) = (a+b)+c
    4. a*(b*c) = (a*b)*c
    5. a*(b+c) = a*b + a*c
    6. a+0 = a
    7. a*1 = a
    8. If a=b, then a+c = b+c and a-c = b-c
    9. -a = (-1)*a

    So, to apply these to your example E = 5*x[-4 + 10*(x - y)] + 7*x (I am just calling the expression 'E' since I want to be able to refer to it by name, perhaps several times). There ARE several different ways to handle E. For example, we can use property 3 (with a = 5*x, b = -4, c = 10*(x-y)) to get E = (5*x)*(-4) + (5*x)*[10*(x-y)] + 7*x. We can use properties 2 and 4 to write (5*x)*(-4) = (-4)*(5*x) = (-4*5)*x = -20*x. We can use property 5 to get (5*x)*[10*(x-y)] = (5*x)*(10*x - 10*y) and use it again to get (5*x)*(10*x) - (5*x)*(10*y), and then use the various properties to get (5*x)*(10*x) = 50*x^2 (where x^2 = x*x) and (5*x)*(10*y) = 50*x*y. Altogether we get E = -20*x + 50*x^2 - 50*x*y + 7*x = 50*x^2 - 50*x*y - 13*x.

    A much quicker way is to observe first that E will finally involve terms in x alone, in x*y and in x*x = x^2. We can just isolate the separate terms one-by-one: the coefficient of x (that is, the number that will multiply x in the answer) is 5*(-4) + 7 = -13. The coefficient of x^2 will be 5*10, and the coefficient of x*y will be 5*(-10) = -50. Admittedly, this method only becomes really useful after you have done lots of examples and gotten comfortable with such problems, perhaps by using the long and laborious methods first.
  6. Aug 7, 2013 #5
    Correct, that is what I meant. I'm writing this before reading the rest of your response to save any distress. I was going to edit the post but unfortunately didn't. My apologies!
  7. Aug 7, 2013 #6


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    You want to think of the rules in terms of something real like a bank account. So -(x + 2) could mean, I owe a total of x+2 dollars. Then -x-2 could mean, I owe x dollars and I owe 2 dollars, so they mean exactly the same thing. It helps to think of a real situation sometimes.
  8. Aug 7, 2013 #7


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    Some of us gained a mechanical-like thinking of simple algebra while we first studied back in Algebra 1.

    Expressions inside grouping symbols are like being in a package. When you see this, 5-(x+2)=5x, think of FIVE and then subtract everything which is in the package found in (________). You are subtracting the entire sum which has been packaged; you subtract the x and you subtract the "2". The left side then is 5-x-2.

    To continue, you have this possible sequence of steps:
    5+(-x)+(-2)=5x, just using the meaning of subtraction so we have only addition of numbers and multiplications;
    5+(-2)+(-x)=5x, because we can change positions of numbers in a sum and this does not change the sum;
    3+(-x)=5x, simple arithmetic, signed numbers;
    3+(-x)+(x)=5x+x, adding additive inverse of -x to both sides of equation;
    3+0=6x, property of additive inverses, and on rightside, combined like terms of x
    3=6x, the zero in the sum is no longer needed;
    1/2 = x, just multiplied left and right sides by multiplicative inverse of 6.

    All those steps are shown for you in order to understand the justification. In practice, we would not usually go through all of that. We would usually be able to just do THIS:
    5-(x+2)=5x, originally given,
  9. Aug 8, 2013 #8
    Thank you, symbolipoint.

    It makes sense. As a result of my inexperience and failed schooling I attempt equations in a linear fashion, that is, I am given the equation and make no attempt at making it easy for myself, or as you guys would simply put - just answering the question.
    What makes the difference for me, with the method you have written, is that you did something I was unaware could be done and that is adding the 'x' to the RHS to make up the 6x. I may have said that incorrectly though that is what I am seeing.
  10. Aug 8, 2013 #9


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    Well, you agree that 5-3=2 and also that 5-3+3=2+3, but the left hand side is now simply 5 so we've essentially removed the -3 on that side.

    Equivalently with algebra, we will instead use letters to represent these random numbers.

    If we have x-3=2 (in this case we know x=5 but let's ignore that for a second)
    then equivalently, we can add 3 to both sides to give x-3+3=2+3, and now you can see that on the left side we have -3+3=0 hence we end up with x=2+3=5.

    Now, of course you can do the exact same thing if we had another variable in place of the 3. If you have x+... = ,,,, then simply take away ... from both sides to end up with x = your answer*

    *only if your answer doesn't also include more x's. You need to get all the x's onto the same side.
  11. Aug 8, 2013 #10
    Thank you, everyone. I'm referring to your posts to help me with questions I'm testing myself on. Then I got this sucker:
    -(7-x)+5=x+7 had me going for a bit and then I checked the answer given...or lack of it.
  12. Aug 8, 2013 #11


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    Can you get rid of the brackets in the expression -(7-x) ?
  13. Aug 8, 2013 #12
    Yes, I spent some time on this equation and when I eventually looked at the answer given and method to find out where I was going wrong and there I found out there isn't a correct answer for 'x'(or so the site says: http://www.intmath.com/basic-algebra/4-solving-equations.php (example 6)). When in doubt I also use this site for help and it found similar issues: http://www.algebrahelp.com/calculators/equation/calc.do?equation=-(7-x)+5=x+7&solvf=AUTO

    I don't have a problem or anything like that, I was just saying that after some time and times when I thought something wasn't quite right it turn't out the equation was a curve ball :).
  14. Aug 8, 2013 #13


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    Left side member shows THE OPPOSITE OF {7-x}, plus 5.
    The left side member then is -7-(-x), plus 5;
    which is -7+x+5.

    Revising the equation accordingly, -7+x+5=x+7.

    Knowing that to solve for x, you need all terms of x on one side of the equation and all constants expression on the other side of the equation. What will happen in this equation? What is the meaning?
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