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Substitution, algebra.

  1. Dec 9, 2012 #1
    1. The problem statement, all variables and given/known data

    5a = 5 - b
    5a = 3 - b



    2. Relevant equations



    3. The attempt at a solution

    I got the solution set to be 1/2, 5/2

    i used substitution for substituting a into b of the second equation, just like they were x's and y's just used a's and b's there is no difference correct?

    a = 1- b/5 from top equation.

    then 5(1-b/5) = 3 - b

    ==5/2

    then for 1/2 I substituted the b= 5/2 back into one of the original equations.
    Does this look correct? Thanks alot.
     
  2. jcsd
  3. Dec 9, 2012 #2

    I like Serena

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    Hi rcmango!

    You had 5(1-b/5) = 3 - b
    If you get rid of the parentheses, you get 5 - b = 3 - b
    Can you solve b from that expression?
    Or what else can you deduce from that?
     
  4. Dec 9, 2012 #3
    Please allow me to translate these equations into a "word problem":

    Five apples cost $5 minus a certain bit of change.
    Five apples cost $3 minus the same bit.

    Does this look as if there could be any fixed price for apples that would make sense in both statements?
     
  5. Dec 9, 2012 #4

    symbolipoint

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    First equation solved for b: b=5-5a
    Substitute into second equation: 5a=3-b,
    5a=3-(5-5a)
    5a=3-5+5a
    add -5a to both sides: 0=-2
    Incorrect statement. No solution.

    You can try a different way to what happens.
     
  6. Dec 9, 2012 #5
    I see now, there are no solutions, anyway you work it out. Thankyou.
     
  7. Dec 9, 2012 #6

    symbolipoint

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    Assuming you are in Introductory Algebra like in high school, another thing you can try is to solve each equation for b. Look at the slope, as if a is the horizontal axis and b is the vertical axis. If the slopes are equal, the lines for each equation are parallel and therefore will not intersect, meaning no point in common, meaning the system of equations has no solution.

    b=-5a+5
    b=-5a+3

    Slope of both equations is -5, so the lines are parallel.
     
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