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Math Help - Finding a value for x

  1. #1
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    Finding a value for x

    Just seem to be struggling to understand what I am missing here?

    I have;

    x - 8 = 3x + 20
    7

    This is what I have done to try and find the value for x

    7(x - 8) = 7(3x + 20)
    7

    56 - 7x = 3x + 20

    -7x -3x = 20 - 56

    -10x = -36
    -10 -10

    x = 3.6

    Substitute value into x

    3.6 - 8 = 3 x 3.6 + 20
    7

    -4.4 = 4.4

    My problem to me is that -4.4 is not equal to 4.4?

    The equation should not have two routes?

    Please advise if there is an alterative technique to solve this type of problem?

    P.S. the 7 above (denominator) should be divided by the numberators on the RHS of the equals sign. This software program seens to move the integer to the LHS
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  2. #2
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    Quote Originally Posted by David Green View Post
    7(x - 8) = 7(3x + 20)
    7

    56 - 7x = 3x + 20
    Why did the '7' magically turn negative on the left-hand side?
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    Why did the '7' magically turn negative on the left-hand side?
    Thank you for your help, but there must be another method to solve this equation which I am not aware of?

    Please advise if you can?
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  4. #4
    Super Member Quacky's Avatar
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    The basic principle is there, the algebraic manipulation is tripping you up.

    Is this your original equation?

    \frac{x-8}{7}=3x+20

    This is what it looks like to me. I assume (from your working) that you meant to present:

    x-8=\frac{3x+20}{7}

    Which of these represents your question?
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  5. #5
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    Quote Originally Posted by Quacky View Post
    The basic principle is there, the algebraic manipulation is tripping you up.

    Is this your original equation?

    \frac{x-8}{7}=3x+20

    This is what it looks like to me. I assume (from your working) that you meant to present:

    x-8=\frac{3x+20}{7}

    Which of these represents your question?
    Sorry the equation you wrote is the wrong way round. The way I presented it was incorrectly positioned on this forum because the forum moves all intergers placed as denominators to the far left hand side for some reason?

    I'll do it again;

    x - 8 is on the left hand side and 3x + 20 divided by 7 is on the right hand side.

    so x - 8 = 3x + 20 / 7.

    multiplying both sides of the equation by 7 gives;

    7(x - 8) = 3x + 20 multiply out the LHS.

    7x - 56 = 3x + 20 Move the -56 to the RHS
    7x = 3x + 20 + 56 Now move the 3x to the LHS
    7x - 3x = 76 Subtract LHS
    4x = 76 Divide both sides by 4
    x = 19

    The original equation; x - 8 = 3x + 20 / 7

    Putting in the x value; 19 - 8 = 3 \cdot x + 20 / 7

    the like values are; 11 = 11

    Whatever the reason I couldn't see it before is just a state of mind?

    Thanks for your help.
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  6. #6
    Super Member Quacky's Avatar
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    Quote Originally Posted by David Green View Post
    Sorry the equation you wrote is the wrong way round. The way I presented it was incorrectly positioned on this forum because the forum moves all intergers placed as denominators to the far left hand side for some reason?

    I'll do it again;

    x - 8 is on the left hand side and 3x + 20 divided by 7 is on the right hand side.

    so x - 8 = 3x + 20 / 7.

    multiplying both sides of the equation by 7 gives;

    7(x - 8) = 3x + 20 multiply out the LHS.

    7x - 56 = 3x + 20 Move the -56 to the RHS
    7x = 3x + 20 + 56 Now move the 3x to the LHS
    7x - 3x = 76 Subtract LHS
    4x = 76 Divide both sides by 4
    x = 19

    The original equation; x - 8 = 3x + 20 / 7

    Putting in the x value; 19 - 8 = 3 \cdot x + 20 / 7

    the like values are; 11 = 11

    Whatever the reason I couldn't see it before is just a state of mind?

    Thanks for your help.
    Originally, you had made a significant error with your multiplication - this just goes to show that sometimes, the only help you need is a fresh sheet of paper and 20 minutes of air
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  7. #7
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    That was a great explanation
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