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Math Help - Grade 8 Algebra

  1. #1
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    Grade 8 Algebra

    Hii!

    I have this other piece of practice work that I don't understand.

    The method I do understand is for calculating the final amount:

    The amount to start with is 21 860
    The interest is 8.5 %
    The years are 7

    I know that you must do:

    P+ (i . P . n )
    = 21 860+ (0.085 . 21 860 . 7)

    That's easy.

    What I'm stuck on is:

    Final amount is 8 775
    Interest is 7%
    Years are 5
    CALCULATE THE AMOUNT YOU STARTED WITH.

    What would I do?

    Thanks!

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  2. #2
    Super Member Quacky's Avatar
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    Re: Grade 8 Algebra

    Go backwards. P+i\cdot{P}\cdot{n}=F, where F is the final amount.

    Substitute the known values into the equation and try to solve for the unknown.
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  3. #3
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    Re: Grade 8 Algebra

    Quote Originally Posted by Quacky View Post
    Go backwards. P+i\cdot{P}\cdot{n}=F, where F is the final amount.

    Substitute the known values into the equation and try to solve for the unknown.
    I don't quite understand.

    If the final amount is 8775 ( A)
    Interest is 0.07 ( i)
    And years 5 ( n)

    How would I substitute?

    Would I go:
    8775- (0.07/ 8775 . 5)
    = (amount we started with which is P)

    Like that?
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  4. #4
    Super Member Quacky's Avatar
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    Re: Grade 8 Algebra

    P+i\cdot{P}\cdot{n}=A

    We have:
    i=0.07
    n=5
    A=8775

    So:

    P+0.07\cdot{P}\cdot{5}=8775

    So:

    P+0.35\cdot{P}=8775

    1.35P=8775
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  5. #5
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    Re: Grade 8 Algebra

    So 1.35 is the amount we started with?
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  6. #6
    Super Member Quacky's Avatar
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    Re: Grade 8 Algebra

    No, we have to solve this equation for P, by dividing both sides by 1.35
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  7. #7
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    Re: Grade 8 Algebra

    But we haven't solved P.

    We have A and i and n
    We are trying to solve P.
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  8. #8
    Super Member Quacky's Avatar
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    Re: Grade 8 Algebra

    Quote Originally Posted by JesseElFantasma View Post
    But we haven't solved P.

    We have A and i and n
    We are trying to solve P.
    I agree. In order to solve an equation for P, you need to isolate P (get P by itself). This is what I've done here.

    Quote Originally Posted by Quacky View Post
    P+i\cdot{P}\cdot{n}=A

    We have:
    i=0.07
    n=5
    A=8775

    So:

    P+0.07\cdot{P}\cdot{5}=8775 Here I've just substituted the known values into the formula.

    So:

    P+0.35\cdot{P}=8775 I've simplified, but otherwise haven't changed anything.

    1.35P=8775 I've collected like terms together. In the step above, I had 1P and 0.35P, so these combine to make 1.35P
    Then, if we divide both sides by 1.35, we get:

    P=\frac{8775}{1.35}=6500
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  9. #9
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    Re: Grade 8 Algebra

    But in
    P + 0.35 x P = 8775
    what happens to the 1st P?
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  10. #10
    Super Member Quacky's Avatar
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    Re: Grade 8 Algebra

    The first is 1\times{P}
    The second is 0.35\times{P}

    Because they are both amounts of P that are being added together, we can combine them as 1.35P in the same way that 3+3+3+3= 4\times{3}
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  11. #11
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    Re: Grade 8 Algebra

    I get it!
    Thank you.

    So for another I would go:

    P + 0.085 . P.7
    P + 0.595 x P
    P + 1.595
    1.595P = 34866.70
    34866.70/ 1.595P
    = 21860
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  12. #12
    Super Member Quacky's Avatar
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    Re: Grade 8 Algebra

    If 34866.70 is your final amount, and the question is asking you to solve for P, then yes - although there are some very tiny notation errors, the method is fine. I'd set it out like this:

    P + 0.085 \cdot{P}\cdot{7}=34866.70

    P + 0.595 \times P=34866.70

    1.595P = 34866.70

    34866.70/ 1.595=P

    = 21860
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