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Thread: Algebra I question.

  1. #1
    Join Date
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    Default Algebra I question.

    Chapter 13, of lesson 5-Question 15.
    We cannot seem to get the correct answer. We are doing the steps, but the book does not explain thoroughly. We've listened to the streaming for this lesson.
    -Amy

    Nine children (four sons five daughters) Five graduated, five married, ten grand babies with 4 currently on the way.
    2017-18 MP K, 3A, 8M, 12th eclectic and MPOA. This fall will start our fifth year using MP.

  2. #2
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    Default Re: Algebra I question.

    Quote Originally Posted by Maggie View Post
    Chapter 13, of lesson 5-Question 15.
    We cannot seem to get the correct answer. We are doing the steps, but the book does not explain thoroughly. We've listened to the streaming for this lesson.
    Hi Amy,
    You are asking about 13.5, #15? To solve this rational equation, you will want to multiply by the least common denominator (of all the denominators). To determine LCD, factor the 2nd denominator: (t-3)(t+3). Since the 1st denominator is (t-3), the LCD is (t-3)(t+3). Multiply both sides of the equation by LCD, which yields:
    5(t+3)-30 = (t-3)(t+3)
    Now multiply out, combine like terms and simplify, putting all non-zero terms on left-hand side:
    t2-5t+6 = 0
    Factor and find two possible solutions for t: t=2 and t=3. Since t=3 would cause original equation to have denominator of zero (cannot divide by zero), the only solution to this equation is t=2. Check by substituting.
    Hope this helps!
    Last edited by Cindy in Indy; 09-26-2017 at 05:57 PM.
    Cindy Davis
    Science and Math teacher at Highlands Latin School - Indianapolis
    ds-23 college graduate: working, reading, writing
    ds-21 college senior: Biology/German double-major, applying to med school
    dd-19 college sophomore: Nursing

  3. #3
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    Default Re: Algebra I question.

    Quote Originally Posted by Cindy in Indy View Post
    Hi Amy,
    You are asking about 13.5, #15? To solve this rational equation, you will want to multiply by the least common denominator (of all the denominators). To determine LCD, factor the 2nd denominator: (t-3)(t+3). Since the 1st denominator is (t-3), the LCD is (t-3)(t+3). Multiply both sides of the equation by LCD, which yields:
    5(t+3)-30 = (t-3)(t+3)
    Now multiply out, combine like terms and simplify, putting all non-zero terms on left-hand side:
    t2-5t+6 = 0
    Factor and find two possible solutions for t: t=2 and t=3. Since t=3 would cause original equation to have denominator of zero (cannot divide by zero), the only solution to this equation is t=2. Check by substituting.
    Hope this helps!
    Thank you so much! It helped her, got another question for you.
    Lesson 13-5, question 31. I got the answer 10,-2 and then I used the quadratic formula and still turned out with 10,-2. I looked up the answer and it was 0.5. Would love any help on this! This is Mercy, Amy's daughter.
    -Amy

    Nine children (four sons five daughters) Five graduated, five married, ten grand babies with 4 currently on the way.
    2017-18 MP K, 3A, 8M, 12th eclectic and MPOA. This fall will start our fifth year using MP.

  4. #4
    Join Date
    Dec 2007
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    295

    Default Re: Algebra I question.

    For 13.5, #31, you should multiply both sides by the LCD, which is (x-3)(x+3).
    When you multiply both sides by that LCD, you get 2(x-3)+5=0.
    This is a linear equation (does not need the quadratic formula to solve) and can be solved like any linear equation:
    2x-6+5=0 --> 2x=1 --> x = 1/2
    This satisfies the original equation, so it is a valid solution.
    Last edited by Cindy in Indy; 10-04-2017 at 08:26 PM.
    Cindy Davis
    Science and Math teacher at Highlands Latin School - Indianapolis
    ds-23 college graduate: working, reading, writing
    ds-21 college senior: Biology/German double-major, applying to med school
    dd-19 college sophomore: Nursing

  5. #5
    Join Date
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    Default Re: Algebra I question.

    Quote Originally Posted by Cindy in Indy View Post
    For 13.5, #31, you should multiply both sides by the LCD, which is (x-3)(x+3).
    When you multiply both sides by that LCD, you get 2(x-3)+5=0.
    This is a linear equation (does not need the quadratic formula to solve) and can be solved like any linear equation:
    2x-6+5=0 --> 2x=1 --> x = 1/2
    This satisfies the original equation, so it is a valid solution.

    Thank you very much! I realized what I did wrong!
    -Amy

    Nine children (four sons five daughters) Five graduated, five married, ten grand babies with 4 currently on the way.
    2017-18 MP K, 3A, 8M, 12th eclectic and MPOA. This fall will start our fifth year using MP.

  6. #6
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    KS
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    Default Re: Algebra I question.

    Cindy, thank you for helping Mercy! I am so far gone from Algebra these days. All extra brain cells seem to be focused on Latin . My college niece and husband both were failing as well.

    Be blessed busy math teacher extrodinaire!
    -Amy

    Nine children (four sons five daughters) Five graduated, five married, ten grand babies with 4 currently on the way.
    2017-18 MP K, 3A, 8M, 12th eclectic and MPOA. This fall will start our fifth year using MP.

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