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Thread: Rearranging equations with multiple fractions

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    Rearranging equations with multiple fractions

    Hi,

    I'm currently studying an engineering degree with the Open University, but have come across this question in one of my revision quizzes which I have no idea how to work out. The way they told me I should have worked it out also makes no sense to me, so I'm hoping someone could talk me through it.

    I've posted the same post on the Open University forums, but had no responses. I've taken a screenshot of it and included it here, as i think the equations are a bit easier to read in the screenshot.

    Thanks in advance!

    Rearranging equations with multiple fractions-math-problem.png
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    Re: Rearranging equations with multiple fractions

    Quote Originally Posted by chrisidas View Post
    I've posted the same post on the Open University forums, but had no responses. I've taken a screenshot of it and included it here, as i think the equations are a bit easier to read in the screenshot. Click image for larger version. 

Name:	math problem.png 
Views:	19 
Size:	22.6 KB 
ID:	37188
    I say that you or whomsoever are taking too many steps to fast.
    The least common denominator is $8(10q-8)$. Take one step at a time.
    $7\cdot 8\cdot(10q-8)-9\cdot(10q-8)=7\cdot 8\cdot(10q-8)y-17\cdot 2\cdot(10q-8)$
    Now go from there.
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    Re: Rearranging equations with multiple fractions

    Forget what "the answer says";
    rearrange equation this way:

    9y/(10q-8) - 7y = 9/8 - 17/4

    How would YOU simplify that?
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    Re: Rearranging equations with multiple fractions

    Thanks for the reply. Yeah, i think they skipped a few steps, and just show you various stages, but the jump was two far for me to follow.

    I'm struggling to follow your answer a bit also, sorry! Not sure if it's because I'm new on here and don't understand the layout fully.

    What do the full stops (periods) represent.
    Where does the 7 at the start of your equation come from?
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    Re: Rearranging equations with multiple fractions

    Quote Originally Posted by DenisB View Post
    Forget what "the answer says";
    rearrange equation this way:

    9y/(10q-8) - 7y = 9/8 - 17/4

    How would YOU simplify that?
    Thanks!

    Any chance you could give me a step by step breakdown of how you'd rearrange it to get Q as the subject?
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    Re: Rearranging equations with multiple fractions

    Quote Originally Posted by chrisidas View Post
    9y/(10q-8) - 7y = 9/8 - 17/4

    Any chance you could give me a step by step breakdown
    of how you'd rearrange it to get Q as the subject?
    Let x = 10q - 8
    9/8 - 17/4 = -25/8

    9y/x = 7y - 25/8

    Multiply by 8:
    72y/x = 56y - 25
    x = 72y/(56y - 25)

    Substitute back in:
    10q - 8 = 72y/(56y - 25)
    10q = 72y/(56y - 25) + 8
    q = [72y/(56y - 25) + 8] / 10

    Best I can do without a blackboard!!
    Last edited by topsquark; Mar 14th 2017 at 05:26 PM.
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    Re: Rearranging equations with multiple fractions

    Quote Originally Posted by chrisidas View Post
    Hi,

    I'm currently studying an engineering degree with the Open University, but have come across this question in one of my revision quizzes which I have no idea how to work out. .
    In the box to be magnified for us to read, it states nothing in the instructions or (partial answer) about solving for a particular variable.It doesn't make sense that there are two
    variables in there.

    So then how would you know which variable to solve for!? It would make (more) sense to me that there is a typo. Either all variables are to be q's or all variables are to be y's
    is what I would lean toward. Then you know that A) you could solve for a particular value of the one variable, or B) perhaps there would be no solution. At this point, I claim
    the statement of the problem as you presented is incomplete and/or flawed.
    Last edited by greg1313; Mar 14th 2017 at 03:50 PM.
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    Re: Rearranging equations with multiple fractions

    Seems like a reasonable question to me. see here
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    Re: Rearranging equations with multiple fractions

    Quote Originally Posted by chrisidas View Post
    Hi,

    I'm currently studying an engineering degree with the Open University, but have come across this question in one of my revision quizzes which I have no idea how to work out. The way they told me I should have worked it out also makes no sense to me, so I'm hoping someone could talk me through it.

    I've posted the same post on the Open University forums, but had no responses. I've taken a screenshot of it and included it here, as i think the equations are a bit easier to read in the screenshot.

    Thanks in advance!

    Click image for larger version. 

Name:	math problem.png 
Views:	19 
Size:	22.6 KB 
ID:	37188
    Let's start with two really basic ideas in algebra.

    a + b = c + d means that two unknown numbers have a sum equal to the sum of another two unknown numbers. In other words, the two sides of the equation represent the same number.

    Now if I multiply (a + b) by e and e is any number except 1, the number represented by e(a + b) is different from a + b, which is the same as c + d so it is absolutely FALSE that e(a + b) = c + d. To maintain equality, I must multiply both sides of the equation by e.

    In short $a + b = c + d \implies e(a + b) = e(c + d).$ First week of algebra. As I said, truly basic stuff.

    Now fractions are a nuisance, and it is frequently useful to get rid of them as quickly as possible. To do that we can use this fundamental propert of numbers

    $b \ne 0 \implies b * \dfrac{a}{b} = a.$

    We actually learn that in arithmetic: $3 * \dfrac{2}{3} = \dfrac{\cancel 3}{1} * \dfrac{2}{\cancel 3} = 2.$

    So the way to get rid of a fraction in an equation is to multiply both sides of the equation by the denominator of that annoying fraction.

    Let's go.

    $\dfrac{9y}{10q - 8} - \dfrac{9}{8} = 7y - \dfrac{17}{4}.$

    To get rid of the fraction $\dfrac{17}{4},$

    I multiply BOTH SIDES of the equation by the relevant fraction's denominator, which is 4.

    $4 * \left (\dfrac{9y}{10q - 8} - \dfrac{9}{8} \right ) = 4 * \left ( 7y - \dfrac{17}{4} \right ) \implies$

    $\dfrac{36y}{10q - 8} - \dfrac{36}{8} = 28y - \dfrac{\cancel 4 * 17}{\cancel 4} \implies \dfrac{36y}{10q - 8} - \dfrac{36}{8} = 28y - 17.$

    All the fractions are gone from the right hand side of the equation. Now let's get rid of (36 / 8). The denominator is 8 so we multiply both sides of the equation by 8.

    $8 * \left (\dfrac{36y}{10q - 8} - \dfrac{36}{8} \right ) = 8 * (28y - 17) \implies$

    $\dfrac{288y}{10q - 8} - \dfrac{\cancel 8 * 36}{\cancel 8} = 224y - 136 \implies \dfrac{288y}{10q - 8} - 36 = 224y - 136.$

    We have one fraction left, namely $\dfrac{288y}{10q - 8}$, which has a denominator of $10q - 8.$

    So we multiply both sides of the equation by that denominator just as we did when the denominators were expressed as numerals.

    $(10q - 8) * \left ( \dfrac{288y}{10q - 8} - 36 \right ) = (10q - 8) * (224y - 136) \implies$

    $\dfrac{\cancel {(10q - 8)} * 288y}{\cancel {10q - 8}} - 36(10q - 8) = 224y(10q - 8) - 136(10q - 8) \implies$

    $288y - 360q + 288 = 2240qy - 1792y - 1360q + 1088.$

    Now you have no fractions to deal with. But you complain that equation is not what your explanation gave. It turns out, however, that we can factor both sides of the equation by 8 and divide both sides by 8 as follows.

    $8 * (36y - 45q + 36) = 8(280qy - 224y - 170q + 136) \implies$

    $\dfrac{\cancel 8 * (36y - 45q + 36)}{\cancel 8} = \dfrac{\cancel 8(280qy - 224y - 170q + 136) }{\cancel 8} \implies$

    $36y - 45q + 36 = 280qy - 224y - 170q + 136.$

    That is very tedious. A somewhat less tedious way is to find the least common multiple of all the fractions in the equation and multiply by that least common multiple. That gets rid of all the fractions in one go.

    Now obviously the least common multiple of 4 and 8 is 8. And the least common multiple of 8 and (10q - 8) is their product of 8(10q - 8).

    $\dfrac{9y}{10q - 8} - \dfrac{9}{8} = 7y - \dfrac{17}{4} \implies$

    $8(10q - 8) * \left (\dfrac{9y}{10q - 8} - \dfrac{9}{8} \right ) = 8(10q * 8) * \left ( 7y - \dfrac{17}{4} \right ) \implies$

    $\dfrac{8 \cancel {(10q - 8)} * 9y}{\cancel {10q - 8}} - \dfrac{\cancel 8(10q - 8)9}{\cancel 8} = 8(10q - 8)(7y) - \dfrac{8(10q - 8)(17)}{4} \implies$

    $72y - 9(10q - 8) = 56y(10q - 8) - \dfrac{\cancel 4 * 2 * (10q - 8) * 17}{\cancel 4} \implies$

    $72y - 90q + 72 = 560qy - 448y - 34(10q - 8) \implies$

    $72y - 90q + 72 = 560qy - 448y - 340q + 272.$

    It is obvious that both sides of the equation can be divided by 2 because every numeral is even giving

    $36y - 45q + 36 = 280qy - 224y - 170q + 186.$

    And we reach the same result.

    Sorry to take baby steps all the way, but skipping steps sometimes leads to confusion.
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    Re: Rearranging equations with multiple fractions

    Quote Originally Posted by JeffM View Post
    $36y - 45q + 36 = 280qy - 224y - 170q + 186.$
    Last amount should be 136.

    Solving for q (expanding my previous results):
    q = 4(13y - 5) / (56y - 25) where y <> 25/56
    Thanks from JeffM
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    Re: Rearranging equations with multiple fractions

    Quote Originally Posted by greg1313 View Post
    In the box to be magnified for us to read, it states nothing in the instructions or (partial answer) about solving for a particular variable.It doesn't make sense that there are two
    variables in there.
    Don't think that matters, Greg.
    Chris is apparently trying to learn how to handle
    equations with fractions.
    Equation could be anything; like:
    a/(b-c) - d/e = f - g/h

    Pick a variable and isolate the sucker
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    Re: Rearranging equations with multiple fractions

    Quote Originally Posted by JeffM View Post
    Let's start with two really basic ideas in algebra.

    a + b = c + d means that two unknown numbers have a sum equal to the sum of another two unknown numbers. In other words, the two sides of the equation represent the same number.

    Now if I multiply (a + b) by e and e is any number except 1, the number represented by e(a + b) is different from a + b, which is the same as c + d so it is absolutely FALSE that e(a + b) = c + d. To maintain equality, I must multiply both sides of the equation by e.

    In short $a + b = c + d \implies e(a + b) = e(c + d).$ First week of algebra. As I said, truly basic stuff.

    Now fractions are a nuisance, and it is frequently useful to get rid of them as quickly as possible. To do that we can use this fundamental propert of numbers

    $b \ne 0 \implies b * \dfrac{a}{b} = a.$

    We actually learn that in arithmetic: $3 * \dfrac{2}{3} = \dfrac{\cancel 3}{1} * \dfrac{2}{\cancel 3} = 2.$

    So the way to get rid of a fraction in an equation is to multiply both sides of the equation by the denominator of that annoying fraction.

    Let's go.

    $\dfrac{9y}{10q - 8} - \dfrac{9}{8} = 7y - \dfrac{17}{4}.$

    To get rid of the fraction $\dfrac{17}{4},$

    I multiply BOTH SIDES of the equation by the relevant fraction's denominator, which is 4.

    $4 * \left (\dfrac{9y}{10q - 8} - \dfrac{9}{8} \right ) = 4 * \left ( 7y - \dfrac{17}{4} \right ) \implies$

    $\dfrac{36y}{10q - 8} - \dfrac{36}{8} = 28y - \dfrac{\cancel 4 * 17}{\cancel 4} \implies \dfrac{36y}{10q - 8} - \dfrac{36}{8} = 28y - 17.$

    All the fractions are gone from the right hand side of the equation. Now let's get rid of (36 / 8). The denominator is 8 so we multiply both sides of the equation by 8.

    $8 * \left (\dfrac{36y}{10q - 8} - \dfrac{36}{8} \right ) = 8 * (28y - 17) \implies$

    $\dfrac{288y}{10q - 8} - \dfrac{\cancel 8 * 36}{\cancel 8} = 224y - 136 \implies \dfrac{288y}{10q - 8} - 36 = 224y - 136.$

    We have one fraction left, namely $\dfrac{288y}{10q - 8}$, which has a denominator of $10q - 8.$

    So we multiply both sides of the equation by that denominator just as we did when the denominators were expressed as numerals.

    $(10q - 8) * \left ( \dfrac{288y}{10q - 8} - 36 \right ) = (10q - 8) * (224y - 136) \implies$

    $\dfrac{\cancel {(10q - 8)} * 288y}{\cancel {10q - 8}} - 36(10q - 8) = 224y(10q - 8) - 136(10q - 8) \implies$

    $288y - 360q + 288 = 2240qy - 1792y - 1360q + 1088.$

    Now you have no fractions to deal with. But you complain that equation is not what your explanation gave. It turns out, however, that we can factor both sides of the equation by 8 and divide both sides by 8 as follows.

    $8 * (36y - 45q + 36) = 8(280qy - 224y - 170q + 136) \implies$

    $\dfrac{\cancel 8 * (36y - 45q + 36)}{\cancel 8} = \dfrac{\cancel 8(280qy - 224y - 170q + 136) }{\cancel 8} \implies$

    $36y - 45q + 36 = 280qy - 224y - 170q + 136.$

    That is very tedious. A somewhat less tedious way is to find the least common multiple of all the fractions in the equation and multiply by that least common multiple. That gets rid of all the fractions in one go.

    Now obviously the least common multiple of 4 and 8 is 8. And the least common multiple of 8 and (10q - 8) is their product of 8(10q - 8).

    $\dfrac{9y}{10q - 8} - \dfrac{9}{8} = 7y - \dfrac{17}{4} \implies$

    $8(10q - 8) * \left (\dfrac{9y}{10q - 8} - \dfrac{9}{8} \right ) = 8(10q * 8) * \left ( 7y - \dfrac{17}{4} \right ) \implies$

    $\dfrac{8 \cancel {(10q - 8)} * 9y}{\cancel {10q - 8}} - \dfrac{\cancel 8(10q - 8)9}{\cancel 8} = 8(10q - 8)(7y) - \dfrac{8(10q - 8)(17)}{4} \implies$

    $72y - 9(10q - 8) = 56y(10q - 8) - \dfrac{\cancel 4 * 2 * (10q - 8) * 17}{\cancel 4} \implies$

    $72y - 90q + 72 = 560qy - 448y - 34(10q - 8) \implies$

    $72y - 90q + 72 = 560qy - 448y - 340q + 272.$

    It is obvious that both sides of the equation can be divided by 2 because every numeral is even giving

    $36y - 45q + 36 = 280qy - 224y - 170q + 186.$

    And we reach the same result.

    Sorry to take baby steps all the way, but skipping steps sometimes leads to confusion.
    Thanks, Massive help! I was struggling with simplifying the fractions, which was causing the problem. This makes much more sense now.

    Thank you to everyone else who also contributed, much appreciated.

    For info, the question was actually to make q the subject of the initial equation (not to actually solve the equation), which i have now been able to do.
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    Re: Rearranging equations with multiple fractions

    For info, the question was actually to make q the subject of the initial equation (not to actually solve the equation), which i have now been able to do.
    ... you casually ask to solve for $q$ in post #5, and then wait out multiple responses to say it was actually the primary goal of this problem?

    If you plan on making future posts on this site, it would be preferable if you post the entire problem with complete directions from the start. Thank you for your cooperation.
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    Re: Rearranging equations with multiple fractions

    Quote Originally Posted by skeeter View Post
    ... you casually ask to solve for $q$ in post #5, and then wait out multiple responses to say it was actually the primary goal of this problem?

    If you plan on making future posts on this site, it would be preferable if you post the entire problem with complete directions from the start. Thank you for your cooperation.
    Yeah, sorry. I was in a bit of a mess with it and didn't realise I didn't include it in the initial question!
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    Re: Rearranging equations with multiple fractions

    Quote Originally Posted by chrisidas View Post

    For info, the question was actually to make q the subject of the initial equation (not to actually solve the equation), which i have now been able to do.
    Hence my post # 7!
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