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$y = 4y + 9$

How do I isolate y?

Can I do

$y = 4y + 9$

$\frac{y}{4y} = 9$ etc

Also some other questions please:

  1. $\frac{5x + 1}{3} - 4 = 5 - 7x$

In the above (1), if I want to remove the '3' from the denominator of the LHS, do I multiply everything on the RHS by 3? What about the '-4' on the LHS, do I do anything to that?

  1. In the above

$\frac{y}{4y}$

If I want to solve it, do I cross a y out from the top, or from the bottom? Does that make sense ? :\

Thanks in advance..I have an exam tomorrow..

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  • $\begingroup$ Hint: There is no reason to divide. Which operation do you have to do instead if you want to bring $4y$ to the left. $\endgroup$ – Ruts Oct 20 '15 at 12:23
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    $\begingroup$ There are answers that discuss how to correctly solve the problem, but it's important to know why what you've done doesn't work. You moved the $4y$ around, but it's unclear how you moved it. As others said, the proper strategy is to subtract $4y$ from both sides, but if you had divided by $4y$, you would have gotten $\frac{y}{4y}=\frac{4y+9}{4y}=1+\frac{9}{4y}$. So if you want to get a $4y$ in the denominator, the right hand side is not $9$, but something more complicated. When dealing with equations, you shouldn't think about "moving things around" because not all move make sense. $\endgroup$ – Aaron Oct 20 '15 at 12:48
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    $\begingroup$ (continued) Instead, you should think about doing things to both sides of the equation, and picking the things to do that simplify things as much as possible. $\endgroup$ – Aaron Oct 20 '15 at 12:50
  • $\begingroup$ @Aaron - yes, agreed. At this point, though, it might be better to avoid a string of equalities (to go along with the notion of a scale/balance introduced in some of the answers) $\endgroup$ – The Chaz 2.0 Oct 20 '15 at 13:07
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    $\begingroup$ @TheChaz2.0 Yes, I only put the string in there because comments don't leave a lot of room. I just wanted to convey the idea that people often present dealing with equations as moving things around, and that language is dangerous until you have internalized what kinds of moves are legal. I've seen plenty of calculus students who only manipulated equations in informal ways in order to get utter nonsense, so I feel passionate that one should avoid the standard informalities and metaphors and idioms until after the ground rules have been clearly set. $\endgroup$ – Aaron Oct 20 '15 at 13:22
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To make $y$ the subject or isolate $y$ try to think of equations as 'weighing scales' you can carry out any operation you like but you must do the same to both sides (to keep it balanced): $$y = 4y + 9$$ $$\implies y-4y=9+4y-4y\tag{1}$$ $$\implies-3y=9$$ $$\implies \frac{-3y}{-3}=\frac{9}{-3}\tag{2}$$ $$\implies y =-3$$ For $(1)$ I subtracted $4y$ from both sides.

For $(2)$ I divided both sides by $-3$.

To move on $$\cfrac{5x+1}{3}-4=5-7x$$ $$\implies \cfrac{5x+1}{3}=9-7x$$ $$\implies 5x+1=3\times(9-7x)$$ $$\implies 5x+1=27-21x$$ $$\implies 26x+1=27$$ $$\implies 26x=27-1$$ $$\implies 26x=26$$ $$\implies x=1$$

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    $\begingroup$ I understand, can you show me the next problem please $\endgroup$ – solarsaler Oct 20 '15 at 12:42
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    $\begingroup$ thank you sir for your time $\endgroup$ – solarsaler Oct 20 '15 at 13:48
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Others have already given you the answer, so I won't reiterate.

Instead let me point out to you how you should think about such questions. It appears to me that you have no firm grasp of what equations are and how to manipulate them. This is not a shame, it's not easy to learn and also not easy to teach.

You're approaching the problem with the mindset "here is the equation, what technique do I need to use?". People trained in mathematics use techniques without thinking much about them, because it has become second nature. However, when you learn about the techniques you should never try to blindly apply techniques without understanding. At best, this leads to a lot of mental overhead because you have to memorize seemingly "random" facts. At worst, it leads to general confusion and to you applying methods when they seem to work but actually don't.

Think of the equation as a scale. Everything you do to one side, you also want to do to the other side. If you compute $\frac{y}{4y}$, you effectively divide the left hand side by $4y$. But does that match up what you're doing to the right hand side? Assuming you have a firm grasp of how to manipulate fractions, you should see that this is not the case.

I hope this helps and doesn't come across as too condescending, I could be judging your level of knowledge wrong.

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  • $\begingroup$ I did mention the scale analogy, but your explanation is good so (+1) $\endgroup$ – BLAZE Oct 20 '15 at 13:00
  • $\begingroup$ @Fryie thank you, I understand what you say and it reinforces the importance of understanding in mathematics thank you sir $\endgroup$ – solarsaler Oct 20 '15 at 13:48
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You have the right idea to get $y$ by itself on the left. However, instead of dividing by $4y$, you would subtract $4y$ from both sides. This will give you

$$y - 4y = 9.$$

Now, if you combine the like $y$-terms on the left. What would $y - 4y$ turn into? You can think of this as: $1y - 4y$.

For your next problem, if you want to multiply both sides by $3$, then you multiply everything by $3$, including the $-4$ on the left-hand side.

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    $\begingroup$ thank you sir for your time $\endgroup$ – solarsaler Oct 20 '15 at 13:48

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