How do I solve this algebra equation given in an economics class?

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To solve the equation q = 120 - (q/2), the algebraic manipulation involves multiplying both sides by 2 to eliminate the fraction, resulting in 2q = 240 - q. Adding q to both sides gives 3q = 240, leading to q = 80. The discussion also highlights a potential typo in the textbook's equation Y = (a - bY)/2b, which simplifies to Y = a/3b. The user expresses confusion over the relationship between the two equations but ultimately seeks clarity on solving the algebraic expression correctly. Understanding the algebraic steps is crucial for solving similar equations in economics.
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Homework Statement



I have to solve this algebra equation for an economics class. The equation I must solve is q=120-(q/2). I could solve this by guessing but I need to figure out the algebra behind it.

Homework Equations



The book gives this equation. Y=(a-bY)/2b and solves it for Y, getting Y=a/3b. I have tried to solve this and never do I end up with a/3b. Also, if you plug any numbers into these two equations, they don't match. I'm beginning to think there is a typo in this book.


The Attempt at a Solution



I have no idea how to solve for this type of equation.

Honestly I have no idea. If I take a number, s
 
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Accidently close the window before I finished my attempted solution. I can see how it is solved by simply guessing and checking, but not using actual algebra. If I put q=120-(q/2) into the form in the book, it is q=(240-q)/2. Here I can see that the answer is 2/3 of 240/2(1). In other words it is 2/3*a/2b which is = to a/3b. I just need to know how to solve this using purely algebra. Thanks for any help.
 
Y = (a-bY)/2b

Multiply both sides by 2b. On the right, this cancels the 1/2b, and so we get

2bY = a - bY

Add bY to both sides. On the right, this cancels the -bY, and so we get

2bY + bY = a

Now notice that two of anything plus one of the same thing equals three of that thing.
In other words, the left side equals 3bY. So we now have

3bY = a

Now, divide both sides by 3b. On the left, this cancels the factor of 3b, and we get

Y = a/3b

Voila!
 
\[<br /> \begin{array}{l}<br /> {\rm The algebra is fairly simple, especially if you studied the course(s)}{\rm .} \\ <br /> q = 120 - (\frac{q}{2}) \\ <br /> 2q = 240 - q \\ <br /> 3q = 240 \\ <br /> q = \frac{{240}}{3} = 80 \\ <br /> \end{array}<br /> \]<br />
 
Apparantly my software typesetter does not give the correct formatting when including text; I said, "The algebra is fairly simple, especially if you studied the course(s)"
 
It's fairly simple to get the spacing right, especially if you studied the manual!
 
HallsofIvy said:
It's fairly simple to get the spacing right, especially if you studied the manual!

The spaces between the words were placed in the editor; the program gives choices for 'text' and 'math'. Part of the file was text and part was math. When the information was placed into the forum message, the text spaces disappeared. I found nothing about this in the help manual. Best method would be to not choose 'text', but to keep everything as 'math', and insert the spaces.
 
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