High School Help with a simple algebra problem

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The algebra problem involves solving the equation t / (2t - s) = 3s for t. The user initially arrives at t = -3s^2/(1 - 6s), while the provided answer is t = 3s^2/(6s - 1). A key point in the discussion is the importance of using parentheses to clarify the terms in the equation, as misinterpretation can lead to confusion. Ultimately, it is noted that both expressions represent the same solution despite the differences in appearance. Clear formatting is essential for accurate mathematical communication.
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Hi

I'm trying to solve;
t / 2t-s = 3s
for t
the answer I get is; t = -3s^2/1-6s. However, the answer given is; 3s^2/6s-1

The steps I use are;
t = 3s(2t - s)
t = 6st - 3s^2
t - 6st = -3s^2
t(1 - 6s) = -3s^2
t = 3s^2 / 1 - 6s

If someone could point out where I'm going wrong that would be great, thank you.
 
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MartinJH said:
Hi

I'm trying to solve;
t / 2t-s = 3s
for t
the answer I get is; t = -3s^2/1-6s. However, the answer given is; 3s^2/6s-1

The steps I use are;
t = 3s(2t - s)
t = 6st - 3s^2
t - 6st = -3s^2
t(1 - 6s) = -3s^2
Here. You switched the sign on the right but not on the left.
t = 3s^2 / 1 - 6s

If someone could point out where I'm going wrong that would be great, thank you.
 
MartinJH said:
I'm trying to solve;
t / 2t-s = 3s
for t
the answer I get is; t = -3s^2/1-6s. However, the answer given is; 3s^2/6s-1
Use parentheses! When you write mathematical expressions on a single line, and have fractions with numerators or denominators that have multiple terms, you absolutely need to use parentheses.

Textbooks can have things like ##\frac t {2t - s}##, where it's clear which terms are in the denominator, but when you write t/2t - s, it really means ##\frac t 2 t - s##, which is probably not what you meant.
Also, what you wrote as 3s^2/1 - 6s would reasonably be interpreted to mean ##3\frac {s^2} 1 - 6s.

I used LaTeX for format the two expressions. If you don't use LaTeX (we have a very good tutorial here -- https://www.physicsforums.com/help/latexhelp/), the equation you started with should be written as t/(2t - s) = 3s, and the solution you got should be written as t = 3s^2/(1 - 6s).
 
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MartinJH said:
Hi

I'm trying to solve;
t / 2t-s = 3s
for t
the answer I get is; t = -3s^2/1-6s. However, the answer given is; 3s^2/6s-1

If someone could point out where I'm going wrong that would be great, thank you.

Those are the same answer.
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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