Solve for x

frctl said:
I'm not sure how it can be simplified further
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Maybe not simplify - convert to whatever form acceptable in your class. E.g. I would do something about the fraction in the denominator.
 
frctl said:
View attachment 16882
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No, this is not correct. You had \(\displaystyle \dfrac{-9x}{5}\). Now if you compute \(\displaystyle \dfrac{\dfrac{-9x}{5}} {\dfrac{9}{5}}\) you get -x, not x.

You should either divide by \(\displaystyle \dfrac{-9}{5}\) or multiply by \(\displaystyle \dfrac{-5}{9}\)
 
Jomo said:
No, this is not correct. You had \(\displaystyle \dfrac{-9x}{5}\). Now if you compute \(\displaystyle \dfrac{\dfrac{-9x}{5}} {\dfrac{9}{5}}\) you get -x, not x.

You should either divide by \(\displaystyle \dfrac{-9}{5}\) or multiply by \(\displaystyle \dfrac{-5}{9}\)
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I think it is -9/5.
 
Jomo said:
Please read post #7 and your reply in post #8 where you said all was good.
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OP divided LHS by -9/5, confirmed by post 11. The negative sign is there, in front of the fraction bar.
 
I will solve this one for you as I feel that you need to have at least one good example to work from .

But first we need to notice something. If you have \(\displaystyle \dfrac{a}{b}\) and you want to multiply that by something to get 1 then you multiply by \(\displaystyle \dfrac{b}{a}\) since \(\displaystyle \dfrac{a}{b}*\dfrac{b}{a} = \dfrac{ab}{ab} = 1\).

For example, \(\displaystyle \dfrac{7}{3}*\dfrac{3}{7}=1\)

Onto you example. You got as far as \(\displaystyle y-8 = \dfrac{-9}{5}x.\)

Now \(\displaystyle (\dfrac{-5}{9})(y-8) = (\dfrac{-5}{9})\dfrac{-9}{5}x = x\).

So \(\displaystyle x = (\dfrac{-5}{9})(y-8)\)
 
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