Simplifying $s=w_1-c$, where $c=frac{1}{2}(w_1-t_1+frac{w_2-t_2}{R})$












2












$begingroup$


I have to solve this easy equation, but can't find the same answer as the correction of the exercise.



We have two equations that I want to simplify (plugging $c$ into $s$)



$$c=dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)$$



$$s=w_1-c$$



I find:



$$s=w_1-dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)=dfrac{1}{2}left(w_1+t_1-dfrac{w_2+t_2}{R}right)$$



The correction actually says:



$$s=w_1-dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)=dfrac{1}{2}left(w_1-t_1-dfrac{w_2-t_2}{R}right)$$



I must be wrong but I don't know why. Could somebody explain this to me?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
    $endgroup$
    – Yadati Kiran
    Jan 5 at 17:53












  • $begingroup$
    What are the solving variables?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 5 at 17:56










  • $begingroup$
    Both answers, as presented here, are incorrect (in different ways). In your answer, the $t_2$ should be negative; in their answer, the $t_1$ should be positive. Please check for typographic errors one way or the other, so that we can be sure we're comparing things correctly.
    $endgroup$
    – Blue
    Jan 5 at 18:09


















2












$begingroup$


I have to solve this easy equation, but can't find the same answer as the correction of the exercise.



We have two equations that I want to simplify (plugging $c$ into $s$)



$$c=dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)$$



$$s=w_1-c$$



I find:



$$s=w_1-dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)=dfrac{1}{2}left(w_1+t_1-dfrac{w_2+t_2}{R}right)$$



The correction actually says:



$$s=w_1-dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)=dfrac{1}{2}left(w_1-t_1-dfrac{w_2-t_2}{R}right)$$



I must be wrong but I don't know why. Could somebody explain this to me?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
    $endgroup$
    – Yadati Kiran
    Jan 5 at 17:53












  • $begingroup$
    What are the solving variables?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 5 at 17:56










  • $begingroup$
    Both answers, as presented here, are incorrect (in different ways). In your answer, the $t_2$ should be negative; in their answer, the $t_1$ should be positive. Please check for typographic errors one way or the other, so that we can be sure we're comparing things correctly.
    $endgroup$
    – Blue
    Jan 5 at 18:09
















2












2








2





$begingroup$


I have to solve this easy equation, but can't find the same answer as the correction of the exercise.



We have two equations that I want to simplify (plugging $c$ into $s$)



$$c=dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)$$



$$s=w_1-c$$



I find:



$$s=w_1-dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)=dfrac{1}{2}left(w_1+t_1-dfrac{w_2+t_2}{R}right)$$



The correction actually says:



$$s=w_1-dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)=dfrac{1}{2}left(w_1-t_1-dfrac{w_2-t_2}{R}right)$$



I must be wrong but I don't know why. Could somebody explain this to me?






algebra-precalculus







share|cite|improve this question











$endgroup$




I have to solve this easy equation, but can't find the same answer as the correction of the exercise.



We have two equations that I want to simplify (plugging $c$ into $s$)



$$c=dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)$$



$$s=w_1-c$$



I find:



$$s=w_1-dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)=dfrac{1}{2}left(w_1+t_1-dfrac{w_2+t_2}{R}right)$$



The correction actually says:



$$s=w_1-dfrac{1}{2}left(w_1-t_1+dfrac{w_2-t_2}{R}right)=dfrac{1}{2}left(w_1-t_1-dfrac{w_2-t_2}{R}right)$$



I must be wrong but I don't know why. Could somebody explain this to me?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 18:09









Henning Makholm

240k17305541




240k17305541










asked Jan 5 at 17:51









Raphaël HuleuxRaphaël Huleux

134




134












  • $begingroup$
    Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
    $endgroup$
    – Yadati Kiran
    Jan 5 at 17:53












  • $begingroup$
    What are the solving variables?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 5 at 17:56










  • $begingroup$
    Both answers, as presented here, are incorrect (in different ways). In your answer, the $t_2$ should be negative; in their answer, the $t_1$ should be positive. Please check for typographic errors one way or the other, so that we can be sure we're comparing things correctly.
    $endgroup$
    – Blue
    Jan 5 at 18:09




















  • $begingroup$
    Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
    $endgroup$
    – Yadati Kiran
    Jan 5 at 17:53












  • $begingroup$
    What are the solving variables?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 5 at 17:56










  • $begingroup$
    Both answers, as presented here, are incorrect (in different ways). In your answer, the $t_2$ should be negative; in their answer, the $t_1$ should be positive. Please check for typographic errors one way or the other, so that we can be sure we're comparing things correctly.
    $endgroup$
    – Blue
    Jan 5 at 18:09


















$begingroup$
Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
$endgroup$
– Yadati Kiran
Jan 5 at 17:53






$begingroup$
Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
$endgroup$
– Yadati Kiran
Jan 5 at 17:53














$begingroup$
What are the solving variables?
$endgroup$
– Dr. Sonnhard Graubner
Jan 5 at 17:56




$begingroup$
What are the solving variables?
$endgroup$
– Dr. Sonnhard Graubner
Jan 5 at 17:56












$begingroup$
Both answers, as presented here, are incorrect (in different ways). In your answer, the $t_2$ should be negative; in their answer, the $t_1$ should be positive. Please check for typographic errors one way or the other, so that we can be sure we're comparing things correctly.
$endgroup$
– Blue
Jan 5 at 18:09






$begingroup$
Both answers, as presented here, are incorrect (in different ways). In your answer, the $t_2$ should be negative; in their answer, the $t_1$ should be positive. Please check for typographic errors one way or the other, so that we can be sure we're comparing things correctly.
$endgroup$
– Blue
Jan 5 at 18:09












1 Answer
1






active

oldest

votes


















1












$begingroup$

You first expand a bit to simplify for the coefficient of $w_1$:



$$w_1-frac{1}{2}left(w_1-t_1+frac{w_2-t_2}{R}right) = w_1-frac{1}{2}w_1-frac{1}{2}left(-t_1+frac{w_2-t_2}{R}right)$$



$$= frac{1}{2}w_1-frac{1}{2}left(-t_1+frac{w_2-t_2}{R}right)$$



Factoring $frac{1}{2}w_1$ by $-frac{1}{2}$, the expression simplifies to:



$$-frac{1}{2}left(-w_1-t_1+frac{w_2-t_2}{R}right)$$



And factoring by $-1$ yields:



$$frac{1}{2}left(color{blue}{+}w_1color{blue}{+}t_1color{blue}{-}frac{w_2-t_2}{R}right)$$



You apparently made a mistake in the numerator of the final fraction. It should be $w_2-t_2$ rather than $w_2+t_2$, but I think it may have been a typo or an accident on your part. The rest is correct.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm sorry I must have some kind of mental block but I still don't get it:
    $endgroup$
    – Raphaël Huleux
    Jan 5 at 18:04






  • 1




    $begingroup$
    No problem, I’ll add an extra step!
    $endgroup$
    – KM101
    Jan 5 at 18:05










  • $begingroup$
    Thanks! The $w_2-t_2$ was not a typo but my actual question. I guess I was just confused about the sign change with the fraction.
    $endgroup$
    – Raphaël Huleux
    Jan 5 at 18:14











Your Answer





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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You first expand a bit to simplify for the coefficient of $w_1$:



$$w_1-frac{1}{2}left(w_1-t_1+frac{w_2-t_2}{R}right) = w_1-frac{1}{2}w_1-frac{1}{2}left(-t_1+frac{w_2-t_2}{R}right)$$



$$= frac{1}{2}w_1-frac{1}{2}left(-t_1+frac{w_2-t_2}{R}right)$$



Factoring $frac{1}{2}w_1$ by $-frac{1}{2}$, the expression simplifies to:



$$-frac{1}{2}left(-w_1-t_1+frac{w_2-t_2}{R}right)$$



And factoring by $-1$ yields:



$$frac{1}{2}left(color{blue}{+}w_1color{blue}{+}t_1color{blue}{-}frac{w_2-t_2}{R}right)$$



You apparently made a mistake in the numerator of the final fraction. It should be $w_2-t_2$ rather than $w_2+t_2$, but I think it may have been a typo or an accident on your part. The rest is correct.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm sorry I must have some kind of mental block but I still don't get it:
    $endgroup$
    – Raphaël Huleux
    Jan 5 at 18:04






  • 1




    $begingroup$
    No problem, I’ll add an extra step!
    $endgroup$
    – KM101
    Jan 5 at 18:05










  • $begingroup$
    Thanks! The $w_2-t_2$ was not a typo but my actual question. I guess I was just confused about the sign change with the fraction.
    $endgroup$
    – Raphaël Huleux
    Jan 5 at 18:14
















1












$begingroup$

You first expand a bit to simplify for the coefficient of $w_1$:



$$w_1-frac{1}{2}left(w_1-t_1+frac{w_2-t_2}{R}right) = w_1-frac{1}{2}w_1-frac{1}{2}left(-t_1+frac{w_2-t_2}{R}right)$$



$$= frac{1}{2}w_1-frac{1}{2}left(-t_1+frac{w_2-t_2}{R}right)$$



Factoring $frac{1}{2}w_1$ by $-frac{1}{2}$, the expression simplifies to:



$$-frac{1}{2}left(-w_1-t_1+frac{w_2-t_2}{R}right)$$



And factoring by $-1$ yields:



$$frac{1}{2}left(color{blue}{+}w_1color{blue}{+}t_1color{blue}{-}frac{w_2-t_2}{R}right)$$



You apparently made a mistake in the numerator of the final fraction. It should be $w_2-t_2$ rather than $w_2+t_2$, but I think it may have been a typo or an accident on your part. The rest is correct.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm sorry I must have some kind of mental block but I still don't get it:
    $endgroup$
    – Raphaël Huleux
    Jan 5 at 18:04






  • 1




    $begingroup$
    No problem, I’ll add an extra step!
    $endgroup$
    – KM101
    Jan 5 at 18:05










  • $begingroup$
    Thanks! The $w_2-t_2$ was not a typo but my actual question. I guess I was just confused about the sign change with the fraction.
    $endgroup$
    – Raphaël Huleux
    Jan 5 at 18:14














1












1








1





$begingroup$

You first expand a bit to simplify for the coefficient of $w_1$:



$$w_1-frac{1}{2}left(w_1-t_1+frac{w_2-t_2}{R}right) = w_1-frac{1}{2}w_1-frac{1}{2}left(-t_1+frac{w_2-t_2}{R}right)$$



$$= frac{1}{2}w_1-frac{1}{2}left(-t_1+frac{w_2-t_2}{R}right)$$



Factoring $frac{1}{2}w_1$ by $-frac{1}{2}$, the expression simplifies to:



$$-frac{1}{2}left(-w_1-t_1+frac{w_2-t_2}{R}right)$$



And factoring by $-1$ yields:



$$frac{1}{2}left(color{blue}{+}w_1color{blue}{+}t_1color{blue}{-}frac{w_2-t_2}{R}right)$$



You apparently made a mistake in the numerator of the final fraction. It should be $w_2-t_2$ rather than $w_2+t_2$, but I think it may have been a typo or an accident on your part. The rest is correct.






share|cite|improve this answer











$endgroup$



You first expand a bit to simplify for the coefficient of $w_1$:



$$w_1-frac{1}{2}left(w_1-t_1+frac{w_2-t_2}{R}right) = w_1-frac{1}{2}w_1-frac{1}{2}left(-t_1+frac{w_2-t_2}{R}right)$$



$$= frac{1}{2}w_1-frac{1}{2}left(-t_1+frac{w_2-t_2}{R}right)$$



Factoring $frac{1}{2}w_1$ by $-frac{1}{2}$, the expression simplifies to:



$$-frac{1}{2}left(-w_1-t_1+frac{w_2-t_2}{R}right)$$



And factoring by $-1$ yields:



$$frac{1}{2}left(color{blue}{+}w_1color{blue}{+}t_1color{blue}{-}frac{w_2-t_2}{R}right)$$



You apparently made a mistake in the numerator of the final fraction. It should be $w_2-t_2$ rather than $w_2+t_2$, but I think it may have been a typo or an accident on your part. The rest is correct.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 5 at 18:11

























answered Jan 5 at 18:02









KM101KM101

5,9251524




5,9251524












  • $begingroup$
    I'm sorry I must have some kind of mental block but I still don't get it:
    $endgroup$
    – Raphaël Huleux
    Jan 5 at 18:04






  • 1




    $begingroup$
    No problem, I’ll add an extra step!
    $endgroup$
    – KM101
    Jan 5 at 18:05










  • $begingroup$
    Thanks! The $w_2-t_2$ was not a typo but my actual question. I guess I was just confused about the sign change with the fraction.
    $endgroup$
    – Raphaël Huleux
    Jan 5 at 18:14


















  • $begingroup$
    I'm sorry I must have some kind of mental block but I still don't get it:
    $endgroup$
    – Raphaël Huleux
    Jan 5 at 18:04






  • 1




    $begingroup$
    No problem, I’ll add an extra step!
    $endgroup$
    – KM101
    Jan 5 at 18:05










  • $begingroup$
    Thanks! The $w_2-t_2$ was not a typo but my actual question. I guess I was just confused about the sign change with the fraction.
    $endgroup$
    – Raphaël Huleux
    Jan 5 at 18:14
















$begingroup$
I'm sorry I must have some kind of mental block but I still don't get it:
$endgroup$
– Raphaël Huleux
Jan 5 at 18:04




$begingroup$
I'm sorry I must have some kind of mental block but I still don't get it:
$endgroup$
– Raphaël Huleux
Jan 5 at 18:04




1




1




$begingroup$
No problem, I’ll add an extra step!
$endgroup$
– KM101
Jan 5 at 18:05




$begingroup$
No problem, I’ll add an extra step!
$endgroup$
– KM101
Jan 5 at 18:05












$begingroup$
Thanks! The $w_2-t_2$ was not a typo but my actual question. I guess I was just confused about the sign change with the fraction.
$endgroup$
– Raphaël Huleux
Jan 5 at 18:14




$begingroup$
Thanks! The $w_2-t_2$ was not a typo but my actual question. I guess I was just confused about the sign change with the fraction.
$endgroup$
– Raphaël Huleux
Jan 5 at 18:14


















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