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












2














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







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  • Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
    – Yadati Kiran
    2 days ago












  • What are the solving variables?
    – Dr. Sonnhard Graubner
    2 days ago










  • 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.
    – Blue
    2 days ago


















2














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









New contributor




Raphaël Huleux is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
    – Yadati Kiran
    2 days ago












  • What are the solving variables?
    – Dr. Sonnhard Graubner
    2 days ago










  • 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.
    – Blue
    2 days ago
















2












2








2







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









New contributor




Raphaël Huleux is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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









New contributor




Raphaël Huleux is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Raphaël Huleux is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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share|cite|improve this question








edited 2 days ago









Henning Makholm

238k16303540




238k16303540






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asked 2 days ago









Raphaël HuleuxRaphaël Huleux

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134




New contributor




Raphaël Huleux is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Raphaël Huleux is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Raphaël Huleux is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
    – Yadati Kiran
    2 days ago












  • What are the solving variables?
    – Dr. Sonnhard Graubner
    2 days ago










  • 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.
    – Blue
    2 days ago




















  • Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
    – Yadati Kiran
    2 days ago












  • What are the solving variables?
    – Dr. Sonnhard Graubner
    2 days ago










  • 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.
    – Blue
    2 days ago


















Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
– Yadati Kiran
2 days ago






Your simplification seems fine excepting $w_2+t_2$ should be $w_2-t_2$.
– Yadati Kiran
2 days ago














What are the solving variables?
– Dr. Sonnhard Graubner
2 days ago




What are the solving variables?
– Dr. Sonnhard Graubner
2 days ago












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.
– Blue
2 days ago






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.
– Blue
2 days ago












1 Answer
1






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oldest

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1














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























  • I'm sorry I must have some kind of mental block but I still don't get it:
    – Raphaël Huleux
    2 days ago






  • 1




    No problem, I’ll add an extra step!
    – KM101
    2 days ago










  • 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.
    – Raphaël Huleux
    2 days ago











Your Answer





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1 Answer
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1














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























  • I'm sorry I must have some kind of mental block but I still don't get it:
    – Raphaël Huleux
    2 days ago






  • 1




    No problem, I’ll add an extra step!
    – KM101
    2 days ago










  • 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.
    – Raphaël Huleux
    2 days ago
















1














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























  • I'm sorry I must have some kind of mental block but I still don't get it:
    – Raphaël Huleux
    2 days ago






  • 1




    No problem, I’ll add an extra step!
    – KM101
    2 days ago










  • 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.
    – Raphaël Huleux
    2 days ago














1












1








1






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














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 2 days ago

























answered 2 days ago









KM101KM101

5,5611423




5,5611423












  • I'm sorry I must have some kind of mental block but I still don't get it:
    – Raphaël Huleux
    2 days ago






  • 1




    No problem, I’ll add an extra step!
    – KM101
    2 days ago










  • 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.
    – Raphaël Huleux
    2 days ago


















  • I'm sorry I must have some kind of mental block but I still don't get it:
    – Raphaël Huleux
    2 days ago






  • 1




    No problem, I’ll add an extra step!
    – KM101
    2 days ago










  • 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.
    – Raphaël Huleux
    2 days ago
















I'm sorry I must have some kind of mental block but I still don't get it:
– Raphaël Huleux
2 days ago




I'm sorry I must have some kind of mental block but I still don't get it:
– Raphaël Huleux
2 days ago




1




1




No problem, I’ll add an extra step!
– KM101
2 days ago




No problem, I’ll add an extra step!
– KM101
2 days ago












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.
– Raphaël Huleux
2 days ago




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.
– Raphaël Huleux
2 days ago










Raphaël Huleux is a new contributor. Be nice, and check out our Code of Conduct.










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