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Mathematics 8 Online
OpenStudy (anonymous):

What is the rule for moving an entire fraction to the other side of the equal sign? in this equation can I subtract an entire fraction 'symbol pi/7' and move it to the other side of the equal sign? T/2 + 'symbol pi'/ 7 = 'symbol pi'/3 + K'symbol pi'

jimthompson5910 (jim_thompson5910):

yes you can subtract \(\Large \frac{\pi}{7}\) from both sides

jimthompson5910 (jim_thompson5910):

this works because despite that being a complicated fraction, it's still a number

OpenStudy (anonymous):

Thank you :)

jimthompson5910 (jim_thompson5910):

you're welcome

OpenStudy (anonymous):

if you dont mind I have a follow up, can I usually just subtract an entire fraction from one side to the other when trying to get a variable (like T) alone? I was trying to do it piece by piece first and had no luck

OpenStudy (anonymous):

Hmmmm...............

jimthompson5910 (jim_thompson5910):

No you would do the whole fraction. Not just the numerator or denominator alone.

jimthompson5910 (jim_thompson5910):

we have this equation \[\Large \frac{T}{2} + \frac{\pi}{7} = \frac{\pi}{3} + K*\pi\]

jimthompson5910 (jim_thompson5910):

Subtract \(\Large \frac{\pi}{7}\) from both sides to get \[\Large \frac{T}{2} + \frac{\pi}{7} = \frac{\pi}{3} + K*\pi\] \[\Large \frac{T}{2} + \frac{\pi}{7}-\frac{\pi}{7} = \frac{\pi}{3} + K*\pi-\frac{\pi}{7}\] \[\Large \frac{T}{2} + \frac{\pi}{7}-\frac{\pi}{7} = \frac{\pi}{3} -\frac{\pi}{7}+ K*\pi\] \[\Large \frac{T}{2} + 0= \frac{\pi}{3} -\frac{\pi}{7}+ K*\pi\] \[\Large \frac{T}{2} = \frac{\pi}{3} -\frac{\pi}{7}+ K*\pi\] \[\Large \frac{T}{2} = \frac{7\pi}{21} -\frac{3\pi}{21}+ K*\pi\] \[\Large \frac{T}{2} = \frac{7\pi-3\pi}{21}+ K*\pi\] \[\Large \frac{T}{2} = \frac{4\pi}{21}+ K*\pi\] do you see how to finish up?

OpenStudy (anonymous):

yes, multiply by 2: lhs cancels leavin T alone, rhs 4pi becomes 8pi/21 + 2Kpi So then in yes I guess in other problems I can always just subtract an entire fraction to the other side cuz its in fact an actual number

jimthompson5910 (jim_thompson5910):

yes that is correct

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