Ask your own question, for FREE!
Algebra 8 Online
OpenStudy (superman499):

2p+5r=q Solve for p

OpenStudy (drummergirl3):

2p + 5r = q 2p = 5r - q p = (5r - q) / 2

OpenStudy (superman499):

ok but i need more help

OpenStudy (superman499):

like with the (5r-q)/2

OpenStudy (drummergirl3):

ok what

OpenStudy (superman499):

the (5r-q) / 2

OpenStudy (firekat97):

to solve for p, you want to isolate p on one side, in this case its probably easier to keep it on the left, so you want to move the 5r over and you do so, by subtracting 5r from the left and the right side (so you don't change the equation)

OpenStudy (firekat97):

but then you are left with a 2p on the left hand side, and you want only p, so this time, you can multiply both sides of the equation with p so that the p's on the left hand side cancel, and you do not change the equation

OpenStudy (superman499):

so would the 5r -q =5rq or somethin else

OpenStudy (superman499):

@FireKat97

OpenStudy (anonymous):

a su ma

OpenStudy (superman499):

do what

OpenStudy (superman499):

\[2p+5r=q \]

OpenStudy (firekat97):

So it's 2p + 5r = q And you want to get the p by itself so you want take the 5r to the other side without changing the equation so you can do this 2p + 5r -5r = q - 5r and on the left hand side the +5r and -5r will cancel so you have effectively taken r to the other side. Also, since you have subtracted 5r from both sides, you didn't not change the equation which is most important

OpenStudy (firekat97):

So now that you have 2p = q - 5r, you have to get rid of the 2 in front of the p to isolate p. Now in the same way, you can do this 2p/2 = (q - 5r)/2 so by diving both sides, you do not change the equation and also on the left hand side the 2/2 will cancel so you have only p left, so you have solved for p, so you are left with p = (q-5r)/2 @superman499

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!