# Algebra Question

#### HollowSuperet36

##### Full Member
What are the steps to solve for T in the following...

T+m2[(2T-mg)/m] = mg

every time I solve it I get the wrong answer.

Thank you much!

T = (3/5)mg

#### MedPR

##### Membership Revoked
Removed
nevermind, that's wrong. Idk!

#### milski

##### 1K member
5+ Year Member
T+m2*((2T-mg)/m)=mg
Tm+m2*(2T-mg)=m^2g
Tm+m2*2T-m2*mg=m^2g
T(m+2m2)=m^2g+m2*mg
T=(m^2g+m2*mg)/(m+2m2)

#### MedPR

##### Membership Revoked
Removed
T+m2*((2T-mg)/m)=mg
Tm+m2*(2T-mg)=m^2g
Tm+m2*2T-m2*mg=m^2g
T(m+2m2)=m^2g+m2*mg
T=(m^2g+m2*mg)/(m+2m2)

That's what I got too, but the answer is T = (3/5)mg.

I think the "m2" is supposed to be 2m? I saw how simplified the answer was and gave up.

#### milski

##### 1K member
5+ Year Member
That's what I got too, but the answer is T = (3/5)mg.

I think the "m2" is supposed to be 2m? I saw how simplified the answer was and gave up.

That works - if m2=2m (that's m sub 2 equals twice m), just plugging in my final result will give you 0.6mg. This thread is more than 9 years old.

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