How do you solve #(t+3)/5=(2t+3)/9#?

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Answer 1 · 2018-06-30T22:48:13

#t=12#

Since #45# is the LCM of the denominators, let's multiply both sides by that.

#(t+3)/5*45=(2t+3)/9*45#

#(t+3)/cancel5*9cancel45=(2t+3)/cancel9*5cancel45#

We're left with

#9(t+3)=5(2t+3)#

We can distribute the constants outside to get

#9t+27=10t+15#

Subtracting #10t# from both sides gives us

#-t+27=15#

Subtracting #27#from both sides, we get

#-t=-12#

Lastly, we can divide by #-1# to get

#t=12#

Hope this helps!