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

Answer 1

Kennedy Adams

#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!

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Answer 2

Eva Abramson

To solve the equation (t+3)/5=(2t+3)/9, you can cross-multiply and then solve for t. First, multiply both sides of the equation by 5 and 9 to eliminate the denominators. This gives you 9(t+3) = 5(2t+3). Expanding the equation, you get 9t + 27 = 10t + 15. Next, subtract 9t from both sides to isolate the variable, resulting in 27 = t + 15. Finally, subtract 15 from both sides to solve for t, giving you t = 12.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.