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!

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