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

Jun 30, 2018

$t = 12$

Explanation:

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

$\frac{t + 3}{5} \cdot 45 = \frac{2 t + 3}{9} \cdot 45$

$\frac{t + 3}{\cancel{5}} \cdot 9 \cancel{45} = \frac{2 t + 3}{\cancel{9}} \cdot 5 \cancel{45}$

We're left with

$9 \left(t + 3\right) = 5 \left(2 t + 3\right)$

We can distribute the constants outside to get

$9 t + 27 = 10 t + 15$

Subtracting $10 t$ 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!