How do you solve $5 (2 t - 1) + 3 = 3 (3 t + 2)$ $5(2t - 1) + 3 = 3(3t + 2)$ ?

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Answer 1 · 2018-03-24T15:31:28

$\Rightarrow t = 8$ $=>t = 8$

We start with:

$5 (2 t - 1) + 3 = 3 (3 t + 2)$ $5(2t - 1) + 3 = 3(3t + 2)$

We distribute the $5$ $5$ into $2 t - 1$ $2t-1$ and the $3$ $3$ into $3 t + 2$ $3t+2$ :

$\Rightarrow 10 t - 5 + 3 = 9 t + 6$ $=>10t-5 + 3 =9t+6$

$\Rightarrow 10 t - 2 = 9 t + 6$ $=>10t-2=9t+6$

Subtracting $9 t$ $9t$ from both sides.

$\Rightarrow t - 2 = 6$ $=>t - 2=6$

Adding $2$ $2$ to both sides.

$\Rightarrow t = 8$ $=>t = 8$

How do you solve 5(2t−1)+3=3(3t+2) 5(2t - 1) + 3 = 3(3t + 2) ?