# How do you solve 7x+6-2(5 + 3t/2) = 5t - 11?

See below:

#### Explanation:

I'm first going to solve this assuming that the first $x$ term should really be a $t$:

$7 t + 6 - 2 \left(5 + 3 \frac{t}{2}\right) = 5 t - 11$

Distribute the 2 across the brackets:

$7 t + 6 - 10 + 3 t = 5 t - 11$

Let's move all the $t$ terms to the left and all the non-$t$ terms to the right:

$7 t + 6 - 10 + 3 t \textcolor{red}{- 6 + 10 - 5 t} = 5 t - 11 \textcolor{red}{- 6 + 10 - 5 t}$

$7 t + 3 t - 5 t = - 11 - 6 + 10$

And now simplify:

$5 t = - 7$

$\frac{5 t}{\textcolor{red}{5}} = - \frac{7}{\textcolor{red}{5}}$

$t = - \frac{7}{5}$

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Now let's assume that we want to solve for $x$ in terms of $t$:

$7 x + 6 - 2 \left(5 + 3 \frac{t}{2}\right) = 5 t - 11$

$7 x + 6 - 10 + 3 t = 5 t - 11$

Let's move all non-$x$ terms to the right:

$7 x + 6 - 10 + 3 t \textcolor{red}{- 6 + 10 - 3 t} = 5 t - 11 \textcolor{red}{- 6 + 10 - 3 t}$

$7 x = 2 t - 7$

$x = \frac{2 t - 7}{7} = \frac{2}{7} t - 1$

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Now let's assume we want to solve for $t$ in terms of $x$:

$7 x = 2 t - 7$
$2 t = 7 x + 7$
$t = \frac{7 x + 7}{2} = \frac{7 \left(x + 1\right)}{2} = \frac{7}{2} \left(x + 1\right)$