# How do solve the following linear system?: 3s + 4 = -4t, 7s + 6t + 11 = 0 ?

Jul 5, 2018

$s = - 2$ and $t = \frac{1}{2}$

#### Explanation:

From first equation, $s = \frac{- 4 t - 4}{3}$.

After plugging value of $s$ into second one,

$7 \cdot \frac{- 4 t - 4}{3} + 6 t + 11 = 0$

$\frac{- 28 t - 28 + 18 t + 33}{3} = 0$

$\frac{5 - 10 t}{3} = 0$

$5 - 10 t = 0$

$10 t = 5$, so $t = \frac{5}{10} = \frac{1}{2}$

Thus, $s = \frac{\left(- 4\right) \cdot \frac{1}{2} - 4}{3} = - 2$

Your answer is $- \frac{28}{10} = t$

#### Explanation:

Here we have two equation
1) $3 s + 4 = - 4 t$
2) $7 s + 6 t = 0$

using equation 2 to find value of s

$7 s + 6 t = 0$
$7 s = - 6 t$
$s = - \frac{6 t}{7}$ ....(equation3)

Now put value of s in eq 1

$3 \left(\frac{- 6 t}{7}\right) + 4 = - 4 t$

$\frac{- 18 t + 28}{7} = - 4 t$

$- 18 t + 28 = - 28 t$

$28 = - 28 t + 18 t$

$28 = - 10 t$

$- \frac{28}{10} = t$

Now put value of t in equation 3

$s = - \frac{6 \left(- \frac{28}{10}\right)}{7}$

$s = 6 \left(\frac{28}{70}\right)$