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

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

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Answer 1 · 2018-07-05T14:35:49

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

Explanation:

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

After plugging value of $s$ $s$ into second one,

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

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

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

$5 - 10 t = 0$ $5-10t=0$

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

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

Answer 2 · 2018-07-05T14:37:57

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

Explanation:

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

using equation 2 to find value of s

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

Now put value of s in eq 1

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

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

$- 18 t + 28 = - 28 t$ $-18t+28=-28t$

$28 = - 28 t + 18 t$ $28= -28t+18t$

$28 = - 10 t$ $28=-10t$

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

Now put value of t in equation 3

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

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

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

2 Answers

Explanation:

Explanation:

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How do solve the following linear system?: 3s+4=−4t,7s+6t+11=03s + 4 = -4t, 7s + 6t + 11 = 0 ?

2 Answers

Explanation:

Explanation:

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