How do you solve $6 (t - 3) = 2 (9 k - 2 t)$ $6( t - 3) = 2( 9k-2t )$ ?

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How do you solve $6 (t - 3) = 2 (9 k - 2 t)$ $6( t - 3) = 2( 9k-2t )$ ?

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Answer 1 · 2017-02-08T21:08:36

$k = \frac{5}{9} t - 1$ $k = 5/9t - 1$ or $t = \frac{9}{5} k + \frac{9}{5}$ $t = 9/5k + 9/5$

Explanation:

$6 (t - 3) = 2 (9 k - 2 t)$ $6(t-3) = 2(9k-2t)$
$3 (t - 3) = 9 k - 2$ $3(t-3) = 9k - 2$
$3 t - 9 = 9 k - 2 t$ $3t - 9 = 9k - 2t$
$5 t - 9 = 9 k$ $5t - 9 = 9k$

Therefore:
$k = \frac{5}{9} t - 1$ $k = 5/9t - 1$ or $t = \frac{9}{5} k + \frac{9}{5}$ $t = 9/5k + 9/5$

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How do you solve $6 (t - 3) = 2 (9 k - 2 t)$ $6( t - 3) = 2( 9k-2t )$ ?

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How do you solve 6(t−3)=2(9k−2t)6( t - 3) = 2( 9k-2t )?

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How do you solve $6 (t - 3) = 2 (9 k - 2 t)$ $6( t - 3) = 2( 9k-2t )$ ?