Make k the subject for 3t = 7k/13 - 17 and 7k = 4k/3t - 11t

1 answer:

5 0

$3t = \frac{7k}{13} - 17 \\3t(1\frac{6}{7}) = 1\frac{6}{7}(\frac{7k}{13} - 17) \\5\frac{4}{7}t = k - 31\frac{4}{7} \\\frac{-k}{-1} = \frac{-5\frac{4}{7} - 31\frac{4}{7}}{-1} \\k = 5\frac{4}{7} + 31\frac{4}{7}$

$\frac{7k}{7} = \frac{\frac{4k}{3t} - 11t}{7} \\k = \frac{4k}{21t} - 1\frac{4}{7}$

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To solve this equation, we can use the quadratic formula. The quadratic formula solves equations of the form ax^2 + bx + c = 0
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