Yahoo Answers: Answers and Comments for The equation is 13k = 6k/15t  7t Make k the subject? [Mathematics]
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From Anonymous
enGB
Mon, 04 Feb 2019 14:55:07 +0000
3
Yahoo Answers: Answers and Comments for The equation is 13k = 6k/15t  7t Make k the subject? [Mathematics]
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From llaffer: Presuming you mean:
13k = 6k/(15t)  7t
Sinc...
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Mon, 04 Feb 2019 14:58:36 +0000
Presuming you mean:
13k = 6k/(15t)  7t
Since "k" is in more than one term, we need to group them together. Start with multiplying both sides by the denominator (15t):
195kt = 6k  105t²
Now move the terms with "k" to one side and all other terms to the other side:
195kt  6k = 105t²
to simplify a little, multiply both sides by 1. I'll flip the two terms around the subtraction on the left side:
6k  195kt = 105t²
And now I can divide both sides by 3 to simplify further before we move on:
2k  65kt = 35t²
Now we can factor out the k:
k(2  65t) = 35t²
and divide:
k = 35t² / (2  65t)

From Krishnamurthy: 13 k = 6 k/15 t  7t
13 k = (2 k t)/5  7 t
65...
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Mon, 04 Feb 2019 17:33:05 +0000
13 k = 6 k/15 t  7t
13 k = (2 k t)/5  7 t
65 k = t (2 k  35)
k(65  2 t) = 35 t
k
= 35 t / (65  2 t)
= 35 t / (2 t  65)

From ?: 13k = 6k/15t  7t
// Multiply through by 15t
...
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Mon, 04 Feb 2019 16:05:03 +0000
13k = 6k/15t  7t
// Multiply through by 15t
13k (15t) = 6k  7t (15t)
195 kt  6k =  105 t²
// Factor out k
k (195t  6) = 105 t²
// Divide through by (195t  6)
k = 105 t² / (195t  6)....................ANS

From Raymond: I assume that the  7t is a separate term
in o...
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Mon, 04 Feb 2019 15:03:35 +0000
I assume that the  7t is a separate term
in other words, your equation is NOT:
13k = 6k / (15t  7t)

13k = (6k/15t)  7t
multiply both sides by (15t)
(13k)(15t) = 6k  (15t)(7t)
clean up
195kt = 6k  105t^2
subtract 6k from both sides
195kt  6k = 105t^2
factor out "k" on the left
k(195t  6) = 105t^2
divide both sides by (195t  6)
k = (105t^2) / (195t  6)

From Como: Presentation is open to doubt due to lack of b...
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Mon, 04 Feb 2019 15:35:08 +0000
Presentation is open to doubt due to lack of brackets.