How do you solve for k in $h + p = 3 (k - 8)$ $h + p = 3 (k-8)$ ?

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Answer 1 · 2016-02-14T00:03:53

You must first distribute.

$h + p = 3 (k - 8)$ $h + p = 3(k - 8)$

$h + p = 3 k - 24$ $h + p = 3k - 24$

$h + p + 24 = 3 k$ $h + p + 24 = 3k$

$\frac{h + p + 24}{3} = k$ $(h + p + 24)/3 = k$

Practice exercises:

a) $z x = y z - y x$ $zx = yz - yx$

b) $\frac{2}{x} + \frac{3}{y} = - \frac{5}{z}$ $2/x + 3/y = -5/z$

Good luck!

How do you solve for k in h + p = 3 (k-8)h+p=3(k−8)h + p = 3 (k-8)?