SOLUTION: If k = (x + 3)/(x - 4) and k^2 - 3k = 28, find x. I say replace every k with (x + 3)/(x - 4). I get this: [(x + 3)/(x - 4)]^2 - 3[(x + 3)/(x

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Question 1207662: If k = (x + 3)/(x - 4) and k^2 - 3k = 28, find x.

I say replace every k with (x + 3)/(x - 4).

I get this:

[(x + 3)/(x - 4)]^2 - 3[(x + 3)/(x - 4)] = 28

Correct?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
that looks reasonable, although it might get messy.

you can also do the following:

in the equation k^2 - 3k = 28, subtract 28 from both sides to get:
k^2 - 3k - 28 = 0
factor this quadratic equation to get:
(k-7) * (k+4) = 0
solve for k to get:
k = 7 or k = -4

in the equation k = (x + 3)/(x - 4), replace k with 7 and solve for x.
after you do that, replace k with - 4 and solve for x.

you should get x = 31/6 or x = 13/5.

it should be able to be solved the way that you showed, but i think solving for k first is probably easier.

Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website!
.
If k = (x + 3)/(x - 4) and k^2 - 3k = 28, find x.
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First solve the quadratic equation for k and find its solutions for k. k^2 - 3k = 28 is equivalent to k^2 - 3k - 28 = 0. Factor left side (k-7)*(k+4) = 0. The solutions are k= 7 and k= -4. Now consider and solve equation = k for two values of k: k= 7 and k= -4. (a) Case k = 7. Now solve equation = 7 Step by step x+3 = 7*(x-4) x+3 = 7x - 28 3 + 28 = 7x - x 31 = 6x x = . (b) Case k = -4. Now solve equation = -4 Step by step x+3 = (-4)*(x-4) x+3 = -4x + 16 x + 4x = 16 - 3 5x = 13 x = . ANSWER. Two solutions are x = and x = .

Solved from the beginning to the end, with complete explanations.

SOLUTION: If k = (x + 3)/(x - 4) and k^2 - 3k = 28, find x. I say replace every k with (x + 3)/(x - 4). I get this: [(x + 3)/(x - 4)]^2 - 3[(x + 3)/(x - 4)] = 28 Correct