For which values of k does the equation (x - 1)/(x - 2) = (x - k)/(x + 6) have no solution for k? Enter all the possible values of separated by commas.

Guest Dec 14, 2021

#1**+1 **

Note: != means not equal to. This would be an equal sign with a slash between it in written form.

1) We have the equation (x-1)/(x-2) = (x-k)/(x+6)

2) We can identitify the non permissible values, aka the values that make the denominator equal to 0 (we cannot divide by 0, gives us an error). Now, if we set both denominators to equal 0 and solve, we will get the following: x-2 = 0; x != 2 and x+6 = 0; x != -6. This means that x cannot equal -6 or 2

3) Okay, now we know what x cannot equal to. But the question asks what are the non permissible values of k. We can solve for x in this case to create an equation that will allow us to find the non-permissible values of k.

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1) We start with the equation (x-1)/(x-2) = (x-k)/(x+6).

2) We can cross multiply which will give us (x+6)(x-1) = (x-2)(x-k)

3) Foil and combine like terms for both sides. x^2 + 5x - 6 = x^2 - xk - 2x + 2k

4) Subtract x^2 from both sides. x^2 + 5x - 6 - x^2 = -xk - 2x + 2k

5) We can then isolate the other x^2 on the left side (+ x^2 - x^2). Then we get 5x - 6 = -xk - 2x + 2k

6) We can add xk to both sides. Then we get 5x - 6 + xk = -2x + 2k

7) Then we can add 2x to both sides. We then get 5x + 2x - 6 + xk = 2k

8) Combine 5x and 2x. We get 7x - 6 + xk = 2k

9) Add 6 to both sides. We get 7x + xk = 2k + 6

10) Factor 7x + xk. We get x(k+7) = 2k + 6. Or you could arrange it like x(7+k) = 2k + 6. Whatever works for you.

11) We can now comfortably solve for x. Divide both sides by (k+7). We get x = 2k+6/k+7.

12) Now we can find the non permissible values of k. Set the denominator to equal to 0. k + 7 = 0 and solve. Subtract 7 from both sides and we get k = -7.

Solution (word it however you want but these would work):

1) k != -7

2) The non permissible values of k for the equation (x-1)/(x-2) = (x-k)(x+6) is k != -7

3) The equation (x-1)/(x-2) = (x-k)(x+6) has no solution for k when k = -7

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AlgebraGuru Dec 14, 2021