SOLUTION: Can you help me solve this equation? The equation below is really a family of equations, because for each value of k we get a different equation with the unknown x. The letter k

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Question 119592: Can you help me solve this equation?
The equation below is really a family of equations, because for each value of k we get a different equation with the unknown x. The letter k is called a parameter for this family. What value should we pick for k to make the given value of x a solution of the resulting equation? (If k can have any value, enter ANY.)
4x + 6k - 27 = kx - k + 1
If x = 5, k =
If x = -3, k =
If x = 7, k =

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
he equation below is really a family of equations, because for each value of k we get a different equation with the unknown x. The letter k is called a parameter for this family. What value should we pick for k to make the given value of x a solution of the resulting equation? (If k can have any value, enter ANY.)
:
4x + 6k - 27 = kx - k + 1
:
Simplify equation and solve for k
4x - 27 - 1 = kx - k - 6k
4x - 28 = k(x - 1 - 6)
4x - 28 = k(x-7)
k = %28%284x-28%29%29%2F%28%28x-7%29%29
:
If x = 5, k =
Substitute 5 for x:
k = %28%284%285%29-28%29%29%2F%28%285-7%29%29
k = %28%2820-28%29%29%2F%28%28-2%29%29
k = %28-8%29%2F%28-2%29
k = +4
:
If x = -3, k =
Substitute -3 for x
k = %28%284%28-3%29-28%29%29%2F%28%28-3-7%29%29
k = %28%28-12-28%29%29%2F%28%28-10%29%29
k = %28%28-40%29%29%2F%28%28-10%29%29
k = +4
:
We can say that x can be any value for k = 4
:
x = 7 will not work in my k= equation (division by 0) however is OK in the original equation.
If x = 7, k =