Question 119592
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 = {{{((4x-28))/((x-7))}}}
:
If x = 5, k =
Substitute 5 for x:
k = {{{((4(5)-28))/((5-7))}}}
k = {{{((20-28))/((-2))}}}
k = {{{(-8)/(-2)}}}
k = +4
:
If x = -3, k =
Substitute -3 for x
k = {{{((4(-3)-28))/((-3-7))}}}
k = {{{((-12-28))/((-10))}}}
k = {{{((-40))/((-10))}}}
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 =