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what value of k will make the system -kx+y=3 and 4x-y=2 a consistent-independent?

Sagot :

Eliminate y:

- kx + y = 3  ⇒  Equation 1
  4x - y =  2  ⇒  Equation 2

-kx : 4x = 3 : 2

-kx (2) = 4x (3)   
-2xk = 12x         

-2xk/-2x = 12x/-2x

k = - 6 

Solve the system, substitute - 6 for k in Equation 1

-(-6)x + y = 3
6x + y = 3
y = -6x + 3  ⇒  Equation 3

Substitute for x by  - 6x + 3 for y in Equation 2:
4x - (-6x + 3) = 2
4x + 6x - 3 = 2
10x = 2 + 5
10x/10 = 5/10
x = 1/2

Solve for y, by substituting 1/2 to x in Equation 3:
y = -6x + 3
y = -6(1/2) + 3
y = - 3 + 3
y = 0

The solution to the system is (1/2, 0).

To check, x = 1/2;   y = 0
Equation 1:  
6x + y = 3
6 (1/2) + 0 = 3
3 + 0 = 3
3 = 3

Equation 2:
4x - y = 2
4 (1/2) - 0 = 2
2 - 0 = 2
2 = 2

Therefore - 6 for k satisfies the system as consistent and independent with only one solution (1/2, 0) which is the point of intersection of the given two equations/graphs.