The required solution in terms of k is (k, (k-2)/2)
From the diagram, we are given the simultaneous equation expressed as:
-4y = -2x + 8 ................ 1
3x - 6y = 6
x - 2y = 2 ......................... 2
From equation 2, x = 2 + 2y ............... 3
Substitute equation 3 into 1 to have:
-4y = -2(2+2y) + 8
-4y = -4 -4y + 8
-4y + 4y = -4 + 8
0y = 4
y = 4/0
y = ∞
This shows that the systems of equations have infinite number of solution.
Let x = k
x = 2 + 2y
k = 2 + 2y
2y = k - 2
y = (k-2)/2
Hence the required solution in terms of k is (k, (k-2)/2)
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