Write a system of linear equations that has only one solution. solve the system of equations in two ways.

X + 5y = 1 . . . . (1)
8x - 2y = 3 . . . .(2)
Solution 1.
From (1), x = 1 - 5y
substituting for x in (2), we have:
8(1 - 5y) - 2y = 3
8 - 40y - 2y = 3
8 - 42y = 3
42y = 5
y = 5/42
x = 1 - 5(5/42) = 1 - 25/42 = 17/42.

Solution 2.
Multiply (1) by 8, to get:
8x + 40y = 8 . . . . . (3)
(3) - (2) = 40y - (-2y) = 8 - 3
42y = 5
y = 5/42
substitute for y into (3), to get:
8x + 40(5/42) = 8
8x + 100/21 = 8
8x = 8 - 100/21 = 68/21
x = (68/21)/8 = 17/42

The answer to your question is b’(16,14)