For a system of linear equations to have a solution, it means that they would cross over at some point, thus if we are looking for a system of linear equations that do not have a solution (ie. they do not cross over), we are looking for two parallel lines.
Now for two lines to be parallel, they must have the same gradient. Thus, we must find the value of a for which both the equations have the same gradient. In order to do this, we should first write both equations in the form y = mx + c, where m is the gradient and c the y-intercept.
1) Equation 1:
(1/2)x - (2/3)y = 7
(3/4)x - y = 21/2 (Multiply both sides by 3/2)
(3/4)x = 21/2 + y (Add y to both sides)
(3/4)x - 21/2 = y (Subtract 21/2 from both sides)
Thus, the first equation may be written as y = (3/4)x - 21/2
2) Equation 2:
ax - 8y = -1
(a/8)x - y = -1/8 (Divide both sides by 8)
(a/8)x = -1/8 + y (Add y to both sides)
(a/8)x + 1/8 = y (Add 1/8 to both sides)
Thus, the second equation may be written as y = (a/8)x + 1/8
Now that we know the equations of the two lines, we can compare their gradients.
Equation 1: m = 3/4
Equation 2: m = a/8
Remember, for the two lines to be parallel, their gradients must be the same. Thus, we must equate the two gradients above to find the value of a:
3/4 = a/8
24/4 = a (Multiply both sides by 8)
6 = a
Therefor, if the system of linear equations has no solution, and a is a constant, the value of a is 6 (answer D).