Monday, July 6, 2015

Systems of linear equations

For the following system of equations, find the value of A which produces no solution for the system.
-4w/3 + 2z = -3
5Aw + 5z/8 = -3

A. 6
B. -12
C. 1/3
D. -1/12

First try to think graphically. Each of the equations above is linear. A solution to a system of two linear equations means that there is a point of intersection between the two lines corresponding to the two equations (there can only be one such point). Having no such point means that the lines are parallel (they never meet at any point). If the lines are parallel, they must have the same slope. To have the same slope, coefficients in front of z in each equation are such that 2 = 5k/8 for some constant k. In other words, the constant k transforms a variable coefficient in one equation into the same exact variable coefficient in the second equation. This means that 2 is a scalar multiple of 5/8 for some k. Solving for k we get k = 16/5. Similarly, coefficients in front of w are such that -4/3 = 5A • k, and using the known k we have -4/3 = 5A • 16/5, which gives A = -1/12. Thus the correct answer is (D).

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