Expert: Lynn Houston - 7/3/2009

Question
1) If x(2x+1)=0 and (x+1/2)(2x-3)=0, then x=?

2) For what values of 'k' will the pair of equations 3x+4y=12 and kx+12y=30 not have a unique solution?

Answer
1) If x(2x+1)=0 and (x+1/2)(2x-3)=0, then x=?
2x^2  +x +   0 = 0  and  
2x^2 -2x - 3/2 = 0  subtract the two equations
3x + 3/2 = 0
3x = -3/2
x = -1/2

An easier way to figure this is to know that in order for the equation to = zero, part of what is being multiplied has to equal 0.  So either x, 2x+1 or x+1/2, 2x-3 has to equal zero.  If x = zero, then the second eq is not true, so you know that 2x+1 and either x+1/2 or 2x-3 has to equal zero.  It's real easy to see that for x+1/2 to = 0, x has to be -1/2.

2) For what values of 'k' will the pair of equations 3x+4y=12 and kx+12y=30 not have a unique solution?

3x+4y=12  (multiply by 3)
9x + 12y = 36
kx  +12y = 30 (subtract equations)
(9-k)x = 6

If k = 9, there is no solution to the eq because you would have
9x+12y = 36 and 9x + 12y = 30