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Why C = t ====================================== a + 5b+3c = 0 -2a + 6b + 2c = 0

ID: 2938346 • Letter: W

Question

Why C = t ====================================== a + 5b+3c      = 0 -2a + 6b + 2c = 0 3a -  b + c      =0 answer is a = -1/2t               b = -1/2t               c = t ==================== please explain step by step.. Why C = t ====================================== a + 5b+3c      = 0 -2a + 6b + 2c = 0 3a -  b + c      =0 answer is a = -1/2t               b = -1/2t               c = t ==================== please explain step by step..

Explanation / Answer

t is the parameter in this solution space, there are an infinitenumber of {a, b,c} that satisfy this system of equations. If youpick c = t = 1, then b = -1/2 and a = -1/2 satisfies the system. Ifyou pick c = t = 2, then b = -1 and a = -1 satisfies the system,etc. Step by step, First, cancel out the a-term in equation 2 by multiplying equation1 by 2 and adding to equation 2, to get: 2a -2a +10b +6b+6c+2c = 0 -> 16b + 8c = 0 Then cancel the a-term in equation 3 by multiplying equation 1 by 3and subtracting: 3a -3a+15b-(-b)+9c-c = 0 -> 16b + 8c = 0 Notice that now the two equations you have are the same. Any b andc that satisfies 2b +c = 0 will do, so choose c = t (this means youleave t as a free parameter and solve for a and b in terms of thevalue of c). Then 2b +t =0 so b = -1/2t. Plugging c =t and b = -1/2t into equation 1, you get a -5/2t+3t = 0 -> a = -1/2t Back in the step where we were solving for b and c, we could havedefined b to be the fixed parameter, as in set b = t instead. Thenyou would have the solution a = t, b = t, c = -2t.