I have just completely forgotten how to solve a system ofequations... I know the

Question

I have just completely forgotten how to solve a system ofequations... I know the answers but I'm always off by a little bitA=1/2, B=1/5 and C=-1/10 1) 2A+B+2C=1 2) 3A+2B-C =2 3) -2A          =-1 I need like explicit step by step directions, I'm sofrustrated I can't get it. I have just completely forgotten how to solve a system ofequations... I know the answers but I'm always off by a little bitA=1/2, B=1/5 and C=-1/10 1) 2A+B+2C=1 2) 3A+2B-C =2 3) -2A          =-1 I need like explicit step by step directions, I'm sofrustrated I can't get it.

Explanation / Answer

1) 2A+B+2C=1 2) 3A+2B-C =2 3) -2A          =-1 Starting with 3), we know that if -2A = -1, then A = 1/2 Plug in A = 1/2 in equation 1) and in equation 2) You get 2(1/2) + B + 2C = 1 Therefore 1 + B + 2C = 1 B + 2C = 0. In equation 2): 3(1/2) + 2B - C = 2 2B - C = 2 - 3/2 2B - C = 1/2 So we know that B + 2C = 0   and 2B - C = 1/2 Solve for C in equation 2). 2B - C = 1/2 C = 2B - 1/2 Then plug in C = 2B - 1/2 into equation 1) B + 2(2B - 1/2) = 0 B + 4B -1 = 0 5B = 1 B = 1/5.
Plug in B = 1/5 into equation 2). C = 2(1/5) - 1/2 C = 2/5 - 1/2 C = 4/10 - 5/10 C = -1/10

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