solve for b; -3a=36 and 10a+3c=9 and 2b+5c=23

Question

it can be solved by elimination or subtitution

Answer 1

first you want to line those equations so you can understand them better,

-3A = 36

10A + 3C = 9

2B + 5C = 23

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divide the first equation by -3 and you will get, A =-12

then you use the -12 instead of A in the second equation and you get

10(-12) + 3C = 9, -120 +C = 9, you take -120 to the other side and it would be like this,

3C = 120 + 9, then you sum 120 and 9 and divide the equation by 3, C = 129/3 = 43

and therefore 2B + 5(43) = 23

2B + 215 = 23

2B = -215 +23

2B = -192

B = -96