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helpp pleasee solve by substitution 3a+b+c=7 a+3b-c=13 b=2a-1
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can someone help explain it please
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we are going to give each equations names to simplify the process. 1. 3a + b + c = 7 2. a+3b-c=13 3. b=2a-1 then plug equation 3 into equation 1 to get, 3a + 2a - 1 + c = 7 5a - 1 + c = 7 c = 8-5a, make this equation 4. then plug equation 4 and 3 into equation 2 to get: a + 3(2a-1) - (8-5a) = 13 a + 6a -3 -8 +5a = 13 12a - 11 = 13 12 a = 24 a = 2. Now go back to equation 4 and plug a in to get: c = 8 - 5(2) c = -2 then plug a into equation 3: b = 2(2) -1 b = 4 -1 b = 3 so the answer is a = 2, b=3, c =-2
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