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Mathematics, 08.12.2020 01:00 nsuleban5016

Jerry solved this equation: 3 ( x βˆ’ 1 4 ) = 13 6 1. 3x βˆ’ 3 4 = 13 6 2. 3x βˆ’ 3 4 + 3 4 = 13 6 + 3 4 3. 3x = 26 12 + 9 12 4. 3x = 35 12 5. ( 3 1 ) 3 1 x = 35 12 ( 3 1 ) 6. x = 105 12 In which step did Jerry make an error? In step 2, he should have subtracted Three-fourths from both sides. In step 3, he should have found an LCD of 10. In step 4, he should have subtracted 9 from 26. In step 5, he should have multiplied both sides by One-third.

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Jerry solved this equation: 3 ( x βˆ’ 1 4 ) = 13 6 1. 3x βˆ’ 3 4 = 13 6 2. 3x βˆ’ 3 4 + 3 4 = 13 6 + 3 4 3...
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