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eawoodall

eawoodall

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eawoodall
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  • A+B+C=9. 100A+10B+1C 100C+10B+1A-396=100A+10B+1C. 99C+0B-99A-396=0. 99C=99A+396. divide by 99. C=A+4. A*3=B. substitute (A)+(3A)+(A+4)=9. 5A+4=9. 5A=5. A=1. 1*3=B. 3=B. C=1+4. C=5. 135. 1+3+5=9. checked.
    in Algebra II Review Word Problem? Comment by eawoodall June 2021
  • A+B+C=9. 100A+10B+1C 100C+10B+1A-396=100A+10B+1C. 99C+0B-99A-396=0. 99C=99A+396. divide by 99. C=A+4. A*3=B. substitute (A)+(3A)+(A+4)=9. 5A+4=9. 5A=5. A=1. 1*3=B. 3=B. C=1+4. C=5. 135. 1+3+5=9. checked.
    in Algebra II Review Word Problem? Comment by eawoodall June 2021
  • A+B+C=9. 100A+10B+1C 100C+10B+1A-396=100A+10B+1C. 99C+0B-99A-396=0. 99C=99A+396. divide by 99. C=A+4. A*3=B. substitute (A)+(3A)+(A+4)=9. 5A+4=9. 5A=5. A=1. 1*3=B. 3=B. C=1+4. C=5. 135. 1+3+5=9. checked.
    in Algebra II Review Word Problem? Comment by eawoodall May 2021

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