SOLUTION: ERNEST INVESTED A PART OF HIS MONEY AT 16% ANNUAL INTEREST. ANOTHER PART WHICH IS P50,000 MORE WAS INVESTED AT 18% ANNUAL INTEREST. IF THE TOTAL AMOUNT INCOME FROM THE TWO INVESTME

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Question 953743: ERNEST INVESTED A PART OF HIS MONEY AT 16% ANNUAL INTEREST. ANOTHER PART WHICH IS P50,000 MORE WAS INVESTED AT 18% ANNUAL INTEREST. IF THE TOTAL AMOUNT INCOME FROM THE TWO INVESTMENTS WAS P94,000, HOW MUCH WAS INVESTED AT EACH RATE?
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
S=amount at 16%; E=amount invested at 18%=S+P50000
0.16S+0.18E=P94000 Substitute for E
0.16S+0.18(S+P50000)=P94000
0.16S+0.18S+P9000=P94000 Subtract P9000 from each side.
0.34S=P85000 Divide each side by 0.34
S=P250000 ANSWER 1: The amount invested at 16% was P250000.
E=S+P50000=P250000+P50000=P300000 ANSWER 2: The amount invested at 18% was P300000.
CHECK:
0.16S+0.18E=P94000
0.16(P250000)+0.18(P300000)=P94000
P40000+P54000=P94000
P94000=P94000