Question 962395: A total investment of $10,200 is made into two savings accounts. One account yields 8% simple interest and the other 10% simple interest. He earns a total of $978.00 interest for the year. How much was invested in the 8% account?
Found 2 solutions by macston, addingup:
Answer by macston(5194)   (Show Source):
You can put this solution on YOUR website! E=amount at 8%; T=amount at 10%
E+T=$10200
T=$10200-E Use this to substitute for T.
0.08E+0.10T=$978 Substitute for T.
0.08E+0.10($10200-E)=$978
0.08E+$1020-0.10E=$978 Subtract $1020 from each side.
-0.02E=-$42 Divide each side by -0.02.
E=$2100 ANSWER: $2100 was invested at 8%.
CHECK:
T=$10200-E
T=$10200-$2100=$8100 $8100 was invested at 10%
0.08E+0.10T=$978
0.08($2100)+0.10($8100)=$978
$168+$810=$978
$978=$978
Answer by addingup(3677)  (Show Source):
You can put this solution on YOUR website! x money pays .08, this is .08x. And what is left of the 10,200 pays .10. So we can say:
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.08x+.10*(10,200-x)= 978 Multiply on left:
.08x+ 1020-.10x= 978 Subtract 1020 on both sides and do subtraction on left:
-.02x= -42 Divide both sides by -.02:
x= -42/-.02
x= 2100 This is the amount that pays 8% And 10,200-2,100= 8,100 pays 10%.
Let's see:
2,100*.08= 168
8,100*.10= 810
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Total......978
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