Question

Part of $6,500 was invested at 10% interest and the rest at 12%. If the annual income from these investments was $710, how much was invested at each rate?10%=12%=

Answer 1

6,500 is divided in

X + Y = 6500

also

X• ( 10% ) + Y• (12% ) = 710

then

X• (1/10 ) + Y• (12/100) = 710

Now solve x,y for this 2x2 system, by substitution

Substitute Y= 6500 - X

X/10 + (6500 - X) •1.2 = 710

multiply by 10

X + (6500 - X)• 12 = 7100

6500•12 - 7100 = 12X - X = 11X

then

70900 = 11x

70900/11= X

6445.5 = X

Now find Y

Y= 6500 - 6445.5 = 54.5

Then answers are

10% = $6445

12% = $54.5