$9500 is invested, part of it at 12%, and part of it at 9%, For a certain year, the total yield is $1032.00. How much was invested at each rate?
How much was invested at 12%
How much was invested at 9%

Question

$9500 is invested, part of it at 12%, and part of it at 9%, For a certain year, the total yield is $1032.00. How much was invested at each rate?
How much was invested at 12%
How much was invested at 9%

anonymous · Answer

System of Equations:

a + b = 9500
.12a + .09b = 1032

From the first equation: 
a = 9500 - b

Plug that value of 'a' into the second equation.
.12(9500 - b) + .09b = 1032

Distribute on the left side:

1140 - .12b + .09b = 1032

anonymous · Answer

Combine like terms:

108 = .03b

Solve for b through division of .03:

b = 3600.

anonymous · Answer

Take that value for 'b' and plug it into the FIRST equation:

a + 3600 = 9500

Solve for 'a' by subtracting both sides by 3600:

a = 5900.

a) $5900 at 12%

b) $3600 at 9%

anonymous · Answer

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