Question 1138255: $6,446 is invested, part at 13% and the rest at 10%. If the interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 10% by $568.65, how much is invested at each rate?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = the amount invested at 13%.
y = the amount invested at 10%.
the interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 10% by 568.65.
the equation for that is .13 * x - .10 * y = 568.65.
since a total of 6446 was invested, the equation for that is x + y = 6446.
in the x + y = 6446 equation, solve for y to get y = 6446 - x.
in the .13 * x - .10 * y = 568.65, replace y with 6446 - x to get:
.13 * x - .10 * (6446 - x) = 568.65.
simplify to get .13 * x - 644.6 + .10 * x = 568.65.
combine like terms to get .23 * x - 644.6 = 568.65.
add 644.6 to both sides of the equation to get .23 * x = 1213.25
solve for x to get x = 1213.25 / .23 = 5275.
since x + y = 6446, this means that y is equal to 1171.
you hve x = 5275 and y = 1171.
.13 * x - .10 * y = .13 * 5275 - .10 * 1171 = 685.75 - 117.1 = 568.65.
solution looks good and is 5275 was invested at 13% and 1171 was invested at 10%.
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