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You invested 21000 in two accounts paying 3% and 8% annual interest, respectively. If the total interest earned for the year was $880, how much was invested at each rate

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To determine how much was invested at each interest rate, we can set up a system of equations using the given information. Let's denote the amount invested at 3% as x, and the amount invested at 8% as y. 1. The total amount invested is $21,000: x + y = $21,000 -- Equation 1 2. The total interest earned is $880, which can be calculated by multiplying the amount invested at 3% by 0.03 (3% expressed as a decimal) and the amount invested at 8% by 0.08 (8% expressed as a decimal), and then adding these two values: 0.03x + 0.08y = $880 -- Equation 2 To solve this system of equations, we can use the substitution method or the elimination method. Let's use the substitution method for this example: From Equation 1, we can solve for x: x = $21,000 - y Substituting this expression for x into Equation 2, we have: 0.03($21,000 - y) + 0.08y = $880 Simplifying the equation: $630 - 0.03y + 0.08y = $880 0.05y = $250 y = $5,000 Now, substitute the value of y into Equation 1 to solve for x: x + $5,000 = $21,000 x = $16,000 Therefore, $16,000 was invested at 3% and $5,000 was invested at 8%.

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