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Mama L [17]
4 years ago
13

you invested 29,000 in two accounts paying 6% and 8% annual interest, respectively. if the total interest earned for the year is

$2060, how much was invested at each rate
Mathematics
1 answer:
yulyashka [42]4 years ago
4 0
You can solve this using a system of equations:
0.06x + 0.08y = 2060 (interest)
x + y = 29,000 (money invested)

Multiply each term in the second equation by -0.06 to cancel out the 0.06x from the first equation.

-0.06x - 0.06y = -1740

Now add this equation to the first equation and solve for y.

 0.06x + 0.08y = 2060
-0.06x - 0.06y = -1740

0.02y = 320
y = 16,000

Since you know y = 16,000 you can substitute it into either of the original equations to solve for x. I picked the second equation because it's simpler, but you could also use the first.

x + 16,000 = 29,000
x = 13,000

So you invested 16,000 at 8% and 13,000 at 6%.

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y = - \frac{1}{2}x + b \\ 1 = 1(- \frac{1}{2}) + b \\ 1 = - \frac{1}{2} + b \\ 1 \frac{1}{2} = b

therefore your y-intercept is equal to 1.5. In order to find the x-intercept we'll take our new equation y = - \frac{1}{2} + 1 \frac{1}{2} and make y = 0, because the line intersects with the x-axis when y is equal to zero.

y = - \frac{1}{2}x + 1 \frac{1}{2} \\ 0 = - \frac{1}{2}x + 1 \frac{1}{2} \\ -1 \frac{1}{2} = - \frac{1}{2}x \\ x =  3

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3 years ago
Answer as fast as possible
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Answer: 2 minutes!

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