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Galina-37 [17]
3 years ago
14

ou invested 7000 between two accounts paying 4% and 9% annual​ interest, respectively. If the total interest earned for the year

was $430 how much was invested at each​ rate? ​$ nothing was invested at and ​$ nothing was invested at
Mathematics
1 answer:
lana [24]3 years ago
7 0

9514 1404 393

Answer:

  • $3000 at 9%
  • $4000 at 4%

Step-by-step explanation:

Let x represent the amount invested at 9%. Then 7000-x was invested at 4% and the interest earned was ...

  9%·x +4%(7000-x) = 430

  5%·x +280 = 430 . . . . . . . . . simplify

  0.05x = 150 . . . . . . . . . . . subtract 280

  x = 3000 . . . . . . . . . . divide by 0.05

$3000 was invested at 9%; $4000 was invested at 4%.

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<em>here's</em><em> your</em><em> solution</em>

<em>=</em><em>></em><em> </em><em>(</em><em>b^</em><em>2</em><em> </em><em>+</em><em> </em><em>9</em><em>a</em><em>b</em><em> </em><em>+</em><em> </em><em>5</em><em>a</em><em>)</em><em> </em><em>-</em><em> </em><em>(</em><em>3</em><em>b</em><em>^</em><em>2</em><em> </em><em>-</em><em> </em><em>2</em><em>5</em><em>a</em><em>b</em><em> </em><em>+</em><em> </em><em>1</em><em>)</em>

<em>=</em><em>></em><em> </em><em>b^</em><em>2</em><em> </em><em>+</em><em> </em><em>9</em><em>a</em><em>b</em><em> </em><em>+</em><em> </em><em>5</em><em>a</em><em> </em><em>-</em><em> </em><em>3</em><em>b</em><em>^</em><em>2</em><em> </em><em>+</em><em> </em><em>2</em><em>5</em><em>a</em><em>b</em><em> </em><em>-</em><em> </em><em>1</em>

<em>=</em><em>></em><em> </em><em>-</em><em> </em><em>2</em><em>b</em><em>^</em><em>2</em><em> </em><em>+</em><em> </em><em>3</em><em>4</em><em>a</em><em>b</em><em> </em><em>+</em><em> </em><em>5</em><em>a</em><em> </em><em>-</em><em> </em><em>1</em>

<em> </em><em> </em><em> </em><em> </em>

<em> </em><em> </em><em>hope</em><em> it</em><em> helps</em>

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