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Zorluk: OrtaLinear Equations in One Variable

An investor divided a total capital of $20,000\$20,000 between two accounts. Account A pays a simple annual interest rate of 6%6\%, and Account B pays a simple annual interest rate of 9%9\%. If the total interest earned from both accounts at the end of one year was $1,470\$1,470, how much money was invested in Account A?

  1. A
    $8,000\$8,000
  2. B
    $9,000\$9,000
  3. C
    $9,800\$9,800
  4. $11,000\$11,000Cevap
  5. E
    $12,000\$12,000

Cevap

The amount invested in Account A was $11,000\$11,000.
Let xx represent the amount invested in Account A. The remaining capital, 20,000−x20,000 - x, is invested in Account B. The total interest earned in one year is given by 0.06x+0.09(20,000−x)=1,4700.06x + 0.09(20,000 - x) = 1,470. Expanding gives 0.06x+1,800−0.09x=1,4700.06x + 1,800 - 0.09x = 1,470, which simplifies to −0.03x=−330-0.03x = -330. Dividing by −0.03-0.03 yields x=11,000x = 11,000. Thus, $11,000\$11,000 was invested in Account A.

Adım Adım Çözüm

1
Define the variable for the unknown quantity.
Let xx be the amount in dollars invested in Account A. Then (20,000−x)(20,000 - x) is the amount invested in Account B.
Expressing both quantities in terms of a single variable allows setting up a single linear equation.
2
Formulate the total interest equation.
0.06x+0.09(20,000−x)=1,4700.06x + 0.09(20,000 - x) = 1,470
Total interest is the sum of interest from Account A (6%6\% of xx) and Account B (9%9\% of 20,000−x20,000 - x).
3
Expand and simplify the algebraic equation.
0.06x+1,800−0.09x=1,470  ⟹  −0.03x+1,800=1,4700.06x + 1,800 - 0.09x = 1,470 \implies -0.03x + 1,800 = 1,470
Distribute 0.090.09 across (20,000−x)(20,000 - x) and combine like terms.
4
Isolate the variable xx.
−0.03x=1,470−1,800  ⟹  −0.03x=−330  ⟹  x=11,000-0.03x = 1,470 - 1,800 \implies -0.03x = -330 \implies x = 11,000
Subtract 1,8001,800 from both sides and divide by −0.03-0.03 to find xx.

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Linear Equations in One Variable
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Linear Equations in One Variable

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