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Solve the following word problem using matrices:

Part of $30,000 is invested at 6% yearly interest, another part at 7%, and the remainder at 8% yearly interest. The total yearly interest income from the three investments adds up to $2200. The sum of the amounts invested at 6% and 7% equals the amount of money invested at 8%. How much is invested at each rate?

(Enter your answer in the form $x,xxx--with the amount invested at 6% first, then 7%, then 8%.)

User MortenB
by
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2 Answers

4 votes

Answer:

25%]

Explanation:

User Vashishth
by
8.0k points
3 votes

Based on the system of equations, the amount invested at 6% is $5,000, the amount invested at 7% is $10,000, and the amount invested at 8% is $15,000.

Total amount invested = $30,000

Let the amount invested at 6% = x

Let the amount invested at 7% = y

Let the amount invested at 8% = z

z = x + y

x + y + z = $30,000 ...Equation 1

0.06x + 0.07y + 0.08z = $2200 ...Equation 2

x + y = z ...Equation 3

With the third equation, express z in terms of x and y: z = x + y.

Substitute z = x + y into the first equation: x + y + (x + y) = 30,000

Simplify: 2x + 2y = 30,000

Divide both sides by 2: x + y = $15,000

Thus, x + y = 15,000

Therefore, the amount invested at 8% = $15,000

Substitute z = 15,000

0.06x + 0.07y + 0.08(15,000) = 2,200

0.06x + 0.07y + 1200 = 2,200

0.06x + 0.07y = 1000

Now we have a system of two equations:

x + y = $15,000

0.06x + 0.07y = $1000

Solving this system of equations will give us the values of x and y, which represent the amounts invested at 6% and 7%, respectively.

Multiply x + y = $15,000 by 0.06:

0.06x + 0.06y = 900

Subtract 0.06x + 0.06y = 900 from 0.06x + 0.07y = $1000:

0.06x + 0.07y = $1000

-

0.06x + 0.06y = 900

0.01y = 100

y = $10,000

Substitute y = $10,000 in 0.06x + 0.07y = $1000:

0.06x + 0.07(10,000) = $1,000

0.06x + 700 = $1,000

0.06x = $300

x = $5,000

Thus, we cann conclude that the amount invested at 6% is $5,000, the amount invested at 7% is $10,000, and the amount invested at 8% is $15,000.

User BearInBox
by
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