SOLUTION: Using a graphing calculator how would you set up the solution to the following word problem: Suppose you have received a total of $1,520 a year in interest from three investments.

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Question 1168300: Using a graphing calculator how would you set up the solution to the following word problem:
Suppose you have received a total of $1,520 a year in interest from three investments. The interest rates for the investments are 5%, 7%, and 8%. The amount is invested at 5% is half of the amount invested at 7%. The amount invested at 7% is $1,500 less than the amount invested at 8% Find the amount of money invested at each rate.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
your equation is 1520 = .05 * x + .07 * y + .08 * z

x is the amount invested at 5%
y is the amount invested at 7%
z is the amount invested at 8%.

you are given that the amount invested at 5% is half the amount invested at 7%.

this makes x = .5 * y.

you are given that the amount invested at 7% is 1500 less than the amount invested at 8%.

this makes y = z - 1500.

solve for z in the equation of y = z - 1500 to get z = y + 1500.

you have x = .5 * y and z = y + 1500.

in the equation of 1520 = .05 * x + .07 * y + .08 * z, replace x with .5 * y and replace z with y + 1500 to get:

1520 = .05 * .5 * y + .07 * y + .08 * (y + 1500)

simplify to get 1520 = .025 * y + .07 * y + .08 * y + .08 * 1500

since .08 * 1500 = 120, then the equation becomes:

1520 = .025 * y + .07 * y + .08 * y + 120

subtract 120 from both sides of the equation to get:

1520 - 120 = .025 * y + .07 * y + .08 * y

simplify and combine like terms to get:

1400 = .175 * y

solve for y to get:

y = 8000.

when y = 8000, x = .5 * y = 4000 and z = y + 1500 = 9500

you get x = 4000, y = 8000, z = 9500

your original equation of 1520 = .05 * x + .07 * y + .08 * z becomes:

1520 = .05 * 4000 + .07 * 8000 + .08 * 9500

simplify to get 1520 = 1520.

this confirms the values of x and y and z are good.

the amount invested at 5% is 4000
the amount invested at 7% is 8000
the amount invested at 8% is 9500

that's your solution.

the amount invested at 5% is half the amount invested at 7%.
the amount invested at 7% is 1500 less than the amount invested at 8%.

the requirements of the problem have been satisfied, so the solution is good.