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Solve the following word problem.
Money is invested at two rates of interest. One rate is 8 % and the other is 2%. If there is $1300 more invested at 8 % than at 2 %. find the amount invested at
each rate if the total annual interest received is $350. Let x = amount invested at 8% and y = amount invested at 2 %. Then the system that models the problem
[x = y + 1300
is
Solve the system by using the method of addition.
0.08x +0.02y = 350,

Sagot :

sqdancefan

Answer:

  • at 8%: $3760
  • at 2%: $2460

Step-by-step explanation:

You are given a system of equations and asked to solve it by the method of addition. That method requires you add a multiple of one equation to the other so that one of the variables is eliminated. In some cases, this is easier if multiples of both equations are added together. The resulting single-variable equation is then solved in the usual way.

__

look

The given system of equations is ...

  • x = y +1300
  • 0.08x +0.02y = 350

We notice the first equation has the variables on opposite sides of the equal sign, and both their coefficients are 1. The second equation has the variables on the same side of the equal sign, and their coefficients are 0.08 and 0.02.

plan

To eliminate a variable by the "addition method," we need to have the variable on the same side of the equal sign with opposite coefficients. Or, we need to have the variable on opposite sides of the equal sign with the same coefficient.

Both of the x-variables are on the left side, so we need opposite coefficients. We can get that by multiplying the first equation by -0.08, or by multiplying the second equation by -12.5. We judge the first of these choices to be easier.

The y-variables are on opposite sides of the equal sign, so we need equal coefficients. We can get that by multiplying the first equation by 0.02, or the second equation by 50.

solution

We choose to multiply the first equation by 0.02, so we can eliminate the y-variable. Here is the result of doing that, then adding the results

  (0.02)(x) +(0.08x +0.02y) = (0.02)(y +1300) +(350)

  0.10x +0.02y = 0.02y +376 . . . . . eliminate parentheses

  0.10x = 376 . . . . . . . . . subtract 0.02y from both sides. y is eliminated

  x = 3760 . . . . . . . . . divide by 0.10

  y = x -1300 = 2460

__

The amount invested at 8% was $3760; the amount invested at 2% was $2460.

_____

Additional comment

We chose to eliminate y for a couple of reasons. x is the amount at the higher rate. We have found that solving for the higher-rate amount usually works best for preventing errors. The other reason is that multiplying by 0.02 results in smaller numbers, which we consider easier to deal with.

Had we multiplied by -0.08 to eliminate x, we would have ...

  -0.08(x) +(0.08x +0.02y) = -0.08(y +1300) +(350)

  0.02y = -0.08y +246

We judge -0.08(1300) +350 harder to calculate mentally, than 0.02(1300) +350.

  0.10y = 246

  y = 2460; x = 2460+1300 = 3760

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