Linear system of equations word problem

Question

a sum of 33,000 was invested in three mutal funds. in one year, the first fund grew bt 4%, the second by 5% and the third by 7%. Total earnings were $1979. The amount invested in the third fund was $1000 less than the combined amount invested in the other two funds. Use a linear system of equations to determine the amount invested in each fund. Use the simple interest formula to calculate each funds growth.

there is one solution: the amoutn of the first part of the investment is$__ the amount of the second part of the investment is$___and the amount of the third part of the investment is$__

please help I appreciate it.

Answer 1

 

   X + Y + Z = 33000

 

   0.04X + 0.05Y + 0.07Z = 1979

 

   Z = X + Y - 1000

 

 

Substituting the 2nd equation into the first equation:

 

   2x + 2y = 34000

 

   x + y = 17000 <--- call this equation ALPHA

 

 

Substituting the 2nd equation into the third equation:

 

  0.04X + 0.05Y + 0.07( X + Y - 1000) = 1979

 

 0.04X + 0.05Y + 0.07X + 0.07Y - 70 = 1979

 

0.11X + 0.12Y = 2049

 

11X + 12Y = 204900

 

Substituting equation ALPHA

 

11X + 12( 17000 - X) = 204900

 

11X + 209000 - 12x = 204900

 

-x = 900

 

x = -900

 

Y = 17000 - x = 17000 - (-900) = 17900

 

Z = 16000

 

 

CHECK:

 

    4% of -900 = -36

 

   5% of 17900 = 895

 

  7% of 16000 = 1120

 

1120 + 895 + -36 = 1979

 

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