SOLUTION: PLEASE HELP ASAP! Solve the following problem by setting up and solving a system of three linear equations in three variables. Part of $2,000 is invested at 12%, another part at

Question 457415: PLEASE HELP ASAP!
Solve the following problem by setting up and solving a system of three linear equations in three variables.
Part of $2,000 is invested at 12%, another part at 13%, and the remainder at 14%. The total yearly income from the three investments is $268. The sum of the amounts invested at 12% and 13% equals the amount invested at 14%. Determine how much is invested at each rate.
12% interest =$?
13% interest =$?
14% interest =$?

Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
12%------------x
13%------------y
14%------------z

x+y+z=2000--------------1
12x+13y+14z=268*100--------------2
x+y-z=0 --------------3

consider equation 1 &2 Eliminate y
Multiply 1 by -13
Multiply 2 by 1
we get
-13x -13y-13z= -26000
12x+13y+14z=26800
Add the two
-x+z= 800 -------------4
consider equation 2 & 3 Eliminate y
-1x=800
Multiply 2 by x=-800
Multiply 3 by-13
we get
12x+13y+14z=26800
-13x-13y+13z=0
Add the two
-1x+27z=26800 -------------5 5
Consider (4) & (5) Eliminate x
Multiply 4 by -1
Multiply (5) by 1
we get
x-z=-800
-x+27z= 26800
Add the two
26z=26000
/26
z=1000----14%
Plug the value of z in 5
-x+27000=26800
-x=-200
x=200----12%
plug value of x & z in 1
200+y+1000=2000
y=-200+ -1000+ 2000
y=800----13%

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