Question 1204205
Thomas has $6000 invested among a checking account paying 2% annual interest, a savings account paying 5% annual interest, and a bond paying 7% annual interest. He earns $355 in annual interest and he has $2300 less invested in his savings account than in his bond. How much does he have invested in each account?


Thomas has  invested in a checking account paying 2% annual interest, Let amount be $x

 A savings account paying 5% annual interest,  let investment be  $y

a bond paying 7% annual interest. Let investment be $z

x+y+z=6000 ----------------------(1)

He earns $355 in annual interest

0.02x+0.05y+0.07z = 355-----------------(2)

he has $2300 less invested in his savings account than in his bond.

z-2300 =y

0.02x+0.05y+0.07z= 355   From (2)

0.02x+0.05(z-2300)+0.07z =355   (substitute y)

0.02x+ 0.05z -115+0.07z=355

0.02x+0.12z = 470-----------------------(3)

multiply by 100

2x +12z = 47000-------------------(4)

x+y+z=6000 
substitute z-2300=y

x+z-2300 +z = 6000

x+2z = 8300 -------------------(5)

Solve (4) &(5)

2x +12z = 47000
x+2z = 8300  multiply by 2 and subtract

2x+4z=16600

8z = 30400
z= 3800

y= z-2300 
3800-2300 = 1500

x= 6000-3800-1500 =700

Thomas has $700 invested in the checking account, $1500 in the savings account, and $3800 in the bond.

CHECK

700*2%+1500*5%+3800*7%= 355