Question

Algebra 2 Word Problem

Answer 1

A bank has loaned out $500,000 part of it at 6% annum and the rest at 11% annum. If the bank receives $43,750 in interest each year how much was loaned at 6%?

Answer 2

If x is how much in the 6% and y at 11%:

x + y = 500,000 since the total is $500K

0.06x + 0.11y = 43,750 since the total interest is $43,750

Answer 3

oh okay

Answer 4

so where do we go from here?

Answer 5

Use elimination or substitution method.

Answer 6

so would the next step be .06(500000)=43750?

Answer 7

No, the next step is to solve the system of equations, using elimination or substitution

x + y = 500,000
0.06x + 0.11y = 43,750

Answer 8

oh okay so it would be 0.06x + 0.11(500000) = 43750?

Answer 9

@DanJS can you help?

Answer 10

yeah , what part you not getting?

Answer 11

when i calculate it, it comes out to -11250, but I thought it couldn't be a negative

Answer 12

Take the two equations...
x + y = 500,000
0.06x + 0.11y = 43,750

Solve the first for either x or y.
y = 500000 - x

Put that value for y into the second equation
0.06x + 0.11*(500000 - x) = 43750

solve for x

Answer 13

Looks like you put in 5000000 for y instead of 500000 - x

Answer 14

yeah, i definitely messed up. thank you.

Answer 15

Or you can multiply the first one by 11 and the second by 100 --

11x + 11y = 500000
6x + 11y = 43750

subtract the second equation from the first equation

5x = 456250

Answer 16

welcome