A worker invested ​$21,000 in two​ accounts, one yielding 7% interest and the other yielding 11%. If he received a total of​ $1830 in interest at the end of the​ year, how much did he invest at 7%

Answer 1

let a and b be the two amounts that add up to 21000

a + b = 21000

Then applying the interest rates

.07a + .11b = 1830 interest after 1 year

These are the two equations to solve for a and b.

a = (1830-.11b)/.07

Then (1830-.11b)/.07+ b = 21000

1830 - .11b + .07b = 1470

-.04b = -360

b = $9000

a = 21000 - 9000

a = $12000

Answer 2

x = amount invest at 7%

y = amount invest at 11%

x + y = $21000

y = $21000 - x (1)

0.07x = amount of interest earned at 7%

0.11y = amount of interest earned at 11%

0.07x + 0.11y = $1830 (2)

substitute equation (1) for y in equation 2 and solve for x and get amount invested at 7%