College Algebra

Question

Sam invested $1800, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $24 more than twice the income from the 6% investment. How much did he invest at each rate?

Answer 1

If X is the amount invested at 6% and y is the amount invested at 8% then:

$1800=X+Y

Subtract X from both sides and Y equals:

Y = $1800 - X

If the amount earned at 8% (y) is $24 more than at 6% (x) then:

$1.08y - (2*1.06x) = $24

Based on our first equation we know Y = 1800 - X and we can substitute the y in the above equation with this formula, as such:

$1.08(1800 -x) - 2(1.06x) = $24

$1944 -1.08x -2.12x = $24

-3.2x=-1920

x=600

If X equals $600 then Y must equal $1200 as per the first equation.