Answer:
$3000 was invested at 11%.
Step-by-step explanation:
The total interest was $426. This was comprised of interest earned at 8% (represented by e) and (separately) interest earned at 11% (represented by v).
Then e + v = $4200 total investment, and
i = $426 = e(0.08)(1 year) + v(0.11)(1 year)
We eliminate the variable e as follows: since e + v = $4200, e = $4200 - v. Thus,
i = $426 = e(0.08)(1 year) + v(0.11)(1 year) becomes:
i = $426 = ($4200 - v)(0.08)(1 year) + v(0.11)(1 year)
This is one equation in one unknown, the amount of $ invested at 11%.
Performing the indicated multiplications:
426 = 4200(0.08) - 0.08v + 0.11v. Simplifying this further, we get:
426 = 336 + 0.03v.
Then 90 = 0.03v, and v = 90 / 0.03 = $3000.
$3000 was invested at 11%.
