Question 677729
Let, x be the amount invested at 9% interest.
(10500 - x) be the amount invested at 12% interest.
From I = Prt, interest from first investment = {{{x * (9/100) * 1}}}
= {{{9x/100}}}
Interest from 2nd investment = {{{(10500-x) * (12/100) * 1}}}
= {{{(12*(10500-x))/100}}}
By condition, {{{9x/100 + (12*(10500-x))/100 = 1140}}}
{{{(9x + 126000 - 12x)/100 = 1140}}}
9x + 126000 - 12x = 1140 * 100
9x - 12 x = 114000 - 126000
-3x = -12000
{{{x = (-12000)/(-3)}}}
x = 4000
Therefore, $4,000 has been invested at 9% interest.
$(10,500 - 4,000) = $6,500 has been invested at 12% interest.