Question 1196477
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(1) by standard algebra....<br>
x = amount invested at 9%
10000-x = amount invested at 14%<br>
The total interest was $1275:<br>
{{{.09(x)+.14(10000-x) = 1275}}}
{{{.09x+1400-.14x = 1275}}}
{{{-.05x = -125}}}
{{{x = -125/-.05 = 2500}}}<br>
ANSWER: x = $2500 at 9%; 10000-x = $7500 at 14%<br>
(2) using logical reasoning and mental arithmetic....<br>
All $10,000 invested at 9% would have yielded $900 interest; all at 14% would have yielded $1400 interest.
Look at the three interest amounts $900, $1275, and $1400 (on a number line, if it helps) and observe/calculate that $1275 is 3/4 of the way from $900 to $1400   (1275-900 = 375; 1400-900 = 500; 375/500 = 3/4).
That means 3/4 of the $10,000 was invested at the higher rate.<br>
ANSWER: $7500 at 14%, the other $2500 at 9%<br>