<PRE><B>Question 1161595</B>
&lt;br&gt;
(1) A traditional algebraic setup for solving....&lt;br&gt;
x = amount at 10%
70000-x = amount at 7%&lt;br&gt;
10% of x, plus 7% of (70000-x), equals 6400:&lt;br&gt;
{{{.10(x)+.07(70000-x) = 6400}}}&lt;br&gt;
Solve using basic algebra....&lt;br&gt;
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(2) A quick and easy way to solve a 2-part &quot;mixture&quot; problem like this, if an algebraic solution is not required....&lt;br&gt;
(1) All $70,000 invested at 10% would yield $7000 interest; all at 7% would yield $4900 interest.&lt;br&gt;
(2) The actual interest, $6400, is 5/7 of the way from $4900 to $7000.  ($4900 to $7000 is a difference of $2100; $4900 to $6400 is a difference of $1500; 1500/2100 = 5/7)&lt;br&gt;
(3) That means 5/7 of the total was invested at the higher rate.&lt;br&gt;
ANSWER: 5/7 of $70,000, or $50,000, at 10%; the other $20,000 at 7%.&lt;br&gt;
CHECK:
.10(50000) + .07(20000) = 5000+1400 = 6400&lt;br&gt;
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