<PRE><B>Question 1185556</B>
&lt;br&gt;
(1) Using a traditional formal algebraic approach (good practice in solving problems using algebra)....&lt;br&gt;
x = amount invested at 8%
7000=x = amount invested at 10%&lt;br&gt;
The total interest was $620:&lt;br&gt;
.08(x)+.10(7000-x)=620&lt;br&gt;
Solve using basic algebra -- probably start by multiplying everything by 100 to clear the decimals....&lt;br&gt;
(2) A quick and easy way to solve any 2-part &quot;mixture&quot; problem like this, if a formal algebraic solution is not required....&lt;br&gt;
$7000 all invested at 8% would yield $560 interest; all invested at 10% would yield $700 interest; the actual interest was $620.&lt;br&gt;
The amount invested at each rate is exactly determined by where the actual interest of $620 lies between the two extremes $560 and $700.&lt;br&gt;
(1) Picture the three interest amounts $560, $620, and $700 on a number line and observe/calculate that $620 is 60/140 = 3/7 of the way from $560 to $700.
(2) That means 3/7 of the total was invested at the higher rate.&lt;br&gt;
ANSWER: 3/7 of the $7000, or $3000, was invested at 10%; the other $4000 at 8%.&lt;br&gt;
CHECK: .10(3000)+.08(4000)=300+320=620&lt;br&gt;
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