Question 176578

Let x=amount invested at 12%
And let y=amount invested at 8%

Now we are told the following:
x+y=$8000---------------------------------eq1
0.12x+0.08y=$840-----------------------------eq2

From eq1, y=$8000-x.  Substitute this into eq2 and we have:

0.12x+0.08($8000-x)=$840  get rid of parens
0.12x+$640-0.08x=$840  subtract $640 from each side
0.12x+$640-$640-0.08x=$840-$640  collect like terms
0.04x=$200  divide each side by 0.04
x=$5000----------------------------------amount invested at 12%
substitute x=$5000 into eq1:
$5000+y=$8000  subtract $5000 from each side
y=$3000-------------------------------------amount invested at 8%

CK
0.12*$5000+0.08*$3000=$840
$600+$240=$840
$840=$840

Another way------------------------only 1 equation
Let x=amount invested at 12%
Then $8000-x=amount invested at 8%
0.12x+0.08($8000-x)=$840------------------same equation as before

Does this help???----------ptaylor