Lane invested $10,000 part at 15% and part at 14%?

Question

If the total interest at the end of the year is 1,430, how much did she invest at 15%?


This problem is driving me insane!!!

Answer 1

We'll divide the investment into parts: x and y.

Now we know that .15x + .14y = 1430
We also know that x + y = 10,000, so

y = 10000 - x; now using the first equation:

0.15x + 0.14(10000 - x) = 1430
0.15x - 0.14x = 30
x = 3000

from the original equation:
y = 10000 - x
= 7000

$3,000 at 15%, $7,000 at 14%

Answer 2

let the invest ment on 15% be x
then the investment on 14% will be 10000-x
income=15x/100+14(10000-x)/100
the equation is
15x/100+1400-14x/100=1430
x/100=30
x=3000
so investment on 15%=3000 and on 14% is 7000

Answer 3

First, make any equation with "x" representing the part at 15% and "10000-x" representing the part at 14%

.15x+.14(10000-x)=1430

solve for x:
.15x+1400-.14x=1430

combine like terms:
.01x=30

divide both sides by .01:
x=3000

she invested $3000 at 15%

you can check it by pluging it into the beginning equation.

Answer 4

Too much for my brain

Answer 5

3000
.15x+.14(10000-X)=1430
.15x+1400-.14X=1430
.01X=30
X=3000
3000*.15=450
7000*.14=980
450+980=1430