A person invests $4000: part of the investment is at 10% interest and the rest at 12% interest. If she received $460 at the end of the year in interest, how much was invested at each rate?(whole number in dollars)

Let x be the amount invested at 10%;

SO we have;

.10x=amount invested at 10%

(4000-x).12=amount invested at 12%

So our equation would be;

.10x+(4000-x).12=460

Distribute the 12%;

.10x+480-.12x=460

Put like terms together so you can see them better;

.10x-.12x+480=460

-.02x+480=460;

subtract 480 from both sides;

-.02x=-20

Multiply each side by -.02;

x=1000

Now that we know what was invested at 10% we can subtract that from $4000 to get what was invested at 12%;

$4000-$1000=$3000

Hope you understand

=)

correct answer is infinitely many solutions

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

solving the above equation:

4x + 6 = -4 +4x +10

combining like terms

4x -4x = -4+10-6

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions