I am needing help on my homework

Question

​$5400

is​ invested, part of it at

10​%

and part of it at

8​%.

For a certain​ year, the total yield is

​$492.00.

How much was invested at each​ rate?

Answer 1

$3000 at 10%

$2400 at 8%

A substitution will work here

Total Principal = 5400

Let x = amount invested at 10%

Let y = amount invested at 8%

So

x + y = 5400

y = 5400 -x

Total yield after one year is $492.00

So we have

0.10x + 0.08y = 492

Substitute 5400 -x for y to get everything in terms of x

0.10x + .08(5400 - x) = 492

0.10x + 432 - 0.08x = 492

Combine like terms

0.02x + 432 = 492

Subtract 432 from both sides of the equation

0.02x = 492 - 432

0.02x = 60

Divide both sides by 0.02

x = 60/.02

x = 3000

y = 5400 -x = 5400 - 3000= 2400

y = 2400

Checking

0.10(3000) n+ 0.08(2400) = 492

300 + 192 = 492

492 = 492

You can also use elimination to solve this problem

I hope you find this useful if you have any questions send me a message.