Mario invests 1,500 in a savings account that earns 2% interest a year. he also plans to set aside $50 cash a month. a: 2100(1.02)x b: 1500(1.02)x+600x c: 500(1.02)x-600x d: 2100 pls !

1500(1.02)^x  +   600x is how much he has in savings at the end of x years where it be in the bank or elsewhere

Step-by-step explanation:

x is in years

Let's just think about the investment of 1500 in an account earning 2% per year.

Before the years even start, you are at 1500 ( present value).

The next year (year 1), it would be 1500*.02+1500=(1500)(1.02).

The next year (year 2), it would be 1500(1.02)(.02)+1500(1.02)=1500(1.02)(1.02).

We keep multiplying factors of (1.02) each time.

So for year x, you would have saved 1500(1.02)^x.

Now we are saving 50 cash per month. Per year this would be 12(50) since there are 12 months in a year.  12(50)=600.  

So the first year you would have 600.

The second year you would have 600(2) or 1200.

The third year you would have 600(3) or 1800.

Let's put this together:

1500(1.02)^x  +   600x

let x and 400 - x be the two numbers.  

==> 0.80x + 0.85(400 - x) = 400 - 68 from the statement of the problem.  

< ==> 0.80x + 0.85(400 - x) = 400 - 68  

==> 340 - 0.05x = 332  

< ==> -0.05x = -8 ==> x = 160, the first number ==> 400 - x = 240, the second number.

the answer to that one is b