Here, you invest x dollars in an account with 5% interest and (x-300) dollars in an account with 8% interest. After 1 year, the interest from the 5% account plus the interest from the 8% account = $301. You should be able to figure out what x is and then get (x-300) also.

a1 = Pat’s investment with 5% interest rate

a2 = Pat’s investment with 8% interest rate

But since “a1” is greater than “a2” by $300,

a1 = x + $300

a2 = x

Let “b” be the interest rates of Pat’s investment

b1 = 5% = 0.05

b2 = 8% = 0.08

To get the earning of an investment, the investment must be multiplied with the interest rate. Let “c” be Pat’s earnings.

c1 = (a1)(0.05) = (x + $300)(0.05)

c2 = (a2)(0.08) = (x)(0.08)

But the total earning is $301, so

c1 + c2 = $301

Substitute the values for “c1” and “c2”,

(x + $300)(0.05) + (x)(0.08) = $301

0.05x + $15 + 0.08x = $301

0.05x + 0.08x = $301 – $15

0.13x = $286

x = $2200

Therefore, $2200 is Pat’s investment which has an interest rate of 8% (which is equal to a2). To get the other investment, substitue “x” to “a1”.

a1 = $2200 + $300

a1 = $2500

Therefore, $2500 is Pat’s investment which has an interest rate of 5%.

To check, substitute “a1” and “a1” to “c1” and “c2” respectively

c1 = (a1)(0.05)

c1 = ($2500)(0.05)

c1 = $125

c2 = (a2)(0.08)

c2 = ($2200)(0.08)

c2 = $176

And since “c1” and “c2” adds to $301, the answer is correct.

We’re told y=x-300

assuming synchronized interest calculation period and no compounding, we’re also told

0.05x + 0.08y=301

so if we substitute (x-300) for y, we get

0.05x + 0.08(x-300)=301

or 0.05x + 0.08x – 24 = 301

so 0.05x + 0.08x = 325

so 0.13x = 325

so 13x=32500

and x=2500

so y=2200