<u>**Answer:**</u>

<em>Mathew invested</em><em>** $600 and $2400**</em><em> in each account.</em>

<u>**Solution:**</u>

From question, the total amount invested by Mathew is $3000. Let **p = $3000.**

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

**Step 1:**

Given that Mathew has paid **3% interest** in first account .Let us calculate the **simple interest** earned in first account for one year,

Where

p = amount invested in first account

n = number of years

r = rate of interest

hence, by using above equation we get as,

----- eqn 1

**Step 2:**

Mathew has paid **8% interest** in second account. Let us calculate the simple interest earned in second account,

**Step 3:**

Mathew has earned **4% interest** on total investment of $3000. Let us calculate the total simple interest (I)

**Step 4:**

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

By substituting eqn 1 , 2, 3 in eqn 4

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable **‘P’ is 2400 **

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = **600 **

Hence, Mathew invested **$600 and $2400** in each account.