<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,
![\text {simple interest}=\frac{\text {pnr}}{100}](https://tex.z-dn.net/?f=%5Ctext%20%7Bsimple%20interest%7D%3D%5Cfrac%7B%5Ctext%20%7Bpnr%7D%7D%7B100%7D)
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,
![I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2](https://tex.z-dn.net/?f=I_%7B2%7D%20%3D%20%5Cfrac%7B%283000-P%29%20%5Ctimes%201%20%5Ctimes%208%7D%7B100%7D%20%5Ctext%20%7B%20------%20eqn%20%7D%202)
Step 3:
Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)
![I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B3000%20%5Ctimes%201%20%5Ctimes%204%7D%7B100%7D%20-----%20eqn%203)
Step 4:
Total simple interest = simple interest on first account + simple interest on second account.
Hence we get,
![I = I_1+ I_2 ---- eqn 4](https://tex.z-dn.net/?f=I%20%3D%20I_1%2B%20I_2%20----%20eqn%204)
By substituting eqn 1 , 2, 3 in eqn 4
![\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B3000%20%5Ctimes%201%20%5Ctimes%204%7D%7B100%7D%20%3D%20%5Cfrac%7BP%20%5Ctimes%201%20%5Ctimes%203%7D%7B100%7D%20%2B%20%5Cfrac%7B%283000-P%29%20%5Ctimes%201%20%5Ctimes%208%7D%7B100%7D)
![\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B12000%7D%7B100%7D%20%3D%20%5Cfrac%7B3%20P%7D%7B100%7D%20%2B%20%5Cfrac%7B%2824000-8%20P%29%7D%7B100%7D)
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.