Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on the total investment. How much money did he put in each account?

Answer:Mathew invested $600 and $2400 in each account.Solution: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 – PStep 1:Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest [tex](I_1)[/tex] earned in first account for one year,[tex]\text {simple interest}=\frac{\text {pnr}}{100}[/tex]Where  p = amount invested in first accountn = number of years  r = rate of interesthence, by using above equation we get [tex](I_1)[/tex] as,  [tex]I_{1}=\frac{P \times 1 \times 3}{100}[/tex] ----- eqn 1Step 2:Mathew has paid 8% interest in second account. Let us calculate the simple interest [tex](I_2)[/tex] earned in second account,[tex]I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2[/tex]Step 3:Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)[tex]I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3[/tex]Step 4:Total simple interest = simple interest on first account + simple interest on second account.Hence we get,[tex]I = I_1+ I_2 ---- eqn 4[/tex]By substituting eqn 1 , 2, 3 in eqn 4[tex]\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}[/tex][tex]\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}[/tex]12000=3P + 24000 - 8P5P = 12000P = 2400Thus, the value of the variable ‘P’ is 2400  Hence, the amount invested in first account = p = 2400The amount invested in second account = 3000 – p = 3000 – 2400 = 600  Hence, Mathew invested $600 and $2400 in each account.