Answer
A) $a+b = 225000, 0.03a+0.04b = 7900$
B) $\$ 115,000$ and $\$110,00$ in each of the accounts
Work Step by Step
Part A
We know that Juan has a total fund of $\$ 225,000$ to invest in two different accounts, say account A and account B. He also wants to earn a total amount of $\$7,900$ per year from each of his invests at an interest rate of $3\%$ and $4\%$ respectively. Let $a$ be the amount invested in account A and $b$ be the amount invested in account B. Then, the sum of the money invested must satisfy the following equation: $$ a+b = 225000.$$ Similarly, the sum of the interest earned each year must satisfy the following equation: $$ 0.03a+0.04b = 7900. $$
Hence, the required system of equation is given by $$\begin{cases}
a+b& = 225000 \\
0.03a+0.04b &= 7900.
\end{cases}$$ Part B
Solve the above equations for $a$ and $b$. Multiply the first equation by $-0.03$ and add the result to the second to eliminate $a$. $$\begin{aligned}
-0.03\cdot (a+b)& = -0.03\cdot 225000 \\
-0.03a-0.03b &= -6750.
\end{aligned}$$ Now add. $$\begin{aligned}
-0.03a-0.03b &= -6750\\
0.03a+0.04b &= 7900.
\end{aligned}$$ Rearrange: $$\begin{aligned}
0.01b &= 1150\\
b & = \frac{1150}{0.01}\\
&= \$115,000.
\end{aligned}$$ Finally, use the first equation to find the value of $a$. $$a= 225000-b = 225000-115000= \$110,000.$$
