Answer
A) $a+b = 2050000, 0.06a+0.075b = 135000$
B) $\$800,000$ and $\$1,250,000$ in each account
Work Step by Step
Part A
Mona needs to invest $\$ 2,500,000 -\$450,0000 = \$2,050,000 $ to invest in two different accounts, say account A and account B. He also wants to earn a total amount of $\$135,000$ per year from each of her invests at an interest rate of $6\%$ and $7.5\%$ 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: $$ a+b = 2050000.$$ Similarly, the sum of the interest earned each year must satisfy the following equation: $$ 0.06a+0.075b = 87500.$$ Hence, the required system of equation is given by $$\begin{cases}
a+b& = 2050000 \\
0.06a+0.075b &= 135000.
\end{cases}$$ Part B
Solve the above equations for $a$ and $b$. Multiply the first equation by $-0.06$ and add the result to the second to eliminate $a$. $$\begin{aligned}
-0.06\cdot (a+b)& = -0.06\cdot 1375000 \\
-0.06a-0.06b &= -123000.
\end{aligned}$$ Now add. $$\begin{aligned}
-0.06a-0.06b &= -123000\\
0.05a+0.075b &= 135000.
\end{aligned}$$ Rearrange: $$\begin{aligned}
0.015b &= 12000\\
b & = \frac{12000}{0.03}\\
&= \$800,000.\end{aligned}$$ Finally, use the first equation to find the value of $a$. $$a= 2050000-b = 2050000-800000= \$1,250,000.$$ She needs to invest $\$800,000$ and $\$1,250,000$ in each account.
