Answer
A) $a+b = 1375000, 0.05a+0.08b = 87500$
B) $\$625,000$ and $\$750,000$ in each account
Work Step by Step
Part A
Henry needs to invest $\$ 1,500,000 -\$125,0000 = \$1,375,000 $ to invest in two different accounts, say account A and account B. He also wants to earn a total amount of $\$87,500$ per year from each of his invests at an interest rate of $5\%$ and $8\%$ 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: $$ a+b = 1375000.$$ Similarly, the sum of the interest earned each year must satisfy the following equation: $$ 0.05a+0.08b = 87500 $$ Hence, the required system of equation is given by
$$\begin{aligned}
a+b& = 1375000 \\
0.05a+0.08b &= 87500.
\end{aligned}$$ Part B
Solve the above equations for a and b. Multiply the first equation by $-0.05$ and add the result to the second to eliminate $a$. $$\begin{aligned}
-0.05\cdot (a+b)& = -0.05\cdot 1375000 \\
-0.03a-0.03b &= -68750\\
\end{aligned}$$. Now add. $$\begin{aligned}
-0.05a-0.05b &= -68750\\
0.05a+0.08b &= 87500.
\end{aligned}$$ Rearrange: $$\begin{aligned}
0.03b &= 18750\\
b & = \frac{18750}{0.03}\\
&= \$625,000.
\end{aligned}$$ Finally, use the first equation to find the value of $a$. $$a= 1375000-b = 1375000-625000= \$750,000.$$
