a = total amount invested at 5%
b = total amount invested at 8%
we know 5% was earned in interest from "a", so that'd be (5/100) * a or 0.05a.
we know 8% was earned in interest from "b", so that'd be (8/100) * a or 0.08b.
we also know that the total for both is 1000, a + b = 100.
[tex]\begin{cases} a + b = 1000\\ 0.05a+0.08b=65 \end{cases}\qquad \qquad \stackrel{\textit{since we know}}{a+b=1000}\implies b = 1000-a \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{0.05a+0.08(100-a)=65}\implies 0.05a+80-0.08a=65 \\\\\\ -0.03a+80=65\implies -0.03a=-15\implies a=\cfrac{-15}{-0.03}\implies \boxed{a = 500} \\\\\\ \stackrel{\textit{we know that}}{b = 1000-a}\implies b=1000-500\implies \boxed{b = 500}[/tex]
Step-by-step explanation:
Let [tex]\textbf{\textit{x}}[/tex] be the amount invested into the account with a 5% annual interest and [tex]\textit{\textbf{y}}[/tex] be the amount invested into the account with an 8% annual interest,
It is stated that Helene invested a total of $1000, which can be written mathematically as
[tex]x \ + \ y \ = \ 1000 \ \ \ \text{----- \ (1)}[/tex] .
Subsequently, the total amount of interest after 1 year is $65. This statement can be changed into the equation below.
[tex]0.05x \ + \ 0.08y \ = \ 65 \ \ \text{----- \ (2)}[/tex] .
First, divide each term in equation (2) by 0.05, yields the expression
[tex]x \ + \ 1.6y \ = \ 1300 \ \ \text{----- \ (3)}[/tex] .
Then, let equation (1) be subtracted from equation (3) term-by-term,
[tex](x \ + \ 1.6y ) \ - \ (x \ + \ y) \ = \ 1300 \ - \ 1000 \\ \\ (x \ - \ x) \ + \ (1.6y \ - \ y) \ = \ 300 \\ \\ \-\hspace{3.15cm} 0.6y \ = \ 300 \\ \\ \-\hspace{3.38cm} \displaystyle\frac{3}{5}y \ = \ 300 \\ \\ \-\hspace{3.65cm} y \ = \ 500[/tex].
Therefore, the amount invested in each account is $500.
