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Helene invested a total of $1,000 in two simple-interest bank accounts. One account paid 8% annual interest; the other paid 9% annual interest. The total amount of interest she earned after one year was $86. Enter and solve a system of equations to find the amount invested in each account. Enter the interest rates in order as given in the problem. (Hint: Change the interest rates into decimals first.)
at 9%.

Answer :

Answer: -600 and 1600 were invested into the two accounts.

Step-by-step explanation:

x = amount invested in the first account.

y = amount invested in the second account.

You start off with this equation

[tex]x + y = 1000[/tex]

[tex]0.08x + 0.09y=86[/tex]

Subtract both sides by y.

[tex]x=1000-y[/tex]

Subsitude 1000 - x for x for the second equation.

[tex]0.08(1000-y)+0.09y=$86[/tex]

Use the distributive property to solve for y.

[tex]80-0.08y+0.09y=86[/tex]

[tex]80-0.01y=86[/tex]

[tex]6=-0.01y[/tex]

[tex]=-600[/tex]

Subsitude -600 for y in the first equation and solve for x.

[tex]x-600=1000[/tex]

[tex]x=1600[/tex]

-600 and 1600 were invested into the two accounts.

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