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Melanie invests a total of $25,500 in two accounts paying 3% and 12% simple interest, respectively. How much was invested in each account if, after one year, the total interest was $2,250.00.

A) Enter an equation that uses the information as it is given that can be used to solve this problem. Use
x
as your variable to represent the amount of money invested in the account paying 3% simple interest.

Sagot :

x = amount invested at 3%

y = amount invested at 12%

we know the total amount invested was 25,500 thus x + y = 25,500.

hmmm how much is 3% of "x"? well (3/100) * x or 0.03x.

how much is 12% of "y"? hmmmm (12/100) * y or 0.12y.

we know after a year the total interest was 2,250, therefore whatever "x" and "y" may be 0.03x + 0.12y = 2,250.

[tex]\begin{cases} x + y &= 25500\\ 0.03x + 0.12y &= 2250 \end{cases}\qquad x + y = 25500\implies y = 25500-x \\\\\\ \stackrel{\textit{substituting in the 2nd equation}}{0.03x+0.12(25500-x)=2250}\implies 0.03x+3060-0.12x=2250 \\\\\\ -0.09x+3060=2250\implies -0.09x=-810\implies x=\cfrac{-810}{-0.09}\implies \boxed{x=9000} \\\\\\ \stackrel{\textit{we know that}}{y = 25500-x}\implies y = 25500 - 9000\implies \boxed{y = 16500}[/tex]

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