An actor invests some money at 5%, and $38000 more than twice the amount at 9%. The total annual interest earned from the investment is $37460 How much did he invest at each amount? Use the sle-step method.CEHe invested at 5% and $at9%

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Answer:

Let the amount invested at 5% be

[tex]=x[/tex]

interest earned will be

[tex]\frac{5}{100}\times x=0.05x[/tex]

38,000 more than twice the amount at 9% will be

[tex]\begin{gathered} \frac{9}{100}(2x+38,000) \\ 0.09(2x+38,000) \end{gathered}[/tex]

The total amou nt amount invested will be represented below as

[tex]\begin{gathered} 0.05x+0.09(2x+38,000)=37460 \\ 0.05x+0.18x+3420=37460 \end{gathered}[/tex]

By simplifying further, we will have

[tex]\begin{gathered} 0.05x+0.18x+3,420=37,460 \\ 0.23x=37460-3420 \\ 0.23x=34040 \\ x=\frac{34040}{0.23} \\ x=148,000 \end{gathered}[/tex][tex]\begin{gathered} 2x+38000 \\ 2(148000)+38000 \\ 296000+38000 \\ =334000 \end{gathered}[/tex]

Hence,

The actor invested $148,000 at 5%

The actor invested $334,000 at 9%