Kendra invested twice as much money at an annual rate of 7% as she did at an annual rate of 5.5%. How much did she invest at the higher rate if her income from these investments for one year totaled $175.50?

Let [tex]x[/tex] be the amount Kendra invested at 5.5%. Since the amount invested at 7% is twice the amount she invested at 5.5%, she invested [tex]2x[/tex] at 7%.

Remember that [tex]I=Prt[/tex] , where:

[tex]I[/tex] is the interest earned

[tex]P[/tex] is the principal or initial investment

[tex]r[/tex] is the interest rate in decimal form

[tex]t[/tex] is the time in years

- For the 5.5% investment:

[tex]P=x[/tex]

[tex]r=\frac{5.5}{100} =0.055[/tex]

[tex]t=1[/tex]

Let's replace the values in our formula, [tex]I=x(0.055)(1)[/tex], which simplified is [tex]I=0.055x[/tex]

- For the 7% investment:

[tex]P=2x[/tex]

[tex]r=\frac{7.7}{100} =0.077[/tex]

[tex]t=1[/tex]

[tex]I=2x(0.077)(1)[/tex]

[tex]I=0.154x[/tex]

Now, we that the the amount of interest earned combining the two investment is $175.50, so:

[tex]0.055x+0.154x=175.50[/tex]

[tex]0.209x=175.50[/tex]

[tex]x=\frac{175.50}{0.209}[/tex]

[tex]x=839.71[/tex]

She invested $839.71 at 5.5%, and since she invested twice that amount at 7%, she invested 2*($839.71 ) = $1,679.42 at 7% (the higher rate).