Alex invested money in two accounts, and after one year, one of them

had earned 13% interest and the other 9%. He had $700 more invested

in the lower interest account and earned a total of 239 in interest after

one year. How much did Alex invest in each account? answer is 800 at 13% and invested 1500 at 9%

.13x + .09(x+700) = 239

x = 800

as above.

Let's solve this word problem step by step.

Let's assume that Alex invested x dollars in the lower interest account and (x + $700) dollars in the higher interest account.

Now, we know that the lower interest account earned 9% interest, which can be written as 0.09x. Similarly, the higher interest account earned 13% interest, which can be written as 0.13(x + $700).

According to the problem, the total interest earned in one year is $239. So, we can write the equation as:

0.09x + 0.13(x + $700) = $239

Let's solve this equation to find the value of x:

0.09x + 0.13x + 0.13($700) = $239
0.09x + 0.13x + $91 = $239
0.22x + $91 = $239
0.22x = $239 - $91
0.22x = $148

To isolate x, divide both sides of the equation by 0.22:

x = $148 / 0.22
x ≈ $672.73 (rounded to the nearest penny)

Now, we know that Alex invested approximately $672.73 in the lower interest account. The higher interest account had $700 more, so the investment in the higher interest account is:

x + $700 ≈ $672.73 + $700 ≈ $1372.73 (rounded to the nearest penny)

Therefore, Alex invested approximately $672.73 in the lower interest account and approximately $1372.73 in the higher interest account.