1. algebra word problem

alice invested a portion of $500 at 4% interest and the remainder at 5%. how much did she invest at each rate if her total income from both investments is$23.28.

2. Originally Posted by watsod8
alice invested a portion of $500 at 4% interest and the remainder at 5%. how much did she invest at each rate if her total income from both investments is$23.28.
Let x = amout invested at 4%

Let 500 - x = amount invested at 5%

.04x + .05(500 - x) = 23.28

Finish up....good luck.

3. Originally Posted by watsod8
alice invested a portion of $500 at 4% interest and the remainder at 5%. how much did she invest at each rate if her total income from both investments is$23.28.
Let x denote the amount which is invested at 4%, then (500-x) are invested at 5%:

$x \cdot \frac4{100} + (500-x) \cdot \frac5{100} = 23.38$

Solve for x. I've got $162 4. Hi, I tried to solve this and I don't get$162 . can someone please check my working and tell me were I have gone wrong .

thank you

$0.4x + 0.5(500-x) = 23.38$

$0.4x + 250-0.5x = 23.38$

$-0.1x + 250 = 23.38$

$-0.1x = -226.62$

$x = 2266.2$

?

5. Originally Posted by earboth
Let x denote the amount which is invested at 4%, then (500-x) are invested at 5%:

$x \cdot \frac4{100} + (500-x) \cdot \frac5{100} = {\color{red}23.38}$

Solve for x. I've got $162 Typo, Earboth, but setup is good. Investment at 4% =$172

6. I don't get $172. were have I gone wrong? 7. Originally Posted by Tweety Hi, I tried to solve this and I don't get$162 . can someone please check my working and tell me were I have gone wrong .

thank you

$0.4x + 0.5(500-x) = 23.38$

First, it's 23.28, not 23.38

Second, 4%=.04 and 5%=.05

$0.4x + 250-0.5x = 23.38$

$-0.1x + 250 = 23.38$

$-0.1x = -226.62$

$x = 2266.2$

?
Corrections in red.