1. ## Interesting Algebra Problem

Hey Everyone,

I am unsure how to solve the following problem, could you please show me how it's done so that I can solve similar problems in the future?

Melissa bought a new car with 12 optional features to the car. Each optional feature costs either $23.99 or$120.84. If she paid $675.28 for all the optional features, how many of each type did she buy? Thanks in advance! 2. let no of optional feature cost$23.99 = x
and no of optional feature cost $120.84 =y therefore according to question x+y=12 ..........(1) 23.99x + 120.84y = 675.28 ..........(2) now continue........ 3. Originally Posted by ramiee2010 let no of optional feature cost$23.99 = x
and no of optional feature cost $120.84 =y therefore according to question x+y=12 ..........(1) 23.99x + 120.84y = 675.28 ..........(2) now continue........ I still don't know, I'm used to having only one variable..not two. 4. Originally Posted by ElectroNerd I still don't know, I'm used to having only one variable..not two. for only one variable, let no of optional feature cost$23.99 = x
therefore no of optional feature cost $120.84 =(12-x) (because total no. of options are 12) 23.99x + 120.84(12-x) = 675.28 .................(1) (Melissa paid = total for optional feature cost$23.99 +total for optional feature cost \$120.84)
now solve for x

5. Ok, $x=8$

Thanks!

### melissa bought a new car with 12 optional features

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