How do I handle decimals and fractions within parentheses in linear equations?

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Bavariadude
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What do I do when a decimal or a fraction is found only within the parentheses in a linear equation? I know about the general removal process; but my question involves equations like these:


17(2.33 - x) - 35(4 - 30x) = 2

7(4/3 - x) + 24(5x - 60) = 31



Should I distribute first and then get rid of the decimal and fraction? Like so:

17(2.33 - x) - 35(4 - 30x) = 2
39.6 - 17x - 140 + 1050x = 2
396 - 170x - 1400 + 10500x = 20
-170x + 10500x = 20 - 396 + 1400
10330x = 1024
x = 1024/10330
x = 512/5165

7(4/3 - x) + 24(5x - 60) = 31
28/3 - 7x + 120x - 1440 = 31
28 - 21x + 360x - 4320 = 93
- 21x + 360x = 93 - 28 + 4320
339x = 4385
x = 4385/339
 

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I didn't check your arithmetic, but your algebra looks good.
 


Thanks for your reply guys.

I wasn't really worried, Tiny-Tim :smile:. It's just that, as I've been preparing, all the equations I came across had the decimals and fractions either outside the parentheses or distributed in a way that made the multiplication process easier, like, for instance:

10(2.3 - x) - 0.1(5x - 30) = 0

It's obvious here that you start by multiplying 10, 0.1 and 0 by 10, which would give you:

100(2.3 - x) - 1(5x - 30) = 0

But I never came across an an equation that only has a decimal or a fraction within parentheses -- at least not with the textbook I'm currently studying.

I have another similar question, if you don't mind :biggrin:. Suppose we get an equation of this sort:

5(2.223 + 4x) - 0.2(4 + 16x) = 67

We multiply 5, -0.2 and 67 by 10 first, then we distribute and multiply by 10 again, right? To get:


50(2.223 + 4x) - 2(4 + 16x) = 670
111.2 + 200x - 8 - 32x = 670
1112 + 2000x - 80 - 320x = 6700
2000x - 320x = 6700 - 1112 + 80
1680x = 5668
x = 1680/5668
x = 420/1417
 
Bavariadude said:
Suppose we get an equation of this sort:

5(2.223 + 4x) - 0.2(4 + 16x) = 67

We multiply 5, -0.2 and 67 by 10 first, then we distribute and multiply by 10 again, right? To get:


50(2.223 + 4x) - 2(4 + 16x) = 670
111.2 + 200x - 8 - 32x = 670
1112 + 2000x - 80 - 320x = 6700
2000x - 320x = 6700 - 1112 + 80
1680x = 5668
x = 1680/5668
x = 420/1417

oh i see …

you're getting rid of the decimals (or fractions) by multiplying, in this case, by powers of 10 …

yes, that's the right process. :smile:

(though there's not much point with decimals … you might as well just keep the original decimals …
but i can see it might be easier with fractions :wink:)