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

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SUMMARY

This discussion focuses on solving linear equations that contain decimals and fractions exclusively within parentheses. Participants confirm that the correct approach involves distributing the terms first and then eliminating decimals or fractions by multiplying through by powers of ten. Examples provided include equations such as 17(2.33 - x) - 35(4 - 30x) = 2 and 5(2.223 + 4x) - 0.2(4 + 16x) = 67, demonstrating the step-by-step process of simplification. The consensus is that while it is possible to eliminate decimals, retaining them can simplify calculations in certain cases.

PREREQUISITES
  • Understanding of linear equations and their properties
  • Familiarity with distribution of terms in algebra
  • Knowledge of multiplying fractions and decimals
  • Ability to manipulate equations to isolate variables
NEXT STEPS
  • Practice solving linear equations with mixed decimals and fractions
  • Learn techniques for simplifying complex algebraic expressions
  • Explore the use of powers of ten in equation manipulation
  • Study the impact of retaining decimals versus converting to whole numbers
USEFUL FOR

Students studying algebra, educators teaching linear equations, and anyone looking to improve their skills in solving equations with decimals and fractions.

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.
 
Welcome to PF!

Hi Bavariadude! Welcome to PF! :smile:

Yup … looks fine! :biggrin:

Was there something worrying you about the process?​
 


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:)
 

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