Solving first and second degree equations

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Homework Help Overview

The discussion revolves around solving first and second degree equations, specifically focusing on the equations 2x - 1 = (1/2)x + 3 and 1/x - 4/x² = 0. Participants are exploring the methods for finding solutions and addressing potential errors in reasoning.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants attempt to solve the equations step-by-step, discussing their calculations and the validity of their solutions. There are questions regarding the correctness of solutions provided by a math teacher, particularly concerning the second equation and the nature of its solutions.

Discussion Status

Some participants express confidence in their solutions, while others highlight discrepancies in the teacher's feedback. The conversation reflects a mix of validation and questioning of the approaches taken, with no clear consensus on the correctness of the teacher's assertions.

Contextual Notes

There is mention of assumptions regarding the variable x, particularly concerning division by zero, which is a critical point in the discussion of the second equation.

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Homework Statement



solve the equations below1.) 2x -1 = (1/2) *x +3

and

2) 1/x - 4/ x2= 0

Homework Equations


3. The Attempt at a Solution [/B]2x -1 = 0,5x +3

2x= 0,5x +4
1,5x = 4
x= 4/ 1,5
x= 8/3

my math teacher told me that the x= 8/6 would have been correct solution to this first one
1/x -4/ x2 = 0

1/x = 4/x2

we enlarge the fraction 1/x

x/ x2 = 4/ x2

multiply both sides by x2

x= 4

singular solution assuming that originally x is unequal to 0 (because cannot divide with 0)

my teacher told me that this latter equation has twin solutions either x=-4 or x= 4

plugging -4 into the original equation, this seems not true at all...

-0,25 -( 0,25) = 0

-0,5 = 0

untrue.
 
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late347 said:

Homework Statement



solve the equations below1.) 2x -1 = (1/2) *x +3

and

2) 1/x - 4/ x2= 0

Homework Equations


3. The Attempt at a Solution [/B]2x -1 = 0,5x +3

2x= 0,5x +4
1,5x = 4
x= 4/ 1,5
x= 8/3

my math teacher told me that the x= 8/6 would have been correct solution to this first one
1/x -4/ x2 = 0

1/x = 4/x2

we enlarge the fraction 1/x

x/ x2 = 4/ x2

multiply both sides by x2

x= 4

singular solution assuming that originally x is unequal to 0 (because cannot divide with 0)

my teacher told me that this latter equation has twin solutions either x=-4 or x= 4

plugging -4 into the original equation, this seems not true at all...

-0,25 -( 0,25) = 0

-0,5 = 0

untrue.
Your solutions are correct.
 
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ehild said:
Your solutions are correct.

Thank you for prompt response. It looks like even math teachers make silly mistakes sometimes. Twice in this problem sheet.

Can you take a gander at my other thread about algebra problem that I had. I just posted a simple word problem in the same precalculus homework math subrofum
 
late347 said:
Thank you for prompt response. It looks like even math teachers make silly mistakes sometimes. Twice in this problem sheet.

Can you take a gander at my other thread about algebra problem that I had. I just posted a simple word problem in the same precalculus homework math subrofum
I see you solved it.
 
Time to find another math teacher. :sorry: :rolleyes: :mad:
 

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