Solving first and second degree equations

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In summary, the conversation discusses solving two equations, one with a singular solution and one with two potential solutions. The correct solutions are provided, but it is noted that the math teacher made errors in the problem sheet. The person also mentions posting another algebra problem on a forum for help.
  • #1
late347
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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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  • #2
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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  • #3
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
 
  • #4
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.
 
  • #5
Time to find another math teacher. :sorry: :rolleyes: :mad:
 

1. What is the difference between first and second degree equations?

First degree equations (also known as linear equations) have variables raised to the power of 1, while second degree equations (also known as quadratic equations) have variables raised to the power of 2.

2. How do I solve a first degree equation?

To solve a first degree equation, you need to isolate the variable on one side of the equation by using inverse operations. This means that you need to perform the opposite operation of what is being done to the variable. For example, if the variable is being added to a number, you need to subtract that same number from both sides of the equation.

3. What are the steps for solving a second degree equation?

The general steps for solving a second degree equation are:

  • 1. Move all terms to one side of the equation, so that one side is equal to 0.
  • 2. Use the quadratic formula or factoring method to solve for the variable.
  • 3. Check your solution(s) by plugging them back into the equation.

4. Can a first degree equation have more than one solution?

Yes, a first degree equation can have one solution, no solution, or infinitely many solutions. This depends on the values of the constants and coefficients in the equation.

5. What happens if I have a variable with an exponent higher than 2 in a second degree equation?

If you have a variable with an exponent higher than 2, then it is not a second degree equation and cannot be solved using the methods for second degree equations. You may need to use more advanced techniques, such as the quadratic formula with radicals, to solve for the variable.

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