Equation solving/solving for x

  • Thread starter Weightofananvil
  • Start date
In summary, the student is super stuck on this homework and is looking for help. They have tried a few different methods, but they are still stuck.
  • #1
Weightofananvil
34
1

Homework Statement


1 / 2x + 1 − 2x / x − 3 = x + 1 / 2x ^ 2 − 5x − 3

solve for x. check solution in original equation

The Attempt at a Solution



I've tried working this out a couple different ways. All I can change it to is 4x^2 - x - 3 / 2x^2 - 5x - 3 = x+1/2x^2-5x-3.

I really don't know where to go or what to do.
I even put the answer into maple and tried to work it backwards.
Super stuck. thanks for the help in advance.
 
Last edited by a moderator:
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  • #2
Maybe it is not written correctly.
 
  • #3
Weightofananvil said:

Homework Statement


1 / 2x + 1 − 2x / x − 3 = x + 1 / 2x ^ 2 − 5x − 3

solve for x. check solution in original equation

The Attempt at a Solution



I've tried working this out a couple different ways. All I can change it to is 4x^2 - x - 3 / 2x^2 - 5x - 3 = x+1/2x^2-5x-3.

I really don't know where to go or what to do.
I even put the answer into maple and tried to work it backwards.
Super stuck. thanks for the help in advance.

I guess you omitted a few couples of parentheses .
What you wrote is equivalent to 0.5 x -2 -3 =x+0.5 x^2-5x-3 - are you sure you wanted to write that?
 
Last edited by a moderator:
  • #4
Weightofananvil said:

Homework Statement


1 / 2x + 1 − 2x / x − 3 = x + 1 / 2x ^ 2 − 5x − 3
As ehild already said, you need quite a few more parentheses here.

What you wrote is properly interpreted to mean this:
##\frac 1 2 x + 1 - \frac{2x}{x} - 3 = x + \frac 1 2 x^2 - 5x - 3##
Weightofananvil said:
solve for x. check solution in original equation

The Attempt at a Solution



I've tried working this out a couple different ways. All I can change it to is 4x^2 - x - 3 / 2x^2 - 5x - 3 = x+1/2x^2-5x-3.
Again, you need some parentheses, plus show your intervening steps in getting to the above.

Regarding parentheses, when you write fractions on a single line, if the numerator or denominator has more than a single term, put parentheses around it.
For example: ##\frac{x^2 + 3x + 2}{x + 1}## should be written as (x^2 + 3x + 2)/(x + 1), if written on a single line.
Weightofananvil said:
I really don't know where to go or what to do.
I even put the answer into maple and tried to work it backwards.
Super stuck. thanks for the help in advance.
 
  • #5
Weightofananvil said:

Homework Statement


1 / 2x + 1 − 2x / x − 3 = x + 1 / 2x ^ 2 − 5x − 3

solve for x. check solution in original equation

The Attempt at a Solution



I've tried working this out a couple different ways. All I can change it to is 4x^2 - x - 3 / 2x^2 - 5x - 3 = x+1/2x^2-5x-3.

I really don't know where to go or what to do.
I even put the answer into maple and tried to work it backwards.
Super stuck. thanks for the help in advance.
Using parentheses, (4x^2- x- 3)/(2x^2- 5x- 3)= (x+ 1)/(2x^2- 5x- 3)
Note that the denominator is 0 for x= 3 and x= -1/2. As long as the denominator is NOT 0, we can multiply on both sides by it to get
4x^2- x- 3= x+ 1 which is the same as the quadratic equation 4x^2- 2x- 4= 0
 
  • #6
From where you are you could put all on the same side since you have a common denominator you can put it away and remains a quadratic equation
 
  • #7
Thanks for the help everyone!
 

Related to Equation solving/solving for x

1. What is the purpose of solving for x in an equation?

Solving for x in an equation helps us find the value of the variable x, which allows us to understand the relationship between the different variables in the equation. It also helps us solve real-world problems and make predictions based on the given information.

2. What are the steps for solving an equation for x?

The basic steps for solving an equation for x are:

  1. Combine like terms on both sides of the equation.
  2. Isolate the term with x on one side of the equation.
  3. Use inverse operations to get rid of any constants next to x.
  4. Divide both sides by the coefficient of x to get the value of x.

3. Can there be more than one solution for an equation with one variable?

Yes, an equation with one variable can have multiple solutions. This means that there can be more than one value of x that satisfies the given equation. For example, the equation x + 5 = 10 has the solution x = 5, but it also has the solution x = 10.

4. How do we know if an equation has no solution?

If an equation has no solution, it means that there is no value of x that satisfies the equation. This can happen when the equation leads to a contradiction, such as 2x + 5 = 2x + 10. In this case, no matter what value of x we choose, the equation will never be true.

5. Can we solve an equation with more than one variable?

Yes, we can solve equations with more than one variable, but only if we have the same number of equations as variables. This is known as a system of equations and it requires a different set of methods for solving, such as substitution or elimination. Solving a system of equations allows us to find the values of all the variables in the equations, not just one.

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