Is the problem 1/x- 3+ 1/x+ 3= 10/x²- 9 or 1/(x-3)+ 1/(x+3)= 10/(x²- 9)? Those are quite different equations! You appear to be doing the second. If you multiply both sides by x²- 9 you get x+ 3+ x- 3= 2x= 10 so x= 5, as you say, not 2.mistalopez said:Homework Statement
(1/x-3)+(1/x+3)=(10/x²-9)
2. The attempt at a solution
(1/x-3)+(1/x+3)=(10/x²-9)
(1/x-3)+(1/x+3)=(10/(x+3)(x-3)) - I factored the x^2-9 to make it easier to multiply by LCD.
x+3+x-3=10 - I multipled both sides by the LCD [(x+3)(x-3)]
2x=10 - Combined like terms
x=5 - Divided both sides by 2.
Am I wrong or is the book wrong for having the answer as x=2 ?
To solve an equation, you need to isolate the variable on one side of the equals sign. This can be done by using inverse operations, such as addition, subtraction, multiplication, and division, to get rid of any constants or coefficients that are attached to the variable. Once you have the variable alone on one side, you can solve for its value.
Checking the solution of an equation is important because it ensures that the value you have found for the variable makes the equation true. It helps to catch any mistakes made during the solving process and confirms that the solution is correct.
To check the solution of an equation, simply substitute the value you found for the variable back into the original equation. If the equation is true, then the solution is correct. If the equation is false, then you need to recheck your work to find any errors.
Yes, an equation can have multiple solutions. This is especially true for equations with variables raised to an even power, such as x^2. In these cases, there may be two solutions for x that make the equation true.
If there is no solution to an equation, it means that there is no value for the variable that makes the equation true. This can happen when the equation is contradictory, such as 2x = 4 and x = 5. In this case, there is no value for x that satisfies both equations, and therefore there is no solution.