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Multiply both sides of the rational equation by the least common denominator.
9
We want to solve the given rational equation.
x-6/x+3+2x/x-3=4x+3/x+3
Let's begin by highlighting all of the different factors in the denominators. This will help us find the least common denominator (LCD).
x-6/x+3+2x/x-3=4x+3/x+3
LHS * (x+3) (x-3)=RHS* (x+3) (x-3)
Distribute (x+3)(x-3)
a/c* b = a* b/c
Cancel out common factors
Simplify quotient
Distribute (x-3)
Add and subtract terms
LHS-3x^2=RHS-3x^2
LHS+3x=RHS+3x
LHS-18=RHS-18
Rearrange equation
We obtained a quadratic equation. Let's identify the values of a, b, and c. x^2-6x-27=0 ⇕ 1x^2+( - 6)x+( -27)=0 We see that a = 1, b = - 6, and c = -27. Next, we will substitute these values into the Quadratic Formula.
Substitute values
- (- a)=a
(- a)^2=a^2
Identity Property of Multiplication
- a(- b)=a* b
Add terms
Calculate root
We will find the values for x by using the positive and the negative signs.
| x=6± 12/2 | |
|---|---|
| x=6+ 12/2 | x=6- 12/2 |
| x=18/2 | x=-6/2 |
| x=9 | x=-3 |
We found that x=9 and x=-3 are possible solutions. Now we have to check them!
We check our solutions to see whether any of them are extraneous. To do so, we will substitute x=9 and x=-3 into the given equation. Let's start with x=9.
x= 9
Multiply
Add and subtract terms
a/b=a * 2/b * 2
Add fractions
Since we obtained a true statement, x=9 is a solution of the equation. Let's now substitute x=-3 into the equation.
x= -3
Note that a fraction with denominator equal to 0 is undefined, so we did not obtain a true statement. Therefore, x=-3 is an extraneous solution.