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Multiply both sides of the rational equation by the least common denominator.
c=1, c=4
We want to solve the given rational equation.
2/c-2=2-4/c
For simplicity, we will start by replacing the variable c with the variable x.
2/c-2=2-4/c c= x ⟶ 2/x-2=2-4/x
Let's begin by highlighting all of the different factors in the denominators. This will help us find the least common denominator (LCD).
LHS * x (x-2)=RHS* x (x-2)
Distribute x(x-2)
a/c* b = a* b/c
Cancel out common factors
Simplify quotient
Subtract term
LHS-2x=RHS-2x
.LHS /2.=.RHS /2.
Rearrange equation
Note that we have a quadratic equation now. Let's identify the values of a, b, and c. x^2-5x+4=0 ⇔ 1x^2 +( -5)x+4=0 We have that a= 1, b= -5, and c=4. Let's substitute these values into the Quadratic Formula and solve for x.
Substitute values
- (- a)=a
(- a)^2=a^2
Identity Property of Multiplication
Multiply
Subtract term
Calculate root
Let's calculate both solutions by using the positive and negative signs.
| x=5± 3/2 | |
|---|---|
| x=5+ 3/2 | x=5- 3/2 |
| x=8/2 | x=2/2 |
| x=4 | x=1 |
Therefore, the solutions are x=4 and x=1. Recall that, at the beginning of the solution, we substituted x for c. Therefore, the solutions to the original equation are c=4 and c=1. x=4 &x=c ⟶ c=4 [0.69em] x=1 &x=c ⟶ c=1 Let's check them to see if we have any extraneous solutions. To do this, we need to substitute 4 and 1 for c in the original equation. Let's start with c=4.
Now, we will check our second solution, c=1.
c= 1
Subtract term
Put minus sign in front of fraction
a/1=a
Subtract term
Neither of our solutions is an extraneous solution. Therefore, the solutions to the given equation are c=4 and c=1.