Multiply the numerator on the left-hand side by the denominator on the right-hand side, and the denominator on the left-hand side by the numerator on the right-hand side.
x=3
Practice makes perfect
We will solve the given equation and then check the solutions.
5/x+1=x+2/x+1
Since both fractions are undefined when x=-1, we can suspect that x=-1 is a possible extraneous solution. Let's solve the given equation!
Solving the Equation
We will solve the rational equation by using the Cross Products Property. If a numerator or denominator contains addition or subtraction, be sure to treat each one as a parenthetical factor in the cross multiplication process.
We obtained a quadratic equation. Let's identify the values of a, b, and c.
x^2-2x-3=0 ⇕ 1x^2+( - 2)x+( - 3)=0
We see that a = 1, b = - 2, and c = - 3. Next, we will substitute these values into the Quadratic Formula.
We will find the values for x by using the positive and the negative signs.
x=2± 4/2
x=2+ 4/2
x=2- 4/2
x=6/2
x=- 2/2
x=3
x=- 1
We found that x=3 and x=- 1 are possible solutions. Now we have to check them!
Checking the Solutions
We check our solutions to see whether any of them are extraneous. To do so, we will substitute x=3 and x=- 1 into the given equation. Let's start with x=3.
