In the equation [tex]\dfrac{10}{x^2-1}=\dfrac{15}{3x-3}[/tex], 1 makes the denominator zero and 1 is the solution of the equation derived from cross multiplication.
Given :
- Sean used cross multiplication to correctly solve a rational equation.
- He found one valid solution and one extraneous solution.
- 1 is the extraneous solution.
Option A)
Given :
[tex]\dfrac{10}{x^2-1}=\dfrac{15}{3x-3}[/tex]
The above expression is not defined for x equal to 1 so, 1 is an extraneous solution.
Simplify the above equation by doing the cross multiplication.
[tex]10(3x-3)=15(x^2-1)[/tex]
[tex]30(x-1)=15(x-1)(x+1)[/tex]
2 = x + 1
x = 1
In the given equation 1 makes the denominator zero and 1 is the solution of the equation derived from cross multiplication.
Therefore, the correct option is A) and therefore, no need to check the other options.
For more information, refer to the link given below:
https://brainly.com/question/11897796
