Respuesta :
The restriction on the variable x is x ≠ -3 .
Restriction in algebraic expression is defined as the value that makes the denominator as 0 or the whole expression as not defined.
For Example: 1/(x+2) , here the restriction is [tex]x\neq -2[/tex] as -2 will make the denominator 0.
In the given question
[tex]=\frac{5x+15}{10x-10} \div\frac{x^{2} +6x+9}{3x^{2} -3}[/tex]
rewriting the expression
[tex]=\frac{5x+15}{10x-10} \times\frac{3x^{2} -3}{x^{2} +6x+9}[/tex]
Simplifying the numerator and denominator we get
[tex]=\frac{5(x+3)}{10(x-1)} \times\frac{3(x^{2} -1)}{x^{2} +3x+3x+9}[/tex] ...( by splitting the middle term )
[tex]=\frac{5(x+3)}{10(x-1)} \times\frac{3(x-1)(x+1)}{(x+3)(x+3)}[/tex] ...( using identity a²-b²=(a+b)(a-b) )
On further simplifying we get ,
[tex]=\frac{5*3*(x+1)}{10(x+3)}[/tex]
[tex]=\frac{3(x+1)}{2(x+3)}[/tex]
For finding restriction;
denominator, 2(x+3)≠0
means (x+3)≠0 ⇒ x ≠ -3.
Therefore , the restriction on the variable x is x ≠ -3 as -3 will make the expression not defined, and the value after division is [tex]\frac{3(x+1)}{2(x+3)}[/tex].
Learn more about restriction here https://brainly.com/question/14905213
#SPJ4
