Respuesta :
[tex]\bf \cfrac{x^2-x-42}{x-6}\implies \cfrac{(x+7)\underline{(x-6)}}{\underline{x-6}}\implies x+7[/tex]
so.. the left-hand-side does indeed simplify to x+7, so the equation does check out.
however, notice something, for the equation of x+7, when x = 6, we get (6) + 7 which is 13.
BUT for the rational, we get   [tex]\bf \cfrac{x^2-x-42}{x-6}\qquad \boxed{x=6}\implies \cfrac{x^2-x-42}{\boxed{6}-6}\implies \stackrel{und efined}{\cfrac{x^2-x-42}{0}}[/tex]
so, even though the siimplification is correct, the rational or original expression is constrained in its domain.
so.. the left-hand-side does indeed simplify to x+7, so the equation does check out.
however, notice something, for the equation of x+7, when x = 6, we get (6) + 7 which is 13.
BUT for the rational, we get   [tex]\bf \cfrac{x^2-x-42}{x-6}\qquad \boxed{x=6}\implies \cfrac{x^2-x-42}{\boxed{6}-6}\implies \stackrel{und efined}{\cfrac{x^2-x-42}{0}}[/tex]
so, even though the siimplification is correct, the rational or original expression is constrained in its domain.
1.  (x − 6)(x + 7) ≠x2 + x − 42 the equation is wrong because of inequality.
2. Â The left-hand side is not defined for x = 0, but the right-hand side is. so this statement is also wrong because it is defined for all the values of x.
3. Â The left-hand side is not defined for x = 6, but the right-hand side is. so this statement is also wrong because it is defined for all the values of x.
4. Â Both equations are correct.
Polynomial
An expression of more than two algebraic terms, especially the sum of several terms that contain different power of the same variable.
Given
[tex]\rm \dfrac{x^{2} +x-42}{x-6} = x+7[/tex]
How to check what is wrong with the following equation?
[tex]\begin{aligned} \rm \dfrac{x^{2} +x-42}{x-6} &= x+7\\\rm \dfrac{(x-6)(x+7)}{x-6} &= x+7\\\end{aligned}[/tex]
1.  (x − 6)(x + 7) ≠x2 + x − 42 the equation is wrong because of inequality.
2. Â The left-hand side is not defined for x = 0, but the right-hand side is. so this statement is also wrong because it is defined for all the values of x.
3. Â The left-hand side is not defined for x = 6, but the right-hand side is. so this statement is also wrong because it is defined for all the values of x.
4. Â Both equations are correct.
More about the Polynomial link is given below.
https://brainly.com/question/17822016
