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. 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. 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. 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. 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. 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. Both equations are correct.
More about the Polynomial link is given below.
https://brainly.com/question/17822016
