Answer:
[tex]\frac{x-3}{x+1}[/tex]
Step-by-step explanation:
Given
[tex]\frac{x^2-10x+21}{x^2-6x-7}[/tex] ← factorise numerator/ denominator
x² - 10x + 21 = (x - 7)(x - 3)
x² - 6x - 7 = (x - 7)(x + 1)
The expression can be written as
= [tex]\frac{(x-7)(x-3)}{(x-7)(x+1)}[/tex] ← cancel the common factor (x - 7) on numerator/denominator
= [tex]\frac{x-3}{x+1}[/tex]
Answer:
[tex]\frac{x-3}{x+1} }[/tex]
Step-by-step explanation:
if you factor both of the functions you will get
[tex]\frac{(x-3)(x-7)}{(x-7)(x+1)}[/tex]
the two (x-7) will cancel out and you will b left with
[tex]\frac{\\(x-3)}{(x+1)}[/tex]
