Excluded values: Division by zero is not allowed.
example:
10. (x^2-5x-14)/(x^2+4x+4)= (x-7)(x+2)/(x+2)^2= (x-7)/(x+2)
x=-2 is excluded, as the denominator would be zero.
7. (y+6)/(y^2+5y-6)= (y+6)/(y+6)(y-1)= 1/(y-1)
y=1 is an excluded value.
but notice on 1, and 2.
1. 36x^3/42x^2=36x/42 No excluded values (the x in denomintor is gone)
can someone please help me with my algebra? you don't have to answer every question but I'm very confused, this is algebra 1b unit 6 sample work
simplify the rational expressions. state any excluded values
1. 36x^3 / 42x^2
2. -70n^2 / 28n
3. 2r-4 / r-2
4. 10 / 10a-10
5. x-4 / 3x^2-12x
6. x-5 / x^2-10x+25
7. y+6 / y^2+5y-6
8. x-4 / x^2-16
9. x^2-25 / x+5
10. x^2-5x-14 / x^2+4x+4
I can simplify but have no idea how to find excluded values and I've looked through textbooks and youtube videos but still don't understand. thanks in advance
2 answers
On #1, x=0 is excluded, because in the original expression that evaluates to 0/0
Excluded values may appear to vanish as you eliminate common factors, but the initial expression and the final result are equivalent only when those eliminated values are excluded.
36x^3 / 42x^2 = 6x/7 only when x≠0
Excluded values may appear to vanish as you eliminate common factors, but the initial expression and the final result are equivalent only when those eliminated values are excluded.
36x^3 / 42x^2 = 6x/7 only when x≠0
3 answers