simplify these:



9-7x x-4
+
x+7 x+7
7x 11x
-
x-7 x-7
9x 14x
-
x+9 x+9
3x-4 x-9
-
x +6 x +6
x-3 x-4
+
2x 6x
11 3
+
x+9 x-2
1 1
-
x²-x x

First one:
Same denominator so ((9-7x)+(x-4))/(x+7)
Final
(5-6x)/(x+7)

Second one:
Again same denominator so ((7x)-(11x))/(x-7)
Final
(-4x)/(x-7)

Third one:
Same denominator again, ((9x)-(14x))/(x+9)
Final
(-5x)/(x+9)

Fourth one:
Same denominator again, ((3x-4)-(x-9))/(x+6)
Final
(2x+5)/(x+6)

Fifth one:
Different denominators here, so we have to make them them same. We do this by multiplying the first fraction by 3/3 to get (3x-9)/(2x), so ((3x-9)+(x-4))/(6x)
Final
(4x-13)/(6x)

Sixth one:
This one is trickier, we have to multiply each by the others denominator, so ((11)*(x-2))/((x+9)*(x-2)) and ((3)*(x+9))/((x-2)*(x+9)) to get
(11x-22)/(x^2+7x-18) and (3x+27)/(x^2+7x-18)
Final
(14x+5)/(x^2+7x-18)

Last one:
You can change the first fraction to 1/(x*(x-1)), then multiple the second fraction by (x-1)/(x-1) to get (x-1)/(x^2-x). Now add them.
Final
(2-x)/(x^2-1)

Hope I helped :)

Its d so thats how it would would back from 1.0 form three four two